Topics

It is understood that a compass needle points along the horizontal component of Earth's magnetic field (a property called declination).

Earth's Geomagnetic North acts as a south pole of a magnet, while its Geomagnetic South acts as a North Pole of a magnet.
As shown in the diagram, the axis of the dipole makes an angle of about \({11.5}^{\circ}\) with earth's rotational axis. The axis of dipole makes an angle of about \({11.5}^{\circ}\) with the earth's rotational axis. And Earth's rotational axis makes an angle of \({23.5}^{\circ}\) with the normal to the plane of earth's orbit about the sun.

Elements of earth's magnetic field:
The earth's magnetic field is characterized by three quantities:
(a) Declination
(b) Inclination or dip
(c) Horizontal component of the field.

Hans Christian Oersted was a professor of science at Copenhagen University. In 1819 he arranged in his home a science demonstration to friends and students. He planned to demonstrate the heating of a wire by an electric current, and also to carry out demonstrations of magnetism, for which he provided a compass needle mounted on a wooden stand.
While performing his electric demonstration, Oersted noted to his surprise that every time the electric current was switched on, the compass needle moved. He kept quiet and finished the demonstrations, but in the months that followed worked hard trying to make sense out of the new phenomenon. And this is what we are going to study now.

We have seen that currents (fundamentally moving charges) are the source of magnetism. This can be readily demonstrated by placing compass needles near a wire. As shown in Figure, all compass needles point in the same direction in the absence of current (in the direction of earth's magnetic field). However, when a strong current passes through (so that earth's magnetic field becomes negligible), the needles will be deflected along the tangential direction of the circular path (Figure).

Figure : Deflection of compass needles near a current-carrying wire

Biot-Savart Law

Currents which arise due to the motion of charges are the source of magnetic fields. When charges move in a conducting wire and produce a current \(I\), the magnetic field at any point \(P\) due to the current can be calculated by adding up the magnetic field contributions, \(d\overrightarrow{B}\), from small segments of the wire \(d\overrightarrow{s}\) (Figure).

Figure : Magnetic field \(d\overrightarrow{B}\) at point \(P\) due to a current-carrying element \(Id\overrightarrow{s}\)

These segments can be thought of as a vector quantity having a magnitude of the length of the segment and pointing in the direction of the current flow. The infinitesimal current source can then be written as Ids \(\overrightarrow{\mathbf{s}}\).

Let \(r\) denote as the distance form the current source to the field point \(P\) and \(\widehat{r}\) the corresponding unit vector. The Biot-Savart law gives an expression for the magnetic field contribution, \(d\overrightarrow{B}\), from the current source, I \(d\overrightarrow{s}\),

\[dB = \frac{\mu_{0}}{4\pi}\frac{Id\overrightarrow{\text{ }s} \times \widehat{r}}{r^{2}}\]

where \(\mu_{0}\) is a constant called the permeability of free space:

\[\mu_{0} = 4\pi \times 10^{- 7}\text{ }T.m/A\text{~}\text{here Tesla (T) is SI unit of}\text{~}\overrightarrow{B}\]

Notice that the expression is remarkably similar to the Coulomb's law for the electric field due to a charge element \(dq\) :

\[d\overrightarrow{E} = \frac{1}{4\pi\varepsilon_{0}}\frac{dq}{r^{2}}\widehat{r}\]

Adding up these contributions to find the magnetic field at the point \(P\) requires integrating over the current source,

\[\overrightarrow{B} = \int_{}^{}\ d\overrightarrow{B} = \frac{\mu_{0}I}{4\pi}\int_{}^{}\ \frac{d\overrightarrow{s} \times \widehat{r}}{r^{2}}\]

The integral is a vector integral, which means that the expression for Bis really three integrals, one for each component of \(\overrightarrow{\mathbf{B}}\). The vector nature of this integral appears in the cross product. Understanding how to evaluate this cross product and then perform the integral will be the key to learning how to use the Biot-Savart law.

Magnetic Field due to a Finite Straight Wire.

A thin, straight wire carrying a current \(I\) is placed along the \(x\)-axis, as shown in Figure. Evaluate the magnetic field at point P due to the segment shown in figure.

Figure 9.1.3 A thin straight wire carrying a current \(I\).

The contribution to the magnetic field due to \(Id\overrightarrow{s}\)

\[d\overrightarrow{\text{ }B} = \frac{\mu_{0}I}{4\pi}\frac{\text{ }d\overrightarrow{\text{ }s} \times \widehat{r}}{r^{2}} = \frac{\mu_{0}I}{4\pi}\frac{dxsin\theta}{r^{2}}\widehat{k}\]

which shows that the magnetic field at \(P\) will point in the \(+ \widehat{k}\) direction, or out of the page.
Simplify and carry out the integration
The variables \(\theta,x\) and \(r\) are not independent of each other. In order to complete the integration, let us rewrite the variables \(x\) and \(r\) in terms of \(\theta\). From Figure, we have

\[\left\{ \begin{matrix} r = a/sin\theta = acosec\theta \\ x = acot\theta \Rightarrow dx = - a{cosec}^{2}\theta d\theta \end{matrix} \right.\ \]

Upon substituting the above expressions, the differential contribution to the magnetic field is obtained as

\[dB = \frac{\mu_{0}I}{4\pi}\frac{\left( - a{cosec}^{2}\theta d\theta \right)sin\theta}{(acosec\theta)^{2}} = - \frac{\mu_{0}I}{4\pi a}sin\theta d\theta\]

Integrating over all angles subtended from \(- \theta_{1}\) to \(\theta_{2}\) ( a negative sign is needed for \(\theta_{1}\) in order to take into consideration the portion of the length extended in the negative \(x\)-axis from the origin), we obtain

\[B = \frac{\mu_{0}I}{4\pi a}\int_{- \theta_{1}}^{\theta_{2}}\mspace{2mu} sin\theta d\theta = \frac{\mu_{0}I}{4\pi a}\left( cos\theta_{2} + cos\theta_{1} \right)\]

The first term involving \(\theta_{2}\) accounts for the contribution from the portion along the \(+ x\) axis, while the second term involving \(\theta_{1}\) contains the contribution from the portion along the \(- x\) axis. The two terms add.

Special cases :

(i) Magnetic field on the perpendicular bisector of a finite straight wire of length \(\mathbf{2}\mathbf{L}\)

In this case where \(\theta_{2} = \theta_{1} = \theta\), the field point \(P\) is located along the perpendicular bisector. If the length of the rod is \(2L\), then \(cos\theta = L/\sqrt{L^{2} + a^{2}}\) and the magnetic field is

\[B = \frac{\mu_{0}I}{2\pi a}cos\theta = \frac{\mu_{0}I}{2\pi a}\frac{\text{ }L}{\sqrt{{\text{ }L}^{2} + a^{2}}}\]

(ii) Magnetic field due to semiinfinite straight wire

Here \(\theta_{1} = 90^{\circ},\theta_{2} = 0^{\circ}\) or \(\theta_{1} = 0^{\circ},\theta_{2} = 90^{\circ}\)

\[B = \frac{\mu_{0}I}{4\pi a}\]

(iii) Magnetic field due to infinite straight wire

Here \(\theta_{1} = \theta_{2} = 0\)

\[B = \frac{\mu_{0}I}{2\pi a}\]

Direction of magnetic field of a straight wire

Note that in this limit, the system possesses cylindrical symmetry, and the magnetic field lines are circular, as shown in Figure.

In fact, the direction of the magnetic field due to a long straight wire can be determined by the right-hand rule (Figure). If you direct your right thumb along the direction of the current in the wire, then the fingers of your right hand curl in the direction of the magnetic field.

Practice Exercise

Q. 1 Find the magnetic field \(B\) at the centre of a rectangular loop of length \(l\) and width \(b\), carrying a current \(i\).
Q. 2 A long wire carrying a current i is bent to form a plane angle \(\alpha\). Find the magnetic field B at a point on the bisector of this angle situated at a distance x from the vertex.
Q. 3 A pair of stationary and infinitely long bent wires are placed in the \(x - y\) plane as shown in figure. The wires carry currents of 10 ampere each as shown.

The segments L and M are along the x -axis. The segments P and Q are parallel to the y -axis such that \(OS = OR = 0.02\text{ }m\). Find the magnitude and direction of the magnetic induction at the origin O .

Q. 4 Two long straight parallel wires are 2 m apart, perpendicular to the plane of the paper. The wire \(A\) carries a current of 9.6 ampere, directed into the plane of the paper. The wire B carries a current such that the magnetic field of induction at the point \(P\) at a distance \((10/11)m\) from the wire \(B\) is zero. Find (a) the magnitude and direction of the current in \(B\), (b) the magnitude of the magnetic field of induction at the point \(S\),

Q. 5 Two large metal sheets carry surface currents as shown in figure. The current through a strip of width \(dl\) is \(Kdl\) where \(K\) is a constant. Find the magnetic field at the points \(P,Q\) and \(R\).

Answers

Q. \(1\ \frac{2\mu_{0}i\sqrt{l^{2} + b^{2}}}{\pi l\text{ }b}\)
Q. \(2\ \frac{\mu_{0}i}{2\pi r}cot\frac{\alpha}{4}\)
Q. \(3\ \text{ }B\) is \(10^{- 4}\text{ }T\) and out of the page
Q. 4 (a) -3 A and opposite to that in A (b) \(1.3 \times 10^{- 6}\text{ }T\)
Q. \(5\ 0,\mu_{0}\text{ }K\) towards right in the figure, 0

\[\begin{matrix} & dl = ad\theta \\ & dB = \frac{\mu_{0}}{4\pi}\frac{I(ad\theta)sin90^{0}}{a^{2}} = \frac{\mu_{0}I}{4\pi a}d\theta \\ & B = \int_{}^{}\ dB = \frac{\mu_{0}I}{4\pi a}\int_{0}^{\beta}\mspace{2mu}\mspace{2mu} d\theta \\ & B = \frac{\mu_{0}I}{4\pi R}(\beta) \end{matrix}\]

Practice Exercise

Q. 1 The wire loop \(PQRSP\) formed by joining two semi-circular wires of radii \(R_{1}\) and \(R_{2}\) carries a current \(I\) as shown in figure. What is the magnetic induction at the centre \(O\) and magnetic moment of the loop in cases (a) and (b)?

(a)

(b)

Q. 2 A current I flows along a thin wire shaped as shown in Fig. The radius of a curved part of the wire is equal to \(R\), the angle \(2\varphi\). Find the magnetic induction of the field at the point \(O\).

Q. 3 Find the magnetic induction of the field at the point O of a loop with current I , whose shape is illustrated

Q. 4 Find the magnetic induction of the field at the point O if a current-carrying wire has the shape shown in Fig. . The radius of the curved part of the wire is R, the linear parts are assumed to be very long.

Q. 5 Two circular coils of radii \(5^{'}0\text{ }cm\) and 10 cm carry equal currents of 2 A . The coils have 50 and 100 turns respectively and are placed in such a way that their planes as well as the centres coincide. Find the magnitude of the 'magnetic field \(B\) at the common centre of the coils if the currents in the coils are (a) in the same sense (b) in the opposite sense.
Q. 6 If the outer coil of the previous problem is rotated through \(90^{\circ}\) about a diameter, what would be the magnitude of the magnetic field \(B\) at the centre?
Q. 7 A non-conducting thin ring of radius \(R\) and charge \(q\) rotates about its axis with an angular velocity. Find the magnetic induction at the centre of the ring.
Q. 8 A circular disk of radius \(R\) with uniform charge density \(\sigma\) rotates with an angular speed \(\omega\). Find the magnetic field at the center of the disk.

