NCERT Class 12 Physics — Moving Charges and Magnetism, Chapter 4 Exercises. All 13 questions solved.

The exercises at the end of Chapter 4 fall into four groups. Questions 4.1 to 4.4 and 4.8 ask for the magnetic field a current produces — at the centre of a coil, beside a long straight wire and inside a solenoid. Questions 4.5 to 4.7 ask for the force a magnetic field exerts on a current. Questions 4.9, 4.10 and 4.13 are about the torque on a current-carrying coil, which is how a moving coil galvanometer works. Questions 4.11 and 4.12 follow a single electron moving through a field. Almost everything runs on three sets of results:

• Fields of steady currents — $B = \dfrac{\mu_0 I}{2\pi r}$ at a distance $r$ from a long straight wire, $B = \dfrac{\mu_0 N I}{2R}$ at the centre of a coil of $N$ turns and radius $R$, and $B = \mu_0 n I$ inside a long solenoid with $n$ turns per unit length, where $\mu_0 = 4\pi\times10^{-7}\ \text{T m A}^{-1}$.
• Magnetic force — on a straight wire, $\vec F = I\vec l\times\vec B$, of magnitude $IlB\sin\theta$; on a moving charge, $\vec F = q\vec v\times\vec B$.
• Torque on a coil — $\vec\tau = \vec m\times\vec B$ with $m = NIA$, so $\tau = NIAB\sin\theta$, where $\theta$ is the angle between the field and the normal to the coil.

Every distance in centimetres is converted to metres, and every field in gauss to tesla, before it goes into a formula.

Key insight. Every magnetic force in this chapter is a cross product, so each question comes down to two things: which angle goes into the sine, and which way the product points. For a wire it is the angle between the wire and the field (4.5, and $90^\circ$ in 4.6); for a coil it is the angle between the field and the coil’s normal (4.9, 4.13). The direction comes from the right-hand rule (4.3, 4.4). And because $q\vec v\times\vec B$ is always perpendicular to $\vec v$, a charge moving across a uniform field is bent into a circle without ever changing speed (4.11, 4.12).

Question 4.1

A circular coil of wire consisting of $100$ turns, each of radius $8.0\ \text{cm}$ carries a current of $0.40\ \text{A}$. What is the magnitude of the magnetic field $\vec B$ at the centre of the coil?

Solution. A single circular turn produces a field $\mu_0 I/2R$ at its centre, directed along its axis. All $100$ turns share the same centre and carry the current the same way round, so their fields point the same way and simply add. With $R = 0.080\ \text{m}$:

$$B = \frac{\mu_0 N I}{2R} = \frac{(4\pi\times10^{-7})(100)(0.40)}{2\times0.080} = \pi\times10^{-4}\ \text{T} \approx 3.1\times10^{-4}\ \text{T}$$

Question 4.2

A long straight wire carries a current of $35\ \text{A}$. What is the magnitude of the field $\vec B$ at a point $20\ \text{cm}$ from the wire?

Solution. The field lines of a long straight wire are circles around it, and Ampere’s circuital law applied to one of radius $r$ gives $B = \mu_0 I/2\pi r$. With $r = 0.20\ \text{m}$:

$$B = \frac{\mu_0 I}{2\pi r} = \frac{(4\pi\times10^{-7})(35)}{2\pi\times0.20} = \frac{2\times10^{-7}\times35}{0.20} = 3.5\times10^{-5}\ \text{T}$$

Question 4.3

A long straight wire in the horizontal plane carries a current of $50\ \text{A}$ in north to south direction. Give the magnitude and direction of $\vec B$ at a point $2.5\ \text{m}$ east of the wire.

Solution. The size of the field depends only on the current and the distance, with $r = 2.5\ \text{m}$:

$$B = \frac{\mu_0 I}{2\pi r} = \frac{2\times10^{-7}\times50}{2.5} = 4\times10^{-6}\ \text{T}$$

For the direction, use the right-hand rule: grip the wire with the thumb pointing along the current, southwards, and the fingers curl round the wire in the direction of the field. The field lines are circles in the vertical plane that is perpendicular to the wire, so at a point level with the wire and beside it the field must be vertical — the only question is up or down. Taking $x$ east, $y$ north and $z$ up, the field at $P$ points along $\hat I\times\hat r$, where $\hat I = -\hat{\jmath}$ is the direction of the current and $\hat r = \hat{\imath}$ points from the wire to $P$: $(-\hat{\jmath})\times\hat{\imath} = +\hat k$. The field at $P$ points vertically upwards.

