Exercises — Moving Charges and Magnetism

1. 12 marksNCERT Exercises, Chapter 4

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

   Hint. Use the field at the centre of a circular coil, remembering the factor of N for the number of turns.

   The magnetic field at the centre of a circular coil of $N$ turns is:

   $B=\frac{{\mu }_{0}NI}{2r}$

   Substituting $N=100$, $I=0.40$ A and $r=0.08$ m:

   $B=\frac{4\pi \times 10^{-7}\times 100\times 0.40}{2\times 0.08}=\frac{5.03\times 10^{-5}}{0.16}$

   $B\approx 3.1\times 10^{-4}\text{T}$

   The factor $N$ appears because the turns are wound closely together, so each contributes the same field and they simply add.

   ✦ B = 3.1 x 10^-4 T

2. 22 marksNCERT Exercises, Chapter 4

   A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?

   Hint. Use the field of a long straight wire, which falls off as 1/r.

   For a long straight wire the field at perpendicular distance $r$ is:

   $B=\frac{{\mu }_{0}I}{2\pi r}$

   Substituting $I=35$ A and $r=0.20$ m, and using $\frac{{\mu }_0}{2\pi }=2\times 10^{-7}$:

   $B=\frac{2\times 10^{-7}\times 35}{0.20}=\frac{7.0\times 10^{-6}}{0.20}=3.5\times 10^{-5}\text{T}$

   Note the field falls off as $1/r$ for a straight wire, in contrast to the $1/r^2$ of a point charge's electric field.

   The field weakens with distance because the same current spreads its influence over an ever larger circumference.

   ✦ B = 3.5 x 10^-5 T

3. 33 marksNCERT Exercises, Chapter 4

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

   Hint. Compute the magnitude first, then apply the right-hand grip rule to fix the direction.

   The magnitude follows from the straight-wire formula:

   $B=\frac{{\mu }_{0}I}{2\pi r}=\frac{2\times 10^{-7}\times 50}{2.5}=4\times 10^{-6}\text{T}$

   For the direction, apply the right-hand grip rule: point the right thumb along the current, which flows from north to south, and the fingers curl in the direction of the field.

   At a point to the east of the wire, the curling fingers point vertically upward.

   Stating the direction is essential here, since the question asks for it explicitly and the magnetic field is a vector.

   ✦ B = 4 x 10^-6 T, directed vertically upward

4. 43 marksNCERT Exercises, Chapter 4

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

   Hint. Again find the magnitude first, then use the grip rule with the thumb pointing west.

   Using the straight-wire result with $I=90$ A and $r=1.5$ m:

   $B=\frac{2\times 10^{-7}\times 90}{1.5}=\frac{1.8\times 10^{-5}}{1.5}=1.2\times 10^{-5}\text{T}$

   Applying the right-hand grip rule with the thumb pointing along the current, from east to west, the fingers curl so that at a point below the wire the field points towards the south.

   A useful check is that directly above the line the field would point north, since the field encircles the wire and reverses on opposite sides of it.

   ✦ B = 1.2 x 10^-5 T, directed towards the south

5. 52 marksNCERT Exercises, Chapter 4

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

   Hint. Use F = B I L sin(theta) and divide through by the length.

   The force on a current-carrying wire in a magnetic field is $F=BIL\sin \theta$, so the force per unit length is:

   $\frac{F}{L}=BI\sin \theta$

   Substituting $B=0.15$ T, $I=8$ A and $\theta =30^{\circ }$, so that $\sin \theta =0.5$:

   $\frac{F}{L}=0.15\times 8\times 0.5=0.6\text{N/m}$

   The force would be zero if the wire were parallel to the field and maximum if perpendicular to it, which is why the sine of the angle appears rather than the cosine.

   ✦ F/L = 0.6 N/m

6. 62 marksNCERT Exercises, Chapter 4

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

   Hint. The wire is perpendicular to the field, so the sine factor is 1.

   Since the wire lies perpendicular to the solenoid's axis, and the field inside a solenoid is along that axis, the angle between the current and the field is $90^{\circ }$ and $\sin \theta =1$.

   $F=BIL\sin \theta =0.27\times 10\times 0.03\times 1$

   $F=8.1\times 10^{-2}\text{N}$

   Remember to convert the length from centimetres to metres before substituting, since leaving it as 3.0 would inflate the answer a hundredfold.

   ✦ F = 8.1 x 10^-2 N

7. 73 marksNCERT Exercises, Chapter 4

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

   Hint. Find the force per unit length between the wires first, then multiply by the section length.

   The force per unit length between two parallel current-carrying wires is:

   $\frac{F}{L}=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}=\frac{2\times 10^{-7}\times 8.0\times 5.0}{0.04}$

   $\frac{F}{L}=\frac{8.0\times 10^{-6}}{0.04}=2.0\times 10^{-4}\text{N/m}$

   For a 10 cm section, that is $L=0.10$ m:

   $F=2.0\times 10^{-4}\times 0.10=2.0\times 10^{-5}\text{N}$

   Because the currents flow in the same direction, the force is attractive. Antiparallel currents would repel instead, which is the opposite of the behaviour of like electric charges.

   ✦ F = 2.0 x 10^-5 N, attractive since the currents are parallel

8. 83 marksNCERT Exercises, Chapter 4

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

   Hint. Find the total number of turns first, then the turns per unit length.