Answers

Q. 1 (a) \(\overrightarrow{B} = \frac{\mu_{0}}{4\pi}\pi I\left\lbrack \frac{1}{R_{1}} - \frac{1}{R_{2}} \right\rbrack\) out of the page and \(\overrightarrow{M} = \frac{1}{2}\pi I\left\lbrack R_{2}^{2} - R_{1}^{2} \right\rbrack\) into the page
(b) \(\overrightarrow{B} = \frac{\mu_{0}}{4\pi}\pi I\left\lbrack \frac{1}{R_{1}} + \frac{1}{R_{2}} \right\rbrack\) into the page and \(\overrightarrow{M} = \frac{1}{2}\pi I\left\lbrack R_{2}^{2} + R_{1}^{2} \right\rbrack\) into the page
Q. \(2\ \text{ }B = (\pi - \varphi + tan\phi)\mu_{0}I/2\pi R\)
Q. \(3\ B = \frac{\mu_{0}I}{4\pi}\left( \frac{2\pi - \varphi}{a} + \frac{\varphi}{b} \right)\)
Q. \(4\ B = \left( \mu_{0}/4\pi \right)(1 + 3\pi/2)I/R\)
Q. \(5\ \) (a) \(8\pi \times 10^{- 4}\text{ }T\) (b) zero
Q. 61.8 mT
Q. \(7\ B = \frac{\mu_{o}q\omega}{4\pi R}\)
Q. \(8\ \text{ }B = \frac{1}{2}\mu_{0}\sigma\omega R\)

Consider a circular loop of radius a carrying a current \(i\). We have to find the magnetic field at a Point \(P\) on the axis of the loop at a distance \(d\) from its centre O . In figure

Figure shows the geometry for calculating the magnet ic field at a point on the axis of a circular current loop a distance x from its center. We first consider the current element at the top of the loop. Here, as everywhere around the loop, \(I\overrightarrow{\mathcal{l}}\) is tangent to the loop and perpendicular to the vector \(\overrightarrow{r}\) from the current element to the field point P . The magnetic field \(d\overrightarrow{B}\) due to this element is in the direction shown in the figure, perpendicular to \(\overrightarrow{r}\) and also perpendicular to \(I\overrightarrow{\mathcal{l}}\) The magnitude of \(d\overrightarrow{B}\) is

\[dB = \frac{\mu_{0}}{4\pi}\frac{Id\mathcal{l}sin90}{r^{2}}\]

When we sum around all the current elements in the loop, the components of \(d\overrightarrow{B}\) perpendicular to the axis of the loop, such as \({dB}_{y}\) in Figure sum to zero, leaving only the components \({dB}_{x}\) that are parallel to the axis. We thus compute only the \(x\) component of the field.
From Figure, we have

\[B_{x} = \int_{}^{}\ dBsin\theta = \int_{}^{}\ sin\theta\frac{\mu_{0}I}{4\pi r^{2}}dl = \frac{\mu_{0}I}{4\pi r^{2}}sin\theta\int_{}^{}\ dl = \frac{\mu_{0}I}{4\pi r^{2}}\frac{R}{r}(2\pi R) = \frac{\mu_{0}I\left( 2\pi R^{2} \right)}{4\pi r^{3}}\]

using the facts that

\[r^{2} = x^{2} + R^{2}\]

we get

\[B = \frac{\mu_{0}I\left( 2\pi a^{2} \right)}{4\pi\left( a^{2} + x^{2} \right)^{3/2}}\]

Note: If loop has N close turns then field becomes N times the field due to one turn.

Magnetic field lines due to a circular current

Magnetic Field due to a Solenoid at a point on its axis

A solenoid is a long cylindrical helix, which is obtained by winding closely a large number of turns of insulated copper wire over an in tube of insulating material. When electric current is passed through it, a magnetic field is produced around and within the solenoid.

Consider a solenoid of length \(L\) consisting of \(N\) turns of wi re carrying a current \(I\). We choose the axis of the solenoid to be along the \(x\) axis, with the left end at \(x = - a\) and the right end at \(x = + b\) as shown in Figure. We will calculate the magnetic field at the origin. The figure shows an element of the solenoid of
length dx at a distance x from the origin. If \(n = N/L\) is the number of turns per unit length, there are ndx turns of wire in this element, with each turn carrying a current I. The element is thus equivalent to a single loop carrying a current \(di = nIdx\). The magnetic field at a point on the x axis due to a loop at the origin carrying a current nI dx is given by Equation with I replaced by nIdx:

\[dB = \frac{\mu_{0}(ndx)iR^{2}}{2\left( R^{2} + x^{2} \right)^{3/2}}\text{~}\text{along the axis}\text{~}\]

Net magnetic induction

\[\begin{matrix} & \ B = \int_{}^{}\ dB = \int_{- l_{2}}^{l_{2}}\mspace{2mu}\mspace{2mu}\frac{\mu_{0}niR^{2}dx}{2\left( R^{2} + x^{2} \right)^{3/2}} \\ & \ \Rightarrow \ B = \frac{\mu_{0}ni}{2}\left\lbrack \frac{a}{\sqrt{R^{2} + l_{1}^{2}}} + \frac{b}{\sqrt{R^{2} + l_{2}^{2}}} \right\rbrack = \frac{\mu_{0}ni}{2}\left\lbrack cos\theta_{1} + cos\theta_{2} \right\rbrack \\ & B = \frac{\mu_{0}ni}{2}\lbrack cos\alpha + cos\beta\rbrack \end{matrix}\]

Ampere's Law :

Like Gauss's law in electrostatics, this law provides us a simple method to find magnetic fields in cases of symmetry
Ampere's law gives another method to calculate the magnetic field due to a given current distribution.

Statement: The circulation \(\text{∮}\ \overrightarrow{B} \cdot d\overrightarrow{l}\) of the resultant magnetic field (of a closed circuit or an infinite wire containing steady current) along a closed path (called amperian path) is equal to \(\mu_{0}\) times the total current crossing the area bounded by the closed curve provided the electric field inside the loop remains constant. Thus,

\[\text{∮}\ \ \overrightarrow{B} \cdot \text{ }d\overrightarrow{l} = \mu_{0}i\]

In figure, the positive side is going into the plane of the diagram so that \(i_{1}\) and \(i_{2}\) are positive and \(i_{3}\) is negative. Thus, the total current crossing the area is \(i_{1} + i_{2} - i_{3}\). Any current outside the area is not included in writing the right - hand side of equation. The magnetic field \(\overrightarrow{B}\) on the left -hand side is the resultant field due to all the currents existing anywhere.

Ampere's law may be derived from the Biot-Savart law and Bio-Savart law may be derived from the Ampere's law. Thus, the two are equivalent in scientific content. However, Ampere's law is useful under certain symmetrical conditions.

Justification of Ampere's law

Let us consider a long straight wire carrying a current I in upward direction. Now take a circular path of radius r symmetric to the wire. Let us now divide a circular path of radius \(r\) into a large number of small length vectors \(\Delta\overrightarrow{s} = \Delta s\widehat{\varphi}\), where \(\widehat{\phi}\) point along the tangential direction with magnitude \(\Delta s\) (Figure).

Figure : Amperian loop

In the limit \(\Delta\overrightarrow{s} \rightarrow \overrightarrow{0}\), we obtain

\[\text{∮}\ \ \overrightarrow{B} \cdot \text{ }d\overrightarrow{\text{ }s} = B\text{∮}\ \ ds = \left( \frac{\mu_{0}I}{2\pi r} \right)(2\pi r) = \mu_{0}I\]

The result above is obtained by choosing a closed path, or an "Amperian loop" that follows one particular magnetic field line. Let's consider a slightly more complicated Amperian loop, as that shown in Figure.

Figure : An Amperian loop involving two field lines

The line integral of the magnetic field around the contour \(abcda\) is

\[{\text{∮}\ \ }_{\text{abcda}\text{~}}\overrightarrow{B} \cdot \text{ }d\overrightarrow{\text{ }s} = {\text{∮}\ \ }_{ab}\overrightarrow{\text{ }B} \cdot \text{ }d\overrightarrow{\text{ }s} + {\text{∮}\ \ }_{bc}\overrightarrow{\text{ }B} \cdot \text{ }d\overrightarrow{\text{ }s} + {\text{∮}\ \ }_{ed}\overrightarrow{\text{ }B} \cdot \text{ }d\overrightarrow{\text{ }s} + {\text{∮}\ \ }_{cd}\overrightarrow{\text{ }B} \cdot \text{ }d\overrightarrow{\text{ }s} = 0 + B_{2}\left( r_{2}\theta \right) + 0 + B_{1}\left\lbrack r_{1}(2\pi - \theta) \right\rbrack\]

where the length of \(arcbc\) is \(r_{2}\theta\), and \(r_{1}(2\pi - \theta)\) for \(arcda\). The first and the third integrals vanish since the magnetic field is perpendicular to the paths of integration. With \(B_{1} = \mu_{0}I/2\pi r_{1}\) and \(B_{2} = \mu_{0}I/2\pi r_{2}\), the above expression becomes

\[{\text{∮}\ \ }_{\text{abcda}\text{~}}\overrightarrow{B} \cdot \text{ }d\overrightarrow{\text{ }s} = \frac{\mu_{0}I}{2\pi r_{2}}\left( r_{2}\theta \right) + \frac{\mu_{0}I}{2\pi r_{1}}\left\lbrack r_{1}(2\pi - \theta) \right\rbrack = \frac{\mu_{0}I}{2\pi}\theta + \frac{\mu_{0}I}{2\pi}(2\pi - \theta) = \mu_{0}I\]

We see that the same result is obtained whether the closed path involves one or two magnetic field lines. As shown above Example, in polar coordinates \((r,\varphi)\) with current flowing in the \(+ z\)-axis, the magnetic
field is given by \(\overrightarrow{B} = \left( \mu_{0}I/2\pi r \right)\widehat{\varphi}\). An arbitrary length element in the polar coordinates can be written as \(:\ d\overrightarrow{s} = dr\widehat{r} + rd\phi\widehat{\phi}\)
which implies

\[{\text{∮}\ \ }_{\text{closed path}\text{~}}\overrightarrow{B} \cdot \text{ }d\overrightarrow{\text{ }s} = {\text{∮}\ \ }_{\text{closed path}\text{~}}\left( \frac{\mu_{0}I}{2\pi r} \right)rd\varphi = \frac{\mu_{0}I}{2\pi}\ {\text{∮}\ \ }_{\text{closed path}\text{~}}d\varphi = \frac{\mu_{0}I}{2\pi}(2\pi) = \mu_{0}I\]

In other words, the line integral of \(\text{∮}\ \overrightarrow{B} \cdot d\overrightarrow{s}\) around any closed Amperian loop is proportional to \(I_{\text{enc}\text{~}}\), the current encircled by the loop.

Figure : An Amperian loop of arbitrary shape.

The generalization to any closed loop of arbitrary shape (see for example, Figure) that involves many magnetic field lines is known as Ampere's law:

\[\text{∮}\ \ \overrightarrow{B} \cdot \text{ }d\overrightarrow{\text{ }s} = \mu_{0}I_{enc}\]

Calculation of magnetic field by using Ampere's Law :

Calculation of magnetic field due to long straight wire.

Figure shows a long, straight current i . We have to calculate the magnetic field at a point P which is at a distance \(r\) from the wire. Figure shows the situation in the plane perpendicular to the wire and passing through \(P\). The current is perpendicular to the plane of the diagram and is coming out of it.

Let us draw a circle passing through the point and with the axis as wire. We put an arrow to show the
positive sense of the circle. The radius of the circle is r. The magnetic field due to the long, straight current at any point on the circle is along the tangent as shown in the figure. Same is the direction of the length-element dl there. By symmetry, all points of the circle are equivalent and hence the magnitude of the magnetic field should be the same at all these points. The circulation of magnetic field along the circle is

\[\text{∮}\ \ \overrightarrow{B} \cdot \overrightarrow{dl} = \text{∮}\ \ Bdl = B\text{∮}\ \ dl = B(2\pi r)\]

The current crossing the area bounded by the circle is

\[\sum_{}^{}\ I_{en} = + I\]

Thus, from Ampere's law,

\[\begin{matrix} & B(2\pi r) = \mu_{0}I \\ \Rightarrow \ & B = \frac{\mu_{0}I}{2\pi r} \end{matrix}\]

Magnetic field due to a current carrying thin long pipe.