Question 4.4

A horizontal overhead power line carries a current of $90\ \text{A}$ in east to west direction. What is the magnitude and direction of the magnetic field due to the current $1.5\ \text{m}$ below the line?

Solution. With $r = 1.5\ \text{m}$:

$$B = \frac{\mu_0 I}{2\pi r} = \frac{2\times10^{-7}\times90}{1.5} = 1.2\times10^{-5}\ \text{T}$$

For the direction, look along the line towards the west, so that the current flows away from you, into the page; north is then on your right and south on your left. The right-hand rule with the thumb pointing into the page makes the fingers curl clockwise, so the field circles the wire clockwise as seen in the figure. At the bottom of a clockwise circle the field points to the left, which here is south. The cross product agrees: with $x$ east, $y$ north and $z$ up, $\hat I = -\hat{\imath}$ and the point is below the wire, $\hat r = -\hat k$, so $\hat I\times\hat r = (-\hat{\imath})\times(-\hat k) = \hat{\imath}\times\hat k = -\hat{\jmath}$, which is south.

Question 4.5

What is the magnitude of magnetic force per unit length on a wire carrying a current of $8\ \text{A}$ and making an angle of $30^\circ$ with the direction of a uniform magnetic field of $0.15\ \text{T}$?

Solution. The force on a straight wire is $\vec F = I\vec l\times\vec B$, of magnitude $IlB\sin\theta$, where $\theta$ is the angle between the wire and the field — exactly the angle the question gives. Only the component of the field perpendicular to the wire pushes on it, which is why the sine appears. Dividing by $l$:

$$\frac{F}{l} = IB\sin\theta = (8)(0.15)\sin30^\circ = 1.2\times\tfrac12 = 0.6\ \text{N m}^{-1}$$

Question 4.6

A $3.0\ \text{cm}$ wire carrying a current of $10\ \text{A}$ is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be $0.27\ \text{T}$. What is the magnetic force on the wire?

Solution. The field inside a solenoid runs along its axis, and the wire lies perpendicular to the axis, so the angle between wire and field is $90^\circ$ and $\sin\theta = 1$. The whole $3.0\ \text{cm}$ is inside, in the uniform field:

$$F = IlB = (10)(0.030)(0.27) = 8.1\times10^{-2}\ \text{N}$$

The force is perpendicular to both the wire and the axis of the solenoid; which of the two ways it points depends on the sense of the current and of the field, and is given by Fleming’s left-hand rule.

Question 4.7

Two long and parallel straight wires $A$ and $B$ carrying currents of $8.0\ \text{A}$ and $5.0\ \text{A}$ in the same direction are separated by a distance of $4.0\ \text{cm}$. Estimate the force on a $10\ \text{cm}$ section of wire $A$.

Solution. Wire $B$ is long, so at the position of $A$, $d = 0.040\ \text{m}$ away, it produces a field $B_B = \mu_0 I_B/2\pi d$, perpendicular to $A$ and the same all along it. A length $L$ of wire $A$ sitting in that field feels a force $I_A L B_B$:

$$F = \frac{\mu_0 I_A I_B}{2\pi d}\,L = \frac{(2\times10^{-7})(8.0)(5.0)(0.10)}{0.040} = 2\times10^{-5}\ \text{N}$$

Parallel currents in the same direction attract, so the force on $A$ is perpendicular to it and directed towards $B$. It is an estimate only in the sense that the wires are treated as infinitely long.

Question 4.8

A closely wound solenoid $80\ \text{cm}$ long has $5$ layers of windings of $400$ turns each. The diameter of the solenoid is $1.8\ \text{cm}$. If the current carried is $8.0\ \text{A}$, estimate the magnitude of $\vec B$ inside the solenoid near its centre.

Solution. Each of the five layers is a solenoid in its own right, and their fields inside add, so what matters is the total number of turns: $5\times400 = 2000$. Spread over $0.80\ \text{m}$, that gives

$$n = \frac{2000}{0.80} = 2500\ \text{turns per metre}$$

The solenoid is $80\ \text{cm}$ long and only $1.8\ \text{cm}$ across, so near its centre it behaves as an ideal long solenoid, where the field is $\mu_0 nI$ and does not depend on the diameter at all:

$$B = \mu_0 n I = (4\pi\times10^{-7})(2500)(8.0) = 8\pi\times10^{-3}\ \text{T} \approx 2.5\times10^{-2}\ \text{T}$$

Question 4.9

A square coil of side $10\ \text{cm}$ consists of $20$ turns and carries a current of $12\ \text{A}$. The coil is suspended vertically and the normal to the plane of the coil makes an angle of $30^\circ$ with the direction of a uniform horizontal magnetic field of magnitude $0.80\ \text{T}$. What is the magnitude of torque experienced by the coil?