Case I : \(r > R\)
\[{\text{∮}\ \overrightarrow{B} \cdot \overrightarrow{dl} = \text{∮}\ B \cdot dl = B\text{∮}\ dl = B(2\pi r) }{\sum I_{en} = + I }{\Rightarrow B(2\pi r) = \mu_{0}I }{\Rightarrow B = \frac{\mu_{0}I}{2\pi r} }\]

Case II : \(r < R\)
\[{\text{∮}\ \overrightarrow{B} \cdot \overrightarrow{dl} = \text{∮}\ Bdl = B\text{∮}\ dl = B(2\pi r) }{\sum I_{en} = 0 }{\Rightarrow B(2\pi r) = \mu_{0}(0) }{\Rightarrow B = 0 }\]

Magnetic field due to current carrying rod having uniform current density.

Case I : \(r > R\)
\[{\text{∮}\ \overrightarrow{B} \cdot \overrightarrow{dl} = \text{∮}\ B \cdot dl = B\text{∮}\ dl = B(2\pi r) }{\sum I_{en} = + I }{\Rightarrow B(2\pi r) = \mu_{0}I }{\Rightarrow B = \frac{\mu_{0}I}{2\pi r} }\]

Case II : \(r < R\)
\[{\text{∮}\ \overrightarrow{B} \cdot \overrightarrow{dl} = \text{∮}\ Bdl = B\text{∮}\ dl = B(2\pi r) }{\sum I_{\text{en}\text{~}} = \frac{I}{R^{2}}r^{2} }{\Rightarrow B(2\pi r) = \mu_{0}\frac{I}{R^{2}}r^{2} }\]

\[\Rightarrow B = \frac{\mu_{0}I}{2\pi R^{2}}r \]

Magnetic Field Due to non uniform current density

Suppose that the current density in a wire of radius a varies with r according to \(J = {Kr}^{2}\), where K is a constant and \(r\) is the distance from the axis of the wire. We have to find the magnetic field at a point distance \(r\) from the axis when (a) \(r < a\) and (b) \(r > a\)

Choose a circular path centred on the axis of the conductor and apply Ampere's law
(a) To find the current passing through the area enclosed by the path integrate

\[dI = JdA = \left( {Kr}^{2} \right)(2\pi rdr)\]

\[\Rightarrow \ I = \int_{}^{}\ dI = K\int_{0}^{'}\mspace{2mu} 2\pi r^{3}dr = \frac{K\pi r^{4}}{2}\]

Since \(\int_{}^{}\ \overrightarrow{B} \cdot d\overrightarrow{\mathcal{l}} = \mu_{\Delta} \mid\)
\[\Rightarrow B2\pi r = \mu_{l} \cdot \frac{\pi{Kr}^{4}}{2}\ \Rightarrow \text{ }B = \frac{\mu_{l}{Kr}^{3}}{4} \](b) If \(r > a\), then net current through the Amperian loop is
\[{I = \int_{a}^{\infty}\mspace{2mu} Kr^{2}2\pi rdr = \frac{\pi Ka^{4}}{2} }{\Rightarrow B = \frac{\mu_{a}{Ka}^{*}}{4r}}\]

Practice Exercise

Q. 1 Inside a long straight uniform wire of round cross-section there is a long round cylindrical cavity whose axis is parallel to the axis of the wire and displaced from the latter by a distance \(\overrightarrow{\mathcal{l}}\). A direct current of density \(\overrightarrow{j}\) flows along the wire. Find the magnetic induction inside the cavity. Consider, in particular, the case \(I = 0\).

Answers

Q. \(1\ B = (1/2)\mu_{0}\lbrack\overrightarrow{j} \times \overrightarrow{\mathcal{l}}\rbrack\), i.e. field inside the cavity is uniform.

A solenoid is a long coil of wire tightly wound in the helical form. Figure shows the magnetic field lines of a solenoid carrying a steady current \(I\). We see that if the turns are closely spaced, the resulting magnetic field inside the solenoid becomes fairly uniform, provided that the length of the solenoid is much greater than its diameter. For an "ideal" solenoid, which is infinitely long with turns tightly packed, the magnetic field inside the solenoid is uniform and parallel to the axis, and vanishes outside the solenoid.

Figure : Magnetic field lines of a solenoid

We can use Ampere's law to calculate the magnetic field strength inside an ideal solenoid. The crosssectional view of an ideal solenoid is shown in Figure. To compute \(\overrightarrow{B}\), we consider a rectangular path of length \(\mathcal{l}\) and width \(w\) and traverse the path in a counterclockwise manner. The line integral of \({\overrightarrow{B}}^{\text{along}\text{~}}\) this loop is

\[\begin{matrix} \text{∮}\ \ \overrightarrow{B} \cdot \text{ }d\overrightarrow{\text{ }s} & \ = {\text{∮}\ \ }_{1}\overrightarrow{\text{ }B} \cdot \text{ }d\overrightarrow{\text{ }s} + {\text{∮}\ \ }_{2}\overrightarrow{\text{ }B} \cdot \text{ }d\overrightarrow{\text{ }s} + {\text{∮}\ \ }_{3}\overrightarrow{\text{ }B} \cdot \text{ }d\overrightarrow{\text{ }s} + {\text{∮}\ \ }_{4}\overrightarrow{\text{ }B} \cdot \text{ }d\overrightarrow{\text{ }s} \\ & \ = 0 + 0 + B\mathcal{l} + 0 \end{matrix}\]

Figure : Amperian loop for calculating the magnetic field of an ideal solenoid.

In the above, the contributions along sides 2 and 4 are zero because \(\overrightarrow{B}\) is perpendicular to \(d\overrightarrow{s}\). In addition, \(\overrightarrow{\mathbf{B}} = \overrightarrow{0}\) along side 1 because the magnetic field is non-zero only inside the solenoid. On the other hand, the total current enclosed by the Amperian loop is \(I_{\text{enc}\text{~}} = n/I\), where \(n\) is the total number of turns per unit length. Applying Ampere's law yields

\[\text{∮}\ \ \overrightarrow{B}.d\overrightarrow{\text{ }s} = B\mathcal{l} = \mu_{0}n\mathcal{l}I\ \text{ }B = \mu_{0}nI\]

Magnetic Field due to Toroid

Consider a toroid which consists of \(N\) turns, as shown in Figure. Find the magnetic field everywhere.

Figure : A toroid with \(N\) turns

One can think of a toroid as a solenoid wrapped around with its ends connected. Thus, the magnetic field is completely confined inside the toroid and the field points in the azimuthal direction (clockwise due to the way the current flows, as shown in Figure 9.4.5.)
Applying Ampere's law, we obtain

\[\text{∮}\ \ \overrightarrow{B} \cdot \text{ }d\overrightarrow{\text{ }s} = \text{∮}\ \ Bds = B\text{∮}\ \ ds = B(2\pi r) = \mu_{0}NI\]

or

\[B = \frac{\mu_{0}NI}{2\pi r}\]

where \(r\) is the distance measured from the center of the toroid. Unlike the magnetic field of a solenoid, the magnetic field inside the toroid is non-uniform and decreases as \(1/r\).

Magnetic Field thin sheet of infinite dimension carrying a current of uniform linear current density ' \(i\) '.

\[\begin{matrix} & \text{∮}\ \ \overrightarrow{B} \cdot \overrightarrow{dl} = \int_{a}^{b}\mspace{2mu}\mspace{2mu}\overrightarrow{B} \cdot \overrightarrow{dl} + \int_{b}^{c}\mspace{2mu}\mspace{2mu} B \cdot \overrightarrow{dl} + \int_{c}^{d}\mspace{2mu}\mspace{2mu}\overrightarrow{B} \cdot \overrightarrow{dl} + \int_{d}^{e}\mspace{2mu}\mspace{2mu}\overrightarrow{B} \cdot \overrightarrow{dl} + \int_{e}^{f}\mspace{2mu}\mspace{2mu}\overrightarrow{B} \cdot \overrightarrow{dl} + \int_{f}^{g}\mspace{2mu}\mspace{2mu}\overrightarrow{B} \cdot \overrightarrow{dl} \\ & \ = \int_{a}^{b}\mspace{2mu}\mspace{2mu} B \cdot cos0 \cdot dl + 0 + 0 + \int_{d}^{e}\mspace{2mu}\mspace{2mu} Bdl + 0 + 0 = B\int_{a}^{b}\mspace{2mu}\mspace{2mu} dl + B\int_{d}^{e}\mspace{2mu}\mspace{2mu} dl = Bl + Bl = 2Bl \\ & \ \sum_{}^{}\ I_{en} = il \\ & \text{∮}\ \ \overrightarrow{B} \cdot \overrightarrow{dl} = \mu_{0}\sum_{}^{}\ I_{en} \\ & \ \Rightarrow 2Bl = \mu_{0}il \\ & \ \Rightarrow B = \frac{\mu_{0}i}{2} \end{matrix}\]

Practice Exercise

Q. 1 We have a long cylindrical shell of non-conducting material which carries a surface charge fixed in place (glued down) of \(\sigma C/m^{2}\), as shown in Figure. The cylinder is suspended in a manner such that it is free to revolve about its axis, without friction. Initially it is at rest. We come along and spin it up until the speed of the surface of the cylinder is \(v_{0}\).

(a) What is the surface current \(K\) on the walls of the cylinder, in \(A/m\) ?
(b) What is magnetic field inside the cylinder?
(c) What is the magnetic field outside of the cylinder? Assume that the cylinder is infinitely long.
Q. 2 Sometimes we show an idealised magnetic field which is uniform in a given region and falls to zero abruptly. One such field is represented in figure. Using Ampere's law over the path PQRS, show that such a field is not possible.

Answers

Q. 1 (a) \(K = \sigma v_{0}\)
\[\begin{matrix} \text{~}\text{(b)}\text{~}\ B = \mu_{0}K = \mu_{0}\sigma v_{0} & \text{, oriented along axis right-handed with respect to spin}\text{~} & \text{~}\text{(c)}\text{~}0 \end{matrix}\]

Suppose we have an infinitesimal current element in the form of a cylinder of cross-sectional area \(A\) and length \(ds\) consisting of \(n\) charge carriers per unit volume, all moving at a common velocity \(\overrightarrow{v}\) along the axis of the cylinder. Let \(I\) be the current in the element, which we define as the amount of charge passing through any cross-section of the cylinder per unit time. From Chapter 6, we see that the current I can be written as

\[nAq|\overrightarrow{v}| = I\]

The total number of charge carriers in the current element is simply \(dN = nAds\), so that the magnetic field \(d\overrightarrow{B}\) due to the \(dN\) charge carriers is given by

\[dB = \frac{\mu_{0}}{4\pi}\frac{(nAq|\overrightarrow{v}|)d\overrightarrow{\text{ }s} \times \widehat{r}}{r^{2}} = \frac{\mu_{0}}{4\pi}\frac{(nAds)q\overrightarrow{v} \times \widehat{r}}{r^{2}} = \frac{\mu_{0}}{4\pi}\frac{(dN)q\overrightarrow{v} \times \widehat{r}}{r^{2}}\]

where \(r\) is the distance between the charge and the field point \(P\) at which the field is being measured, the unit vector \(\widehat{r} = \overrightarrow{r}/r\) points from the source of the field (the charge) to \(P\). The differential length vector \(d\overrightarrow{s}\) is defined to be parallel to \(\overrightarrow{v}\). In case of a single charge, \(dN = 1\), the above equation becomes

\[\overrightarrow{B} = \frac{\mu_{0}}{4\pi}\frac{q\overrightarrow{v} \times \widehat{r}}{r^{2}}\]

Practice Exercise

Q. 1 An electron carrying a charge \(e = - 1.6 \times 10^{- 19}C\) moves in a straight line at a speed \(v = 3 \times 10^{7}\text{ }m/s\). What is the magnitude and direction of the magnetic field caused by the electron at a point \(P,10\text{ }mm\) ahead of the electron and 20 mm away from its line of motion, as illustrated in figure?

Answers

Q. \(1\ 8.8 \times 10^{- 14}\text{ }T\)

Consider a particle of charge \(q\) and moving at a velocity \(\overrightarrow{v}\). Experimentally we have the following observations:
(1) The magnitude of the magnetic force \({\overrightarrow{F}}_{B}\) exerted on the charged particle is proportional to both \(v\) and \(q\).
(2) The magnitude and direction of \({\overrightarrow{F}}_{B}\) depends on \(\overrightarrow{v}\) and \(\overrightarrow{B}\).
(3) The magnetic force \({\overrightarrow{F}}_{B}\) vanishes when \(\overrightarrow{v}\) is parallel to \(\overrightarrow{B}\). However, when \(\overrightarrow{v}\) makes an angle \(\theta\) with \(h_{\overrightarrow{B}}\), the direction of \({\overrightarrow{F}}_{B}\) is perpendicular to the plane formed by and \(\overrightarrow{v}\) and \(\overrightarrow{B}\) and the magnitude of \({\overrightarrow{F}}_{B}\) is proportional to \(sin\theta\).
(4) When the sign of the charge of the particle is switched from positive to negative (or vice versa), the direction of the magnetic force also reverses.

Figure : The direction of the magnetic force

The above observations can be summarized with the following equation:

\[{\overrightarrow{F}}_{B} = q\overrightarrow{v} \times \overrightarrow{B}\]

The above expression can be taken as the working definition of the magnetic field at a point in space.
The magnitude of \({\overrightarrow{F}}_{B}\) is given by

\[F_{B} = |q|vBsin\theta\]

The SI unit of magnetic field is the tesla (T):

\[1\text{~}\text{tesla}\text{~} = 1\text{ }T = 1\frac{\text{~}\text{Newton}\text{~}}{(\text{~}\text{Coulomb}\text{~})(\text{~}\text{meter}\text{~}/\text{~}\text{second}\text{~})} = 1\frac{\text{ }N}{C \cdot \text{ }m/s} = 1\frac{\text{ }N}{\text{ }A \cdot \text{ }m}\]

Another commonly used non-SI unit for \(\overrightarrow{B}\) is the gauss ( G ), where \(1\text{ }T = 10^{4}G\)

Note that \({\overrightarrow{F}}_{B}\) is always perpendicular to \(\overrightarrow{v}\) and \(\overrightarrow{B}\), and cannot change the particle's speed \(v\) (and thus the kinetic energy). In other words, magnetic force cannot speed up or slow down a charged particle. Consequently, Magnetic field does no work on charged particle:

\[dW = {\overrightarrow{F}}_{B} \cdot \text{ }d\overrightarrow{\text{ }s} = q(\overrightarrow{v} \times \overrightarrow{B}) \cdot \overrightarrow{v}dt = q(\overrightarrow{v} \times \overrightarrow{v}) \cdot \overrightarrow{B}dt = 0\]

The direction of \(\overrightarrow{v}\) however, can be altered by the magnetic force

Illustration :

Two very long, straight, parallel wires carry steady currents I and -I respectively. The distance between the wires is \(d\). At a certain instant of time, a point charge \(q\) is at a point equidistant from the two wires, in the plane of the wires.Find magnitude of the force due to the magnetic field acting on the charge .

Sol. Since the currents are flowing in the opposite directions, the magnetic field at a point equidistant from the two wires will be zero. Hence, the force acting on the charge at this instant will be zero.

Practice Exercise

Q. 1 A circular loop of radius 20 cm carries a current of 10 A . An electron crosses the plane of the loop with a speed of \(2.0 \times 106\text{ }m/s\). The direction of motion makes an angle of \(30^{\circ}\) with the axis of the circle and passes through its centre. Find the magnitude of the magnetic force on the electron at the instant it crosses the plane.
Q. 2 Two protons move parallel to each other with an equal velocity \(v\). Find the ratio of forces of magnetic and electrical interaction of the protons.

Answers

Q. \(116\pi \times 10^{- 19}N\ \) Q. \(2\ (v/c)^{2}\)

There are three possible paths in which a charged particle may move in presence of uniform magnetic field which is uniform in space.
(a) Straight line path
(b) Circular path
(c) Helical path

We shall see them one by one.

Straight line path

When the charged particle projected in the direction of or opposite to uniform magnetic field, magnetic field exerts no force hence it will travel along straight line with fixed speed.

Circular path

If a particle of mass \(m\) moves in a circle of radius \(r\) at a constant speed \(v\), what acts on the particle is a radial force of magnitude \(F = {mv}^{2}/r\) that always points toward the center and is perpendicular to the velocity of the particle.

In previous section, we have also shown that the magnetic force \({\overrightarrow{F}}_{B}\) always points in the direction perpendicular to the velocity \(\overrightarrow{v}\) of the charged particle and the magnetic field \(\overrightarrow{B}\). Since \({\overrightarrow{F}}_{B}\) can do not work, it can only change the direction of \(\overrightarrow{v}\) but not its magnitude. What would happen if a charged particle moves through a uniform magnetic field \(\overrightarrow{B}\) with its initial velocity \(\overrightarrow{v}\) at a right angle to \(\overrightarrow{B}\) ? For simplicity, let the charge be \(+ q\) and the direction of \(\overrightarrow{B}\) be into the page. It turns out that \({\overrightarrow{F}}_{B}\) will play the role of a centripetal force and the charged particle will move in a circular path in a counterclockwise direction, as shown in Figure.

\[qvB = \frac{mv^{2}}{r}\]

the radius of the circle is found to be

\[r = \frac{mv}{qB}\]

The period \(T\) (time required for one complete revolution) is

\[T = \frac{2\pi r}{v} = \frac{2\pi}{v}\frac{mv}{qB} = \frac{2\pi\text{ }m}{qB}\]

Similarly, the angular speed of the particle can be obtained as

\[\omega = 2\pi f = \frac{v}{r} = \frac{qB}{m}\]

Practice Exercise

Q. 1 Particle \(A\) with charge \(q\) and mass \(m_{A}\) and particle \(B\) with charge \(2q\) and mass \(m_{B}\), are accelerated from rest by a potential difference \(\Delta V\), and subsequently deflected by a uniform magnetic field into semicircular paths. The radii of the trajectories by particle \(A\) and \(B\) are \(R\) and \(2R\), respectively. The direction of the magnetic field is perpendicular to the velocity of the particle. What is their mass ratio?
Q. 2 Suppose the entire \(x\) - \(y\) plane to the right of the origin \(O\) is filled with a uniform magnetic field \(\overrightarrow{B}\) pointing out of the page, as shown in Figure.

Two charged particles travel along the \(x\) axis in the positive \(x\) direction, each with speed \(v\), and enter the magnetic field at the origin \(O\). The two particles have the same charge \(q\), but have different masses, \(m_{1}\) and \(m_{2}\). When in the magnetic field, their trajectories both curve in the same direction, but describe semicircles with different radii. The radius of the semi-circle traced out by particle 2 is exactly twice as big as the radius of the semi-circle traced out by particle 1 .
(a) Is the charge \(q\) of these particles such that \(q > 0\), or is \(q < 0\) ?
(b) What is the ratio \(m_{2}/m_{1}\) ?
Q. 3 A particle of mass \(m\) and positive charge \(q\), moving with a uniform velocity \(v\), enters a magnetic field \(B\) as shown in figure (a) Find the radius of the circular arc it describes in the magnetic field. (b) Find the angle subtended by the arc at the centre. (c) How long does the particle stay inside the magnetic field? (d) Solve the three parts of the above problem if the charge \(q\) on the particle is negative.

Q. 4 A particle of charge is moving with a velocity \(\overrightarrow{v}\). It then enters midway between two long plates ( \(\mathcal{l} \rightarrow \infty\) ) where there exists a uniform magnetic field pointing into the page, as shown in Figure.Assume the space to be electric field and gravity free.Take \(d = \frac{mv}{qB}\)

(a) Is the trajectory of the particle deflected upward or downward?
(b) Compute the distance between the left end of the plate and where the particle strikes.
Q. 5 A proton accelerated by a potential difference \(V = 500kV\) flies through a uniform transverse magnetic field with induction \(B = 0.51\text{ }T\). The field occupies a region of space \(d = 10\text{ }cm\) in thickness (Fig.). Find the angle a through which the proton deviates from the initial direction of its motion.

Answers

Q. \(1\ \frac{{\text{ }m}_{A}}{m_{B}} = \frac{1}{8}\)
Q. 2 (a) - ve (b) \(2:1\)
Q. 3 (a) \(\frac{mv}{qB}\)
(b) \(\pi - 2\theta\)
(c) \(\frac{m}{qB}(\pi - 2\theta)\)
(d) \(\frac{mv}{qB},\pi + 2\theta,\frac{\text{ }m}{qB}(\pi + 2\theta)\)
Q. 4 (a) downward (b) \(d\frac{\sqrt{3}}{2}\)
Q. \(5\ a = \sin^{- 1}\left( dB\sqrt{\frac{q}{2mV}} \right) = 30^{\circ}\)

In this situation velocity of the particle can be resolved into two components.
(1) \(\ v_{||} \rightarrow\) projection of velocity parallel to the magnetic fd .
(2) \(\ v_{\bot} \rightarrow\) projection of velocity perpendicular to the magnetic fd .

Due to \(v_{\bot}\) motion of the charged particle will be uniform circular in the plane perpendicular to the fd . whereas
\(v_{||}\)remain unaffected as it is perpendicular to magnetic force.
As a whole its motion is helical with constant pitch. radius of the helix will be

\[r = \frac{{mv}_{\bot}}{qB}\]

pitch of the helix will be

\[p = \frac{{mv}_{D}}{qB}\]

Practice Exercise

Q. 1 An electron accelerated by a potential difference \(V = 1.0kV\) moves in a uniform magnetic field at an angle \(\alpha = 30^{\circ}\) to the vector B whose modulus is \(B = 29mT\). Find the pitch of the helical trajectory of the electron.
Q. 2 Can a charged particle be speed up through a uniform magnetic field?
Q. 3 If no work can be done on a charged particle by the magnetic field, how can the motion of the particle be influenced by the presence of a field?

Answers

Q. \(1\ \Delta l = 2\pi\sqrt{2mV/{eB}^{2}}\ cos\alpha = 2.0\text{ }cm\)
Q. 2 No
Q. 3 By changing its direction of motion

In the presence of both electric field \(\overrightarrow{E}\) and magnetic field \(\overrightarrow{B}\), the total force on a charged particle is

\[\overrightarrow{F} = q(\overrightarrow{E} + \overrightarrow{v} \times \overrightarrow{B})\]

This is known as the Lorentz force.

Velocity Selector:

By combining the two fields, particles which move with a certain velocity can be selected. This was the principle used by J. J. Thomson to measure the charge-to-mass ratio of the electrons. In Figure the schematic diagram of Thomson's apparatus is depicted.

Figure : Thomson's apparatus

The electrons with charge \(q = - e\) and mass \(m\) are emitted from the cathode C and then accelerated toward slit A . Let the potential difference between A and C be \(V_{A} - V_{C} = \Delta V\). The change in potential energy is equal to the external work done in accelerating the electrons: \(\Delta U = W_{\text{ext}\text{~}} = q\Delta V = - e\Delta V\). By energy conservation, the kinetic energy gained is \(\Delta K = - \Delta U = mv^{2}/2\). Thus, the speed of the electrons is given by

\[v = \sqrt{\frac{2e\Delta\text{ }V}{\text{ }m}}\]

If the electrons further pass through a region where there exists a downward uniform electric field, the electrons, being negatively charged, will be deflected upward. However, if in addition to the electric field, a magnetic field directed into the page is also applied, then the electrons will experience an additional downward magnetic force \(- e\overrightarrow{v} \times \overrightarrow{B}\). When the two forces exactly cancel, the electrons will move in a
straight path. From Eq., we see that when the condition for the cancellation of the two forces is given by \(eE = evB\). which implies \(\ \mathbf{v} = \frac{E}{B}\)
In other words, only those particles with speed \(v = E/B\) will be able to move in a straight line. Combining
the two equations, we obtain \(\frac{e}{m} = \frac{E^{2}}{2(\Delta V)B^{2}}\)
By measuring \(E,\Delta V\) and \(B\), the charge-to-mass ratio can be readily determined. The most precise measurement to date is \(e/m = 1.758820174(71) \times 10^{11}C/kg\).