Solution. A coil carrying a current is a magnetic dipole of moment $m = NIA$, directed along its normal, and in a uniform field it feels a torque $\tau = mB\sin\theta$, where $\theta$ is the angle between the field and the normal — the angle the question gives. The area is $A = (0.10)^2 = 1.0\times10^{-2}\ \text{m}^2$:

$$\tau = NIAB\sin\theta = (20)(12)(1.0\times10^{-2})(0.80)\sin30^\circ = 1.92\times\tfrac12 = 0.96\ \text{N m}$$

The coil hangs vertically in a horizontal field, so this torque turns it about the vertical, towards the position where its normal lies along the field.

Question 4.10

Two moving coil meters, $M_1$ and $M_2$ have the following particulars:

$$R_1 = 10\ \Omega,\quad N_1 = 30,\quad A_1 = 3.6\times10^{-3}\ \text{m}^2,\quad B_1 = 0.25\ \text{T}$$

$$R_2 = 14\ \Omega,\quad N_2 = 42,\quad A_2 = 1.8\times10^{-3}\ \text{m}^2,\quad B_2 = 0.50\ \text{T}$$

(The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of $M_2$ and $M_1$.

Question 4.10 (a)

Solution. In a moving coil meter the magnetic torque $NIAB$ is balanced by the spring’s restoring torque $k\phi$, so the deflection per unit current — the current sensitivity — is

$$\frac{\phi}{I} = \frac{NAB}{k}$$

The spring constants are the same, so $k$ cancels from the ratio:

$$\frac{(\phi/I)_2}{(\phi/I)_1} = \frac{N_2A_2B_2}{N_1A_1B_1} = \frac{(42)(1.8\times10^{-3})(0.50)}{(30)(3.6\times10^{-3})(0.25)} = \frac{3.78\times10^{-2}}{2.70\times10^{-2}} = 1.4$$

Question 4.10 (b)

Solution. A voltage $V$ across the meter drives a current $I = V/R$ through it, so the deflection per unit voltage is the current sensitivity divided by the resistance:

$$\frac{\phi}{V} = \frac{NAB}{kR}$$

The ratio is therefore the answer to (a) multiplied by $R_1/R_2$:

$$\frac{(\phi/V)_2}{(\phi/V)_1} = 1.4\times\frac{R_1}{R_2} = 1.4\times\frac{10}{14} = 1$$

$M_2$ is more sensitive to current, but its larger resistance takes all of that advantage away when it is used to measure a voltage.

Question 4.11

In a chamber, a uniform magnetic field of $6.5\ \text{G}$ ($1\ \text{G} = 10^{-4}\ \text{T}$) is maintained. An electron is shot into the field with a speed of $4.8\times10^{6}\ \text{m s}^{-1}$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. ($e = 1.6\times10^{-19}\ \text{C}$, $m_e = 9.1\times10^{-31}\ \text{kg}$)

Note on the question. The book prints $e = 1.5\times10^{-19}\ \text{C}$ here. That is a misprint for the electron’s charge, $1.6\times10^{-19}\ \text{C}$, and the book’s own answers for this question and the next (4.2 cm and 18 MHz) are the values $1.6\times10^{-19}\ \text{C}$ gives. With the misprinted value the radius would come out as $4.5\ \text{cm}$.

Solution. Why a circle. The magnetic force on the electron, $\vec F = -e\,\vec v\times\vec B$, is always perpendicular to its velocity. A force at right angles to the motion does no work, so the electron’s speed never changes; it only turns the velocity. Because the speed and the field are both constant and $\vec v$ is perpendicular to $\vec B$, the force has the constant magnitude $evB$ and always points sideways to the motion — exactly what a uniform circular motion needs as its centripetal force. The electron has no velocity component along the field, so the path is a flat circle rather than a helix.