Mass Spectrometer:

Various methods can be used to measure the mass of an atom. One possibility is through the use of a mass spectrometer. The basic feature of a Bainbridge mass spectrometer is illustrated in Figure. A particle carrying a charge \(+ q\) is first sent through a velocity selector.

Figure : A Bainbridge mass spectrometer

The applied electric and magnetic fields satisfy the relation \(E = vB\) so that the trajectory of the particle is a straight line. Upon entering a region where a second magnetic field \({\overrightarrow{B}}_{0}\) pointing into the page has been applied, the particle will move in a circular path with radius \(r\) and eventually strike the photographic
plate. Using Eq., we have
Since \(v = E/B\), the mass of the particle can be written as

\[\begin{matrix} & r = \frac{mv}{{qB}_{0}} \\ & \text{ }m = \frac{{qB}_{0}r}{v} = \frac{{qB}_{0}Br}{E} \end{matrix}\]

Hall's Effect:

In 1879, Edwin H. Hall, then a 24-year-old graduate student at the Johns Hopkins University, showed that they can drift electrons in copper wire in presence of magnetic field. This Hall effect allows us to find " if charge carriers in a conductor are positively or negatively charged.
" the number of charge carriers per unit volume of the conductor.
Consider a strip of current carrying wire kept in external magnetic field. Let the wire has width d , Crosssectional area A , and charge carriers per unit volume as n .

Figure (a) shows a copper strip of width d , carrying a current i whose conventional direction is from the top of the figure to the bottom. The charge carriers are electrons and, as we know, they drift (with drift speed vd) in the opposite direction, from bottom to top. At the instant shown in figure, an external
magnetic field B , pointing into the plane of the figure, has just been turned on. We see that a magnetic force will act on each drifting electron, pushing it towards the right edge of the strip.

(a)

(b)

(c)

As time goes on, electrons move to the right, mostly piling up on the right edge of the strip, leaving uncompensated positive charges in fixed positions at the left edge. The separation of positive charges on the left edge and negative charges on the right edge produces an electric field E within the strip, pointing from left to right in Fig. b. This field exerts an electric force FE on each electron, tending to push it to the left. Thus, this electric force on the electrons, which opposes the magnetic force on them, begins to build up.
Equilibrium quickly develops in which the electric force on each electron has increased enough to match the magnetic force. When this happens, as Fig. b shows, the force due to B and the force due to E are in balance. The drifting electrons then move along the strip toward the top of the page at velocity vd with no further collection of electrons on the right edge of the strip and thus no further increase in the electric fieldE.

\[\begin{array}{r} eE\ = ev_{d}B\#(1) \\ v_{d}\ = \frac{J}{ne} = \frac{i}{neA}\#(2) \end{array}\]

A Hall potential difference \(V\) is associated with the electric field across strip width \(d\).

\[\begin{array}{r} V = Ed\#(3) \end{array}\]

From (1), (2) and (3)
\[{\frac{E}{B} = \frac{V}{Bd} = \frac{i}{\text{~}\text{neA}\text{~}} }{\therefore n = \frac{idB}{eAV} }\]& \(v_{d} = \frac{i}{neV} = \frac{V}{Bd}\)
By connecting a voltmeter across the width, we can measure the potential difference between the two edges of the strip. Moreover, the voltmeter can tell us which edge is at higher potential. For the situation of Fig. b, we would find that the left edge is at higher potential, which is consistent with our assumption that the charge carriers are negatively charged.

It is also possible to use the Hall effect to measure directly the drift speed vd of the charge carriers, which you may recall is of the order of centimeters per hour. In this clever experiment, the metal strip is moved mechanically through the magnetic field in a direction opposite that of the drift velocity of the charge carriers. The speed of the moving strip is then adjusted until the Hall potential difference vanishes.
At this condition, with no Hall effect, the velocity of the charge carriers with respect to the laboratory frame must be zero, so the velocity of the strip must be equal in magnitude but opposite the direction of the velocity of the negative charge carriers.
For a moment, let us make the opposite assumption, that the charge carriers in current \(i\) are positively charged (Fig. c).
Convince yourself that as these charge carriers move from top to bottom in the strip, they are pushed to the right edge by \(F_{B}\) and thus that the right edge is at higher potential. Because that last statement is contradicted by our voltmeter reading, the charge carriers must be negatively charged.

Practice Exercise

Q. 1 A proton goes un-deflected in a crossed electric and magnetic field (the fields are perpendicular to each other) at a speed of \(2.0 \times 10^{5}\text{ }m/s\). The velocity is perpendicular to both the fields. When the electric field is switched off, the proton moves along a circle of radius 4.0 cm . Find the magnitudes of the electric and the magnetic fields. Take the mass of the proton \(1.6 \times 10^{- 27}\text{ }kg\).

Answers

Q. \(11.0 \times 10^{4}\text{ }N/C,0.05\text{ }T\)

We have just seen that a charged particle moving through a magnetic field experiences a magnetic force \({\overrightarrow{F}}_{B}\). Since electric current consists of a collection of charged particles in motion, when placed in a magnetic field, a current-carrying wire will also experience a magnetic force.

Consider a long straight wire suspended in the region between the two magnetic poles. The magnetic field points out the page and is represented with dots \(( \cdot )\). It can be readily demonstrated that when a downward current passes through, the wire is deflected to the left. However, when the current is upward, the deflection is rightward, as shown in Figure.

To calculate the force exerted on the wire, consider a segment of wire of length \(\mathcal{l}\) and cross-sectional area \(A\), as shown in Figure. The magnetic field points into the page, and is represented with crosses (X).

The charges move at an average drift velocity \({\overrightarrow{v}}_{d}\). Since the total amount of charge in this segment is \(Q_{\text{tot}\text{~}} = q(nA\mathcal{l})\), where \(n\) is the number of charges per unit volume, the total magnetic force on the segment is

\[{\overrightarrow{F}}_{B} = Q_{tot}{\overrightarrow{v}}_{d} \times \overrightarrow{B} = qnA\mathcal{l}\left( {\overrightarrow{v}}_{d} \times \overrightarrow{B} \right) = I(\overrightarrow{\mathcal{l}} \times \overrightarrow{B})\]

where \(I = {nqv}_{d}A\), and \(\overrightarrow{\mathcal{l}}\) is a length vector with a magnitude \(\mathcal{l}\) and directed along the direction of the electric current.

Special Case-1:

Wire of arbitrary shape placed in uniform magnetic field

For a wire of arbitrary shape, the magnetic force can be obtained by summing over the forces acting on the small segments that make up the wire. Let the differential segment be denoted as \(d\overrightarrow{s}\) (Figure).

The magnetic force acting on the segment is : \(d{\overrightarrow{F}}_{B} = Id\overrightarrow{s} \times \overrightarrow{B}\)
Thus, the total force is : \(\ {\overrightarrow{F}}_{B} = I\int_{a}^{b}\mspace{2mu} d\overrightarrow{s} \times \overrightarrow{B}\) where \(a\) and \(b\) represent the endpoints of the wire.
As an example, consider a curved wire carrying a current \(I\) in a uniform magnetic field \(\overrightarrow{B}\), as shown in Figure.

Using the magnetic force on the wire is given by

\[{\overrightarrow{F}}_{B} = I\left( \int_{a}^{b}\mspace{2mu}\mspace{2mu}\text{ }d\overrightarrow{\text{ }s} \right) \times \overrightarrow{B} = I\overrightarrow{\mathcal{l}} \times \overrightarrow{B}\]

where \(\overrightarrow{\mathcal{l}}\) is the length vector directed from \(a\) to \(b\). However, if the wire forms a closed loop of arbitrary shape (Figure), then the force on the loop becomes

\[{\overrightarrow{F}}_{B} = I(\text{∮}\ \ \text{ }d\overrightarrow{\text{ }s}) \times \overrightarrow{B}\]

Special Case-2:

Magnetic Force on a closed loop in uniform magnetic field

Figure : A closed loop carrying a current \(I\) in a uniform magnetic field.

Since the set of differential length elements \(d\overrightarrow{s}\) form a closed polygon, and their vector sum is zero, i.e. \(\text{∮}\ d\overrightarrow{s} = 0\). The net magnetic force on a closed loop is \({\overrightarrow{F}}_{B} = \overrightarrow{0}\).

Practice Exercise

Q. 1 Consider a closed semi-circular loop lying in the \(xy\) plane carrying a current \(I\) in the counterclockwise direction, as shown in Figure. Find the magnetic Force on a Semi-Circular Loop

Q. 2 A conducting rod having a mass density \(\lambda kg/m\) is suspended by two flexible wires in a uniform magnetic field \(\overrightarrow{B}\) which points out of the page, as shown in Figure.

If the tension on the wires is zero, what are the magnitude and the direction of the current in the rod?
Q. 3 A rod with a mass \(m\) and a radius \(R\) is mounted on two parallel rails of length \(a\) separated by a distance \(\mathcal{l}\), as shown in the Figure. The rod carries a current \(I\) and rolls without slipping along the rails which are placed in a uniform magnetic field \({\overrightarrow{B}}^{2}\) directed into the page. If the rod is initially at rest, what is its speed as it rolls off the rails?

Q. 4 A strong magnet is placed under a horizontal conducting ring of radius r that carries current I as shown in figure. If the magnetic field B makes an angle with the vertical at the ring's location, what are the magnitude and direction of the resultant force on the ring?

Answers

Q. 12 IRB into the page
Q. \(2I = \frac{mg}{B\mathcal{l}} = \frac{\lambda g}{B}\)
Q. \(3\ v = \sqrt{\frac{4I\mathcal{l}Ba}{3\text{ }m}}\)
Q. \(4\ F_{y} = (Bsin\theta)I(2\pi r)\)

We have already seen that a current-carrying wire produces a magnetic field. In addition, when placed in a magnetic field, a wire carrying a current will experience a net force. Thus, we expect two current-carrying wires to exert force on each other. Consider two parallel wires separated by a distance \(a\) and

carrying currents \(I_{l}\ ^{1}\) and \(I_{2}\ ^{2}\) in the \(+ x\)-direction, as shown in Figure.

The magnetic force, \({\overrightarrow{F}}_{12}\), exerted on wire 1 by wire 2 may be computed as follows: Using the result from
the previous example, the magnetic field lines due to \(I_{2}\ ^{2}\) going in the \(+ x\)-direction are circles concentric with wire 2 , with the field \({\overrightarrow{B}}_{2}\) pointing in the tangential direction. Thus, at an arbitrary point \(P\) on wire 1 , we have \({\overrightarrow{B}}_{2} = - \left( \mu_{0}I_{2}/2\pi a \right)\widehat{j}\), which points in the direction perpendicular to wire 1 , as depicted in Figure. Therefore,

\[{\overrightarrow{F}}_{12} = I_{1}\overrightarrow{\mathcal{l}} \times {\overrightarrow{B}}_{2} = (\mathcal{l}\widehat{i}) \times \left( - \frac{\mu_{0}I_{2}}{2\pi a}\widehat{j} \right) = - \frac{\mu_{0}I_{1}I_{2}\mathcal{l}}{2\pi a}\widehat{k}\]

Clearly \({\overrightarrow{F}}_{12}\) points toward wire 2 . The conclusion we can draw from this simple calculation is that two parallel wires carrying currents in the same direction will attract each other. On the other hand, if the currents flow in opposite directions, the resultant force will be repulsive.

Definition of ampere

Consider two parallel wires separated by 1 m and carrying a current of 1 A each. Then \(i_{1} = i_{2} = 1\text{ }A\) and \(d = 1\text{ }m\), so that from equation

\[\frac{dF}{\text{ }dl} = 2 \times 10^{- 7}\text{ }N/m\]

This is used to formally define the unit 'ampere' of electric current. If two parallel, long wires, kept 1 m apart in vacuum, carry equal currents in the same direction and there is a force of attraction of \(2 \times 10^{- 7}\) newton per metre of each wire, the current in each wire is said to be 1 ampere.

Practice Exercise

Q. 1 If a current is passed through a spring, does the spring stretch or compress?
Q. 2 A rectangular loop of length \(\mathcal{l}\) and width \(w\) carries a steady current \(I_{l}\). The loop is then placed near an finitely long wire carrying a current \(I_{2}\), as shown in Figure. What is the magnetic force experienced by the loop due to the magnetic field of the wire?

Q. 3 Find the magnitude and direction of a force vector acting on a unit length of a thin wire, carrying a current \(I\) at a point \(O\), if the wire is bent as shown in with curvature radius \(R\)

Q. 4 Two long thin parallel conductors of the shape shown in Fig. carry direct currents \(I_{1}\) and \(I_{2}\). The separation between the conductors is a , the width of the right-hand conductor is equal to \(b\). With both conductors lying in one plane, find the magnetic interaction force between them reduced to a unit of their length.

Answers

Q. 1 compress
Q. \(2 = \frac{\left. \ \left. \ \mu_{0} \right|_{1} \right|_{2}\mathcal{l}}{2\pi}\left\lbrack \frac{1}{a} - \frac{1}{(a + w)} \right\rbrack\)
Q. \(3{\text{ }F}_{\text{unit}\text{~}} = m_{0}I^{2}/4R\)
Q. \(4F_{1} = \frac{\mu_{0}}{4\pi}\frac{2I_{1}I_{2}}{b}ln(1 + b/a)\)

Magnetic field (at large distances) due to current in a circular current loop is very similar in behavior to the electric field of an electric dipole. We know that the magnetic field on the axis of a circular loop, of a radius \(R\), carrying a steady current \(I\)

\[B = \frac{\mu_{0}I\left( 2\pi a^{2} \right)}{4\pi\left( a^{2} + x^{2} \right)^{3/2}}\]

its direction is along the axis and given by the right-hand thumb rule Here, \(x\) is the distance along the axis from the centre of the loop.
For \(x \gg R\), we may drop the \(R^{2}\) term in the denominator. Thus

\[B = 2\left( \frac{\mu_{0}}{4\pi} \right)\left( \frac{IA}{x^{3}} \right)\]

Where

\[A = \pi R^{2} = \text{~}\text{area of the loop}\text{~}\]

The expression is very similar to an expression obtained earlier for the electric field of a dipole. The similarity may be seen if we we can define the magnetic dipole moment \(\overrightarrow{\mu}\) as

\[\overrightarrow{\mu} = I\overrightarrow{\text{ }A}\]

The direction of \(\overrightarrow{\mu}\) is the same as the area vector \(\overrightarrow{A}\) (perpendicular to the plane of the loop) and is determined by the right-hand rule (Figure). The SI unit for the magnetic dipole moment is ampere-meter \(\ ^{2}\left( A \cdot m^{2} \right).\).

Practice Exercise

Q. 1 Find the magnetic moment of a thin round loop with current if the radius of the loop is equal to \(R = 100\text{ }mm\) and the magnetic induction at its centre is equal to \(B = 6.0\mu\text{ }T\).
Q. 2 A thin insulated wire forms a plane spiral of \(N = 100\) tight turns carrying a current \(I = 8\text{ }mA\). The radii of inside and outside turns (Fig.) are equal to \(a = 50\text{ }mm\) and \(b = 100\text{ }mm\).

Find the magnetic moment of the spiral with a given current.
Q. 3 We want to estimate the magnetic dipole moment associated with the motion of an electron as it orbits a proton. We use a "semi-classical" model to do this. Assume that the electron has speed \(v\) and orbits a proton (assumed to be very massive) located at the origin. The electron is moving in a right-handed sense with respect to the \(z\)-axis in a circle of radius \(r = 0.53\text{Å}\), as shown in Figure. Note that 1 \(\text{Å} = 10^{- 10}\text{ }m\).

(a) The inward force \(m_{e}v^{2}/r\) required to make the electron move in this circle is provided by the Coulomb attractive force between the electron and proton ( \(m_{e}\) is the mass of the electron). Using this fact, and the value of \(r\) we give above, find the speed of the electron in our "semi-classical" model.
(b) Given this speed, what is the orbital period \(T\) of the electron?
(c) What current is associated with this motion? Think of the electron as stretched out uniformly around the circumference of the circle. In a time \(T\), the total amount of charge \(q\) that passes an observer at a point on the circle is just \(e\)
(d) What is the magnetic dipole moment associated with this orbital motion? Give the magnitude and direction. The magnitude of this dipole moment is one Bohr magneton \(\mu_{B}\)
Q. 4 A non-conducting thin disc of radius R charged uniformly over one side with surface density \(\sigma\) rotates about its axis with an angular velocity \(\omega\). Find:
(b) the magnetic moment of the disc.

Answers

Q. \(1p_{m} = 2\pi R^{3}\text{ }B/\mu_{0} = 30\text{ }mA \cdot {\text{ }m}^{2}\)
Q. \(2p_{m} = 1/3\pi IN\left( a^{2} + ab + b^{2} \right)\)
Q. 3 (a) \(2.18 \times 10^{6}\text{ }m/s\);
(b) \(1.52 \times 10^{- 16}\text{ }s\) (c) 1.05 mA . Big!;
(d) \(9.27 \times 10^{- 24}\) along the . \(z\)-axis.
Q. \(4p_{m} = 1/4\pi\sigma\omega R^{4}\)

What happens when we place a rectangular loop carrying a current \(I\) in the \(xy\) plane and switch on a uniform magnetic field \(\overrightarrow{B} = Bi\) which runs parallel to the plane of the loop, as shown in Figure (a)?

Figure : (a) A rectangular current loop placed in a uniform magnetic field.
(b) The magnetic forces acting on sides 2 and 4.

From Eq., we see the magnetic forces acting on sides 1 and 3 vanish because the length vectors \({\overrightarrow{\mathcal{l}}}_{1} = - bi\) and \({\overrightarrow{\mathcal{l}}}_{3} =\) bi are parallel and anti-parallel to \(\overrightarrow{B}\) and their cross products vanish. On the other hand, the magnetic forces acting on segments 2 and 4 are non-vanishing:

\[\left\{ \begin{matrix} {\overrightarrow{F}}_{2} = I( - a\widehat{j}) \times (B\widehat{i}) = IaB\widehat{k} \\ {\overrightarrow{F}}_{4} = I(a\widehat{j}) \times (B\widehat{i}) = - IaB\widehat{k} \end{matrix} \right.\ \]

with \({\overrightarrow{F}}_{2}\) pointing out of the page and \({\overrightarrow{F}}_{4}\) into the page. Thus, the net force on the rectangular loop is

\[{\overrightarrow{F}}_{net} = {\overrightarrow{F}}_{1} + {\overrightarrow{F}}_{2} + {\overrightarrow{F}}_{3} + {\overrightarrow{F}}_{4} = \overrightarrow{0}\]

as expected. Even though the net force on the loop vanishes, the forces \({\overrightarrow{F}}_{2}\) and \({\overrightarrow{F}}_{4}\) will produce a torque which causes the loop to rotate about the \(y\)-axis (Figure). The torque with respect to the center of the loop is

\[\begin{matrix} & \overrightarrow{\tau} = \left( - \frac{b}{2}\widehat{i} \right) \times {\overrightarrow{F}}_{2} + \left( \frac{b}{2}\widehat{i} \right) \times \left( IaB\widehat{k} + \left( \frac{b}{2}\widehat{i} \right) \times ( - IaB\widehat{k}) \right) \\ & \ = \left( \frac{IabB}{2} + \frac{IabB}{2} \right)\widehat{j} = IabB\widehat{j} = IAB\widehat{j} \end{matrix}\]

where \(A = ab\) represents the area of the loop and the positive sign indicates that the rotation is clockwise about the \(y\)-axis. It is convenient to introduce the area vector \(\overrightarrow{A} = A\widehat{n}\) where \(\widehat{n}\) is a unit vector in the direction normal to the plane of the loop. The direction of the positive sense of \(\widehat{n}\) is set by the conventional right-hand rule. In our case, we have \(\widehat{n} = + \widehat{k}\). The above expression for torque can then be rewritten as

\[\overrightarrow{\tau} = I\overrightarrow{\text{ }A} \times \overrightarrow{B}\]

Notice that the magnitude of the torque is at a maximum when \(\overrightarrow{B}\) is parallel to the plane of the loop (or perpendicular to).
Consider now the more general situation where the loop (or the area vector \(\overrightarrow{A}\) ) makes an angle \(\theta\) with respect to the magnetic field.

Figure : Rotation of a rectangular current loop

From Figure, the lever arms and can be expressed as:

\[{\overrightarrow{r}}_{2} = \frac{b}{2}( - sin\theta\widehat{i} + cos\theta\widehat{k}) = - {\overrightarrow{r}}_{4}\]

and the net torque becomes

\[\begin{matrix} & \overrightarrow{\tau} = {\overrightarrow{r}}_{2} \times {\overrightarrow{F}}_{2} + {\overrightarrow{r}}_{4} = 2{\overrightarrow{r}}_{2} \times {\overrightarrow{F}}_{2} = 2 \cdot \frac{\text{ }b}{2}( - sin\theta\widehat{i} + cos\theta\widehat{k}) \times (IaBk) \\ & \text{~}\text{jIabB}\text{~}sin\theta\widehat{j} = I\overrightarrow{\text{ }A} \times \overrightarrow{B} \end{matrix}\]

For a loop consisting of \(N\) turns, the magnitude of the toque is

\[\tau = NIABsin\theta\]

The quantity NI \(\overrightarrow{A}\) is called the magnetic dipole moment \(\overrightarrow{\mu}\)

\[\overrightarrow{\mu} = NI\overrightarrow{\text{ }A}\]

Using the expression for \(\overrightarrow{\mu}\), the torque exerted on a current-carrying loop can be rewritten as

\[\overrightarrow{\tau} = \overrightarrow{\mu} \times \overrightarrow{B}\]

The above equation is analogous to \(\overrightarrow{\tau} = \overrightarrow{p} \times \overrightarrow{E}\) in Eq., the torque exerted on an electric dipole moment \(\overrightarrow{p}\) in the presence of an electric field \(\overrightarrow{E}\).

Configuration energy of current loop in uniform magnetic field.

Recalling that the potential energy for an electric dipole is \(U = - \overrightarrow{p} \cdot \overrightarrow{E}\) [see Eq.], a similar form is expected for the magnetic case. The work done by an external agent to rotate the magnetic dipole from an angle \(\theta\) to \(\theta\) is given by

\[\begin{matrix} & W_{ext} = \int_{\theta_{0}}^{\theta}\mspace{2mu}\mspace{2mu}\left( \mu Bsin\theta^{'} \right)d\theta^{'} = \mu B\left( cos\theta_{0} - cos\theta \right) \\ & \ = \Delta U = U - U_{0} \end{matrix}\]

Once again, \(W_{\text{ext}\text{~}} = - W\), where \(W\) is the work done by the magnetic field. Choosing \(U_{0} = 0\) at \(\theta_{0} = \pi/\) 2 , the dipole in the presence of an external field then has a potential energy os

\[U = - \mu Bcos\theta = - \overrightarrow{\mu} \cdot \overrightarrow{B}\]

The configuration is at a stable equilibrium when \(\overrightarrow{\mu}\) is aligned parallel to \(\overrightarrow{B}\), making U a minimum with \(U_{\text{min}\text{~}} = - \mu B\). On the other hand, when \(\overrightarrow{\mu}\) and \(\overrightarrow{B}\) are anti-parallel, \(U_{\text{max}\text{~}} = + \mu B\) is a maximum and the system is unstable.