Radius. Setting the magnetic force equal to the centripetal force, $evB = m_e v^2/r$, gives $r = m_e v/eB$. First convert the field: $B = 6.5\ \text{G} = 6.5\times10^{-4}\ \text{T}$. Then

$$r = \frac{m_e v}{eB} = \frac{(9.1\times10^{-31})(4.8\times10^{6})}{(1.6\times10^{-19})(6.5\times10^{-4})} = \frac{4.37\times10^{-24}}{1.04\times10^{-22}} = 4.2\times10^{-2}\ \text{m}$$

Question 4.12

In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

Solution. One revolution is a distance $2\pi r$ covered at speed $v$, so the period is $T = 2\pi r/v$. Putting in $r = m_e v/eB$ from Question 4.11, the speed cancels:

$$T = \frac{2\pi m_e}{eB}, \qquad \nu = \frac{1}{T} = \frac{eB}{2\pi m_e} = \frac{(1.6\times10^{-19})(6.5\times10^{-4})}{2\pi\times9.1\times10^{-31}} = 1.8\times10^{7}\ \text{Hz}$$

This is the cyclotron frequency. It does not depend on the speed: a faster electron moves in a proportionally larger circle ($r \propto v$), so the longer path and the higher speed cancel and each revolution takes the same time. This holds as long as the speed is small enough for the mass to stay constant, which it is at $4.8\times10^6\ \text{m s}^{-1}$.

Question 4.13

(a) A circular coil of $30$ turns and radius $8.0\ \text{cm}$ carrying a current of $6.0\ \text{A}$ is suspended vertically in a uniform horizontal magnetic field of magnitude $1.0\ \text{T}$. The field lines make an angle of $60^\circ$ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning. (b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

Question 4.13 (a)

Solution. The arrangement is the same as in Question 4.9, with $\theta = 60^\circ$ now given directly as the angle between the field and the normal. The field turns the coil with a torque $NIAB\sin\theta$, and the counter torque needed to hold it still must be equal and opposite. The area is $A = \pi(0.080)^2 = 2.01\times10^{-2}\ \text{m}^2$:

$$\tau = NIAB\sin\theta = (30)(6.0)(2.01\times10^{-2})(1.0)\sin60^\circ = 3.62\times0.866 = 3.1\ \text{N m}$$

Question 4.13 (b)

Solution. No. The torque $\vec\tau = NI\vec A\times\vec B$ depends on the coil only through its area and the direction of its normal, not its shape. One way to see why: divide any flat loop into a grid of tiny rectangular loops, each carrying the same current round it. Along every inner edge two neighbouring loops carry equal and opposite currents, which cancel, leaving just the current round the outer boundary. Each tiny loop feels a torque proportional to its own area, and the torques all point the same way, so the total is proportional to the total area.

Common mistakes

• Question 4.1: using the straight-wire formula. The field at the centre of a coil is $\mu_0 NI/2R$, not $\mu_0 I/2\pi r$. The two look alike, and mixing them up gives an answer $\pi$ times too small; forgetting the factor $N$ makes it $100$ times too small.
• Questions 4.3 and 4.4: giving a horizontal field in 4.3 or a vertical one in 4.4. The field lines circle the wire, so beside a horizontal wire the field is vertical and below it the field is horizontal. Students who picture the field running along or away from the wire get the direction wrong before the right-hand rule is even applied.
• Question 4.5: using $\cos30^\circ$. Only the part of the field perpendicular to the wire exerts a force, so it is $\sin30^\circ$, giving $0.6\ \text{N m}^{-1}$ rather than $1.04\ \text{N m}^{-1}$.
• Question 4.8: counting one layer only, or using the diameter. All five layers add, so $n = 2000/0.80 = 2500\ \text{m}^{-1}$. Using $400$ turns gives a field five times too small, and the $1.8\ \text{cm}$ diameter is there only to show that the solenoid is long; it does not enter $B = \mu_0 nI$.
• Questions 4.9 and 4.13: using the cosine of the given angle. Both questions give the angle between the field and the normal, and the torque is $NIAB\sin\theta$ for that angle. Treating it as the angle with the plane of the coil swaps sine and cosine, giving $1.66\ \text{N m}$ in 4.9 and $1.8\ \text{N m}$ in 4.13.
• Question 4.10(b): assuming the more current-sensitive meter is also more voltage-sensitive. Voltage sensitivity is current sensitivity divided by $R$, and $M_2$’s larger resistance cancels its advantage exactly, leaving a ratio of $1$.
• Question 4.11: leaving the field in gauss. $6.5\ \text{G}$ must become $6.5\times10^{-4}\ \text{T}$ before it goes into $r = mv/eB$; forgetting makes the orbit $10^4$ times too small.
• Question 4.12: thinking a faster electron goes round more often. It moves faster but in a proportionally larger circle, so the frequency $eB/2\pi m$ does not change.

Practise next

• Chapter 5, Magnetism and Matter — the current loop of Questions 4.9 and 4.13 reappears as a magnetic dipole, and $m = NIA$ carries over to solenoids and bar magnets.
• Chapter 1, Electric Charges and Fields — the torque on an electric dipole, $\tau = pE\sin\theta$, is the electric counterpart of the torque on a coil, and worth revising alongside it.