Practice Exercise

Q. 1 A current loop consists of a semicircle of radius \(R\) and two straight segments of length \(\mathcal{l}\) with an angle \(\theta\) between them. The loop is then placed in a uniform magnetic field pointing to the right, as shown in Figure.

(a) Find the net force on the current loop.
(b) Find the net torque on the current loop.
Q. 2 Figure shows a wooden cylinder with a mass m and a length L with N turns of wire wrapped around it longitudinally, so that the plane of the wire loop contains the axis of the cylinder. What is the least current through the loop that will prevent the cylinder from rolling down a plane inclined at an angle to the horizontal, in the presence of a vertical, uniform magnetic field B , if the plane of the windings is parallel to the inclined plane?

Q. 3 A copper wire of density \(\rho\) with cross-sectional area \(S\) bent to make three sides of a square can turn about a horizontal axis \(\mathbf{OO}^{'}\) (Fig.). The wire is located in uniform vertical magnetic field. Find the magnetic induction if on passing a current I through the wire the latter deflects by an angle \(\theta\)

Answers

Q. 1
Q. \(2\ i = \frac{mg}{2NBL}\)
Q. \(3\text{ }B = (2rgS/I)tan\theta\)

The main parts of a moving-coil galvanometer are shown in figure.

The current to be measured is passed through the galvanometer. As the coil is in the magnetic field \(\overrightarrow{B}\) of the permanent magnet, a torque \(\overrightarrow{\Gamma} = ni\overrightarrow{A} \times \overrightarrow{B}\) acts on the coil. Here \(n =\) number of turns, \(i =\) current in the coil \(\overrightarrow{A} =\) area-vector of the coil and \(\overrightarrow{B} =\) magnetic field at the site of the coil. This torque deflects the coil from its equilibrium position.
The pole pieces are made cylindrical. As a result, the magnetic field at the arms of the coil remains parallel to the plane of the coil everywhere even as the coil rotates. The deflecting torque is then \(\Gamma = niAB\). As the upper end of the suspension strip \(W\) is fixed, the strip gets twisted when the coil rotates. This produces a restoring torque acting on the coil. If the deflection of the coil is \(\theta\) and the torsional constant of the suspension strip is k , the restoring torque is \(k\theta\). The coil will stay at a deflection \(\theta\) where

\[\begin{matrix} & niAB = k\theta \\ \text{~}\text{or,}\text{~}\ & i = \frac{k}{nAB}\theta \end{matrix}\]

Hence, the current is proportional to the deflection. The constant \(\frac{k}{nAB}\) is called the galvanometer constant.

We define the current sensitivity of the galvanometer as the deflection per unit current. From Eq. this current sensitivity is.

\[\frac{\phi}{I} = \frac{NAB}{k}\]

A convenient way for the manufacturer to increase the sensitivity is to increase the number of turns N . We choose galvanometers having sensitivities of value, required by our experiment.

We define the voltage sensitivity as the deflection per unit volt of applied potential difference

\[\frac{\phi}{I} = \left( \frac{NAB}{k} \right)\frac{I}{V} = \left( \frac{NAB}{k} \right)\frac{1}{R}\]

An interesting point to note is that increasing the current sensitivity may not necessarily increase the voltage sensitivity. If \(N \rightarrow 2\text{ }N\), i.e., we double the number of turns, then

\[\frac{\phi}{I} \rightarrow 2\frac{\phi}{I}\]

Thus, the current sensitivity doubles. However, the resistance of the galvanometer is also likely to double, since it is proportional to the length of the wire. In eq. \(N \rightarrow 2\text{ }N\), and \(R \rightarrow 2R\), thus the voltage sensitivity,

\[\frac{\phi}{V} \rightarrow \frac{\phi}{V}\]

remains unchanged.

Practice Exercise

Q. 1 Two moving coil meters, \(M_{1}\) and \(M_{2}\) have the following particulars :

\[\begin{matrix} & R_{1} = 10\Omega,{\text{ }N}_{1} = 30 \\ & {\text{ }A}_{1} = 3.6 \times 10^{- 3}{\text{ }m}^{2},{\text{ }B}_{1} = 0.25\text{ }T \\ & R_{2} = 14\Omega,{\text{ }N}_{2} = 42 \\ & {\text{ }A}_{2} = 1.8 \times 10^{- 3}{\text{ }m}^{2},{\text{ }B}_{2} = 0.50\text{ }T \end{matrix}\]

(The spring constants are identical for the two meters)
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of \(M_{2}\) and \(M_{1}\).

Answers

Q. \(1\ \) (a) 1.4 (b) 1

Q. 1 A current I flows in a long straight wire with cross-section having the form of a thin half-ring of radius R (Fig.). Find the induction of the magnetic field at the point O .

Sol. \(\ dI = \frac{1}{\pi R}(Rd\theta) = \frac{1}{\pi}d\theta\)
\[{dB = \frac{\mu_{0}dI}{2\pi R} = \frac{\mu_{0}I}{2\pi^{2}R}d\theta }{B = \int dBcos\theta = \frac{\mu_{0}I}{2\pi^{2}R}\int_{- \frac{\pi}{2}}^{+ \frac{\pi}{2}}\mspace{2mu} cos\theta d\theta = \frac{\mu_{0}I}{\pi^{2}R} }\]

\[B = \mu_{0}I/\pi^{2}R \]Q. 2 In the figure shown the magnetic field at the point P.

Sol. Consider the figure

\[{\overrightarrow{B}}_{p} = \left( {\overrightarrow{B}}_{1} \right)_{p} + \left( {\overrightarrow{B}}_{2} \right)_{p} + \left( {\overrightarrow{B}}_{3} \right)_{p} + \left( {\overrightarrow{B}}_{4} \right)_{p} + \left( {\overrightarrow{B}}_{5} \right)_{p}\]

where \(\left( {\overrightarrow{B}}_{1} \right)_{p} = \frac{\mu_{0}i}{4\pi\left( \frac{3a}{2} \right)}( - \widehat{j})\) (semi-infinite wire)

\[\begin{matrix} & \left( {\overrightarrow{B}}_{2} \right)_{p} = \frac{\mu_{0}i}{4\left( \frac{3a}{2} \right)}( + \widehat{k}) \\ & \left( {\overrightarrow{B}}_{3} \right)_{p} = 0 \\ & \left( {\overrightarrow{\text{ }B}}_{4} \right)_{p} = \frac{\mu_{0}i}{4\left( \frac{a}{2} \right)}( - \widehat{k}) \\ \Rightarrow \ & \left( {\overrightarrow{B}}_{5} \right)_{p} = \frac{\mu_{0}i}{4\pi\left( \frac{a}{2} \right)}( - \widehat{j}) \\ \Rightarrow \ & {\overrightarrow{B}}_{p} = \frac{\mu_{0}i}{2a}\left\lbrack - \left( \frac{1}{3\pi} + \frac{1}{\pi} \right)\widehat{j} - \left( 1 - \frac{1}{3} \right)\widehat{k} \right\rbrack \\ \Rightarrow \ & {\overrightarrow{B}}_{p} = \frac{2\mu_{0}i}{3a}\left\lbrack \frac{1}{\pi}\widehat{j} - \widehat{k} \right\rbrack \\ \Rightarrow \ & {\overrightarrow{B}}_{p} = \frac{\mu_{0}i}{3\pi a}\sqrt{1 + \pi^{2}}. \end{matrix}\]

Q. 3 A thin insulated wire forms a plane spiral of N tight turns carrying a current I . The radii of inside and outside turns (Fig.) are equal to \(a\) and \(b\). Find the magnetic induction at the centre of the spiral;

Sol. \(\ B_{\text{one}\text{~}} = \frac{\mu_{0}I}{4\pi r}(2\pi) = \frac{\mu_{0}I}{2r}\)

\[\begin{matrix} & dN = \frac{N}{b - a}dr \\ & dB = B_{\text{one}\text{~}}dN = \frac{\mu_{0}NI}{2(b - a)}\frac{dr}{r} \\ & B = \int_{}^{}\ dB = \frac{\mu_{0}NI}{2(b - a)}\int_{a}^{b}\mspace{2mu}\mspace{2mu}\frac{dr}{r} = \frac{\mu_{0}NI}{2(b - a)}ln\frac{b}{a} \end{matrix}\]

Q. 4 A disc of radius R rotates at an angular velocity \(\omega\) about the axis perpendicular to its surface and passing through its centre. If the disc has a uniform surface charge density \(\sigma\), find the magnetic induction on the axis of rotation at a distance \(x\) from the centre.

Sol. Consider a ring of radius r and width dr .
Charge on the ring, \(dq = (2\pi rdr)\sigma\)
Current due to ring is \(dI = \frac{dq}{T}\)

\[= \frac{\omega dq}{2\pi} = \sigma\omega rdr\]

Magnetic field due to ring at point P is

\[dB = \frac{\mu_{0}{dlr}^{2}}{2\left( r^{2} + x^{2} \right)^{3/2}}\]

\[\begin{array}{r} \text{~}\text{or}\text{~}\ B = \int_{}^{}\ dB = \frac{\mu_{0}\sigma\omega}{2}\int_{0}^{R}\mspace{2mu}\mspace{2mu}\frac{r^{3}dr}{\left( r^{2} + x^{2} \right)^{3/2}}\#(i) \end{array}\]

Putting \(r^{2} + x^{2} = t^{2}\) and 2r dr \(= 2tdt\) and integrating (i) we get

\[B = \frac{\mu_{0}\sigma\omega}{2}\left\lbrack \frac{R^{2} + 2x^{2}}{\sqrt{R^{2} + x^{2}}} - 2x \right\rbrack.\]

Q. 5 A circular loop of radius R is bent along a diameter and given a shape as shown in figure. One of the semi-circles (KNM) lies in the x-z plane and the other one (KLM) in the y-z plane with their centres at origin. Current \(I\) is flowing through each of the semi-circles as shown in figure.

A particle of charge q is released at the origin with a velocity \(\overrightarrow{V} = - V_{0}\widehat{i}\). Find the instantaneous force \(\overrightarrow{F}\) on the particle. Assuem that space is gravity free.

Sol. Magnetic field at the centre of a circular wire of radius R carrying a current I is given by

\[B = \frac{\mu_{0}I}{2R}\]

In this problem, currents are flowing in two semi-circles, KLM in the \(y - z\) plane and KNM in the \(x - z\) plane. The centres of these semi-circles coincide with the origin of the Cartesian system of axes.

\[\therefore\ {\overrightarrow{B}}_{KLM} = \frac{1}{2}\left( \frac{\mu_{0}I}{2R} \right)( - \widehat{i})\]

\[{\overrightarrow{B}}_{KNM} = \frac{1}{2}\left( \frac{\mu_{0}I}{2R} \right)( - \widehat{j})\]

The total magnetic field at the origin is

\[B_{0} = \frac{\mu_{0}I}{4R}( - \widehat{i} + \widehat{j})\]

It is given that a particle of charge \(q\) is released at the origin with a velocity \(V = - V_{0}\widehat{i}\). The instantaneous force acting on this particle is given by
\[{f = q\lbrack\overrightarrow{V} \times \overrightarrow{B}\rbrack }{= q\left( - V_{0}\widehat{i} \right) \times \left\lbrack \frac{\mu_{0}I}{4R}(\widehat{i} + \widehat{j}) \right\rbrack }{= \left( \frac{qV_{0}\mu_{0}I}{4R} \right)\lbrack( - \widehat{i}) \times ( - \widehat{i} + \widehat{j}) = \frac{qV_{0}\mu_{0}I}{4R}( - \widehat{k})}\].
Q. 6 In the figure a charged sphere of mass \(m\) and charge \(q\) starts sliding from rest on a vertical fixed circular track of radius R from the position shown. There exists a uniform and constant horizontal magnetic field of induction B. The maximum force exerted by the track on the sphere.

Sol. \(\ F_{m} = qvB,a\) and directed radially outward.

\[\begin{matrix} \therefore & N - mgsin\theta + qvB = \frac{{mv}^{2}}{R} \\ \Rightarrow & \text{ }N = \frac{{mv}^{2}}{R} + mgsin\theta - qvB \end{matrix}\]

Hence at \(\theta = \pi/2\)

\[\Rightarrow \ N_{\max} = \frac{2mgR}{R} + mg - qB\sqrt{2gR} = 3mg - qB\sqrt{2gR}.\]

Q. 7 A straight segment OC (of length L meter) of a circuit carrying a current 1 amp is placed along the x -axis. Two infinitely long straight wire \(A\) and \(B\), each extending \(z = - \infty\) to \(+ \infty\) are fixed at \(y = -\) a metre and \(y = +\) a metre respectively, as shown in the figure. If the wires \(A\) and \(B\) each carry a current \(I\) amp into the place of the paper, obtain the expression for the force acting on segment OC. What will be the force on OC if the current in the wire B is reversed?

Sol. Magnetic field \(B_{A}\) produced at \(P(x,0,0)\) due to wire, \(B_{A} = \mu_{0}I/2\pi R,B_{B} = \mu_{0}I/2\pi R\). Components of \(B_{A}\) and \(B_{B}\) along \(x\)-axis cancel, while those along \(y\)-axis add up to give total field.

\[B = 2\left( \frac{\mu_{0}I}{2\pi R} \right)cos\theta = \frac{2\mu_{0}I}{2\pi R} \cdot \frac{x}{R} = \frac{\mu_{0}I}{\pi}\frac{x}{\left( a^{2} + x^{2} \right)}\text{~}\text{(along}\text{~} - y\text{~}\text{direction)}\text{~}\]

The force dF acting on the current element is \(d\overline{F} = I(d\overline{I} \times B\overline{B})\)

\[\begin{matrix} \therefore & dF = \frac{\mu_{0}I^{2}}{\pi}\frac{xdx}{a^{2} + x^{2}}\left\lbrack \because sin90^{\circ} = 1 \right\rbrack \\ \Rightarrow & F = \frac{\mu_{0}I^{2}}{\pi}\int_{0}^{L}\mspace{2mu}\mspace{2mu}\frac{xdx}{a^{2} + x^{2}} = \frac{\mu_{0}I^{2}}{2\pi}ln\frac{a^{2}{\text{ }L}^{2}}{a^{2}} \end{matrix}\]

If the current in B is reversed, the magnetic field due to the two wires would be only along x - direction and the force on the current along \(x\)-direction will be zero.
Q. 8 Shown in the figure is a very long semicylindrical conducting shell of radius \(R\) and carrying a current \(i\) along its length. An infinitely long straight current carrying conductor is lying along the axis of the semicylinder. If

the current flowing through the straight wire is \(i_{0}\), then find the froce on the semicylinder.

Sol. The net magnetic force on the conducting wire

\[\begin{matrix} & \ = F = \int_{}^{}\ 2dFcos\theta \\ & \ \Rightarrow \ \text{ }F = \int_{}^{}\ 2\left\lbrack \frac{\mu_{0}(di)i_{0}}{2\pi R} \right\rbrack cos\theta \\ & \ \Rightarrow \ F = \frac{\mu_{0}i_{0}}{\pi R}\int_{}^{}\ dicos\theta \end{matrix}\]

when \(di = \frac{i}{\pi R} \times Rd\theta = \frac{id\theta}{\pi}\)

\[\begin{matrix} & \ \Rightarrow \ F = \frac{\mu_{0}i_{0}}{\pi R}\int_{}^{}\ \frac{(id\theta)cos\theta}{\pi} \\ & \ \Rightarrow \ F = \frac{\mu_{0}i_{0}i^{\pi/2}}{\pi^{2}R}\int_{0}^{2}\mspace{2mu}\mspace{2mu} cos\theta d\theta = \frac{\mu_{0}i_{0}i}{\pi^{2}R}. \end{matrix}\]

Q. 9 A wire loop carrying a current I is placed in the \(x - y\) plane as shown in figure. (a) If a particle with charge \(q\) and mass \(m\) is placed at the centre P and given a velocity v along NP find its instantaneous acceleration. (b) If an external uniform magnetic induction \(\overrightarrow{B} = Bi\) is applied, find the force and torque acting on the loop.

Sol. (a) As in case of current -carrying straight conductor and arc, the magnitude of \(B\) is given by

\[B_{1} = \frac{\mu_{0}I}{4\pi\text{ }d}(sin\alpha + sin\beta)\]

And \(\ B_{2} = \frac{\mu_{0}I\phi}{4\pi r}\)
So in accordance with right hand screw rule,

\[\left( {\overrightarrow{B}}_{w} \right) = \frac{\mu_{0}}{4\pi}\frac{1}{(acos60)} \times 2sin60( - \widehat{k})\]

and \(\ (\overrightarrow{B})_{MN} = \frac{\mu_{0}}{4\pi}\frac{I}{a} \times \left( \frac{2}{3}\pi \right)( - \widehat{k})\)

and hence net \(\overrightarrow{B}\) at P due to the given loop

\[\begin{array}{r} \overrightarrow{B} = {\overrightarrow{B}}_{w} + {\overrightarrow{B}}_{A}\ \Rightarrow \ \overrightarrow{\text{ }B} = \frac{\mu_{0}}{4\pi}\frac{2I}{a}\left\lbrack \sqrt{3} - \frac{\pi}{3} \right\rbrack( - \widehat{k})\#(i) \end{array}\]

Now as force on charged particle in a magnetic fields is given by

\[\overrightarrow{F} = q(\overrightarrow{v} \times \overrightarrow{B})\]

so here, \(\ \overrightarrow{F} = qvBsin90^{\circ}\) along PF
i.e. \(\ \overrightarrow{F} = \frac{\mu_{0}}{4\pi}\frac{2qvI}{a}\left\lbrack \sqrt{3} - \frac{\pi}{3} \right\rbrack\) along PF
and so \(\overrightarrow{a} = \frac{\overrightarrow{F}}{m} = 10^{- 7}\frac{2qvI}{a}\left\lbrack \sqrt{3} - \frac{\pi}{3} \right\rbrack\) along PF
(b) As \(d\overrightarrow{F} = Id\overrightarrow{L} \times \overrightarrow{B}\), so \(\overrightarrow{F} = \int_{}^{}\ Id\overrightarrow{L} \times \overrightarrow{B}\)

As here \(I\) and \(\overrightarrow{B}\) are constant

\[\overrightarrow{F} = I\lbrack\complement\rbrack d\overrightarrow{\text{ }L}\rbrack \times \overrightarrow{B} = 0\ \lbrack\text{~}\text{as}\text{~∮}\ \ d\overrightarrow{\text{ }L} = 0\rbrack\]

Further as area of coil,

\[\overrightarrow{S} = \left\lbrack \frac{1}{3}\pi a^{2} - \frac{1}{2} \cdot 2asin60^{\circ} \times acos60^{\circ} \right\rbrack\widehat{k} = a^{2}\left\lbrack \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right\rbrack\widehat{k}\]

So \(\ \overrightarrow{M} = I\overrightarrow{S} = {Ia}^{2}\left\lbrack \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right\rbrack\widehat{k}\)
and hence \(\ \overrightarrow{\tau} = \overrightarrow{M} \times \overrightarrow{B} = {Ia}^{2}\text{ }B\left\lbrack \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right\rbrack(\widehat{k} \times \widehat{i})\)
i.e. \(\ \overrightarrow{\tau} = I^{2}B\left\lbrack \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right\rbrack\widehat{j}N - m\ \) as \((\widehat{k} \times \widehat{i} = \widehat{j})\).
Q. 10 A coil of radius \(R\) carries current \(I\). Another concentric coil of radius ( \(r \ll R\) ) carries current \(i\). Planes of two coils are mutually perpendicular and both the coils are free to rotate about common diameter. Find maximum kinetic energy of smaller coil when both the coils are released, masses of coils are \(M\) and \(m\) respectively.

Sol. If a magnetic dipole having moment \(M\) be rotated through angle \(\theta\) from equilibrium position in a uniform magnetic field \(B\), work done on it is \(W = MB(1 - cos\theta)\). This work is stored in the system in the form of energy. When system is release, dipole starts to rotate to occupy equilibrium position and teh energy converts into kinetic energy and kinetic energy of the system is

maximum when stored energy is completely released.

Magnetic induction, at centres due to current in larger coil is \(B = \frac{\mu_{0}I}{2R}\)
Magnetic dipole moment of smaller coil is \(i\pi r^{2}\).

Initially planes of two coils are mutually perpendicular, therfore \(\theta\) is \(90^{\circ}\) or energy of the system is

\[\begin{matrix} & U = \left( i\pi r^{2} \right)B\left( 1 - cos90^{\circ} \right) \\ & U = \frac{\mu_{0}Ii\pi r^{2}}{2R} \end{matrix}\]

When coils are release, both the coils start to rotate about their common diameter and their kinetic energies are maximum when they become coplanar.
Moment of inertia of larger coil about axis of rotation is \(I_{1} = \frac{1}{2}mR^{2}\)
and that of smaller coil is \(I_{2} = \frac{1}{2}mr^{2}\)
Since, two coils rotate due to their mutual interaction only, therefore, if one coil rotates clockwise then the other rotates anticlockwise.

Let angular velocities of larger and smaller coils be numerically equal to \(\omega_{1}\) and \(\omega_{2}\) respectively when they become coplaner,
According to law of conservation of angular momentum,

\[I_{1}\omega_{1} = I_{2}\omega_{2}\]

and according to law of conservation of energy,

\[\frac{1}{2}I_{1}\omega_{1}^{2} + \frac{1}{2}I_{2}\omega_{2}^{2} = U\]

From above equations, maximum kinetic energy of smaller coil,

\[\begin{matrix} & \frac{1}{2}I_{2}\omega_{2}^{2} = \frac{UI_{1}}{I_{1} + I_{2}} \\ & \ = \frac{\mu_{0}\pi lMRr^{2}}{2\left( MR^{2} + mr^{2} \right)} \end{matrix}\]

Q. 11 What is the work done in transferring the wire from position (1) to position (2)?

Sol. The loop can be considered as the combination of teh number of elementary loops.

The net current in the dotted wires is 0 as current in the neighboring loops flowing through the same wire are opposite in direction.
consider an elementary loop of width \(dr\) at a distance \(r\) from the wire
The ' \(d\mu\) ' magnetic moment of the elemental loop

\[= I_{2}ldr\]

The B at that point due to straight wire

\[\begin{matrix} & \ = \mu_{0}I_{1}/2\pi r \\ & dU = - B \cdot \text{ }d\mu = - \frac{\mu_{0}I_{1}}{2\pi r}I_{2}ldr(cos\pi) \end{matrix}\]

[As \(d\mu\) is anti-parallel to B.]

\[U_{1} = \int_{}^{}\ du = \frac{\mu_{0}I_{1}I_{2}l}{2\pi}\int_{a}^{b}\mspace{2mu}\frac{1}{r}dr = \frac{\mu_{0}I_{1}I_{2}l}{2\pi}ln\left( \frac{\text{ }b}{a} \right)\]

==

By symmetry, \(U_{2} = - U_{1}\)

\[\Rightarrow \ - \Delta U = \text{~}\text{work done}\text{~} = - \left( U_{2} - U_{1} \right) = 2\frac{\mu_{0}I_{1}I_{2}l}{2\pi}ln\frac{\text{ }b}{a}.\]

The work done in transferring the wire from position 1 to \(2 = \frac{\mu_{0}I_{1}I_{2}l}{\pi}ln\frac{\text{ }b}{a}\).