   The total number of turns is the product of the layers and the turns per layer:

   $N=5\times 400=2000\text{turns}$

   The number of turns per unit length is therefore:

   $n=\frac{N}{L}=\frac{2000}{0.80}=2500\text{turns/m}$

   The field near the centre of a long solenoid is:

   $B={\mu }_{0}nI=4\pi \times 10^{-7}\times 2500\times 8.0$

   $B\approx 2.5\times 10^{-2}\text{T}$

   The diameter of 1.8 cm is not needed, because the field inside a long solenoid is uniform and independent of its radius.

   ✦ B = 2.5 x 10^-2 T

9. 93 marksNCERT Exercises, Chapter 4

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

   Hint. Use the torque on a current loop, with the angle measured between the normal and the field.

   The torque on a current-carrying coil in a uniform magnetic field is:

   $\tau =NIAB\sin \theta$

   where $\theta$ is the angle between the normal to the coil and the field, which is exactly what the question specifies.

   The area of the square coil is $A=(0.10{)}^2=0.01$ m${}^2$.

   $\tau =20\times 12\times 0.01\times 0.80\times \sin 30^{\circ }$

   $\tau =20\times 12\times 0.01\times 0.80\times 0.5=0.96\text{N m}$

   Had the angle been given between the coil's plane and the field, it would first need converting, since the normal is perpendicular to the plane.

   ✦ Torque = 0.96 N m

10. 104 marksNCERT Exercises, Chapter 4

    Two moving coil meters M1 and M2 have: $R_1=10\Omega$, $N_1=30$, $A_1=3.6\times 10^{-3}$ m${}^2$, $B_1=0.25$ T; and $R_2=14\Omega$, $N_2=42$, $A_2=1.8\times 10^{-3}$ m${}^2$, $B_2=0.50$ T. The spring constants are identical. Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.

    Hint. Write down both sensitivity formulas and note that the identical spring constants cancel in the ratios.

    For a moving coil galvanometer the two sensitivities are:

    $\text{Current sensitivity}=\frac{NBA}{k},\text{Voltage sensitivity}=\frac{NBA}{kR}$

    Since the spring constants $k$ are identical, they cancel from both ratios.

    (a) Ratio of current sensitivities:

    $\frac{N_2{B}_2{A}_2}{N_1{B}_1{A}_1}=\frac{42\times 0.50\times 1.8\times 10^{-3}}{30\times 0.25\times 3.6\times 10^{-3}}=\frac{0.0378}{0.0270}=1.4$

    (b) The voltage sensitivity carries an extra factor of $1/R$, so the ratio is the current-sensitivity ratio multiplied by $\frac{R_1}{R_2}$:

    $1.4\times \frac{10}{14}=1.0$

    So M2 is 1.4 times as sensitive to current, yet the two meters are equally sensitive to voltage, because M2's higher resistance exactly offsets its advantage.

    ✦ (a) Current sensitivity ratio = 1.4 (b) Voltage sensitivity ratio = 1.0

11. 114 marksNCERT Exercises, Chapter 4

    In a chamber a uniform magnetic field of 6.5 G ($1$ G $=10^{-4}$ T) is maintained. An electron is shot into the field with a speed of $4.8\times 10^6$ m s${}^{-1}$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (Take $e=1.5\times 10^{-19}$ C, $m_e=9.1\times 10^{-31}$ kg.)

    Hint. The magnetic force is always perpendicular to the velocity, which is the defining condition for circular motion.

    Why the path is circular. The magnetic force on a moving charge is $\vec{F}=q\vec{v}\times \vec{B}$, which is always perpendicular to the velocity.

    A force perpendicular to the velocity can change the direction of motion but never its magnitude, so the speed stays constant while the direction turns steadily. Since the field is uniform, the force has constant magnitude too, giving constant speed with constant centripetal force — precisely the conditions for uniform circular motion.

    Radius. Equating the magnetic force to the required centripetal force:

    $qvB=\frac{m{v}^2}{r}\Rightarrow r=\frac{mv}{qB}$

    With $B=6.5\text{G}=6.5\times 10^{-4}$ T:

    $r=\frac{9.1\times 10^{-31}\times 4.8\times 10^6}{1.5\times 10^{-19}\times 6.5\times 10^{-4}}=\frac{4.37\times 10^{-24}}{9.75\times 10^{-23}}$

    $r\approx 4.5\times 10^{-2}\text{m}=4.5\text{cm}$

    ✦ The magnetic force is always perpendicular to the velocity, so the speed is unchanged and the motion is uniform circular; r = 4.5 x 10^-2 m, about 4.5 cm

12. 123 marksNCERT Exercises, Chapter 4

    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.

    Hint. Derive the frequency from the radius expression and see which quantities survive.

    The period is the circumference divided by the speed, $T=\frac{2\pi r}{v}$, and substituting $r=\frac{mv}{qB}$ gives:

    $T=\frac{2\pi m}{qB}\Rightarrow f=\frac{qB}{2\pi m}$

    The speed $v$ has cancelled out entirely.

    Substituting the values:

    $f=\frac{1.5\times 10^{-19}\times 6.5\times 10^{-4}}{2\pi \times 9.1\times 10^{-31}}=\frac{9.75\times 10^{-23}}{5.72\times 10^{-30}}$

    $f\approx 1.7\times 10^7\text{Hz}=17\text{MHz}$

    No, the frequency does not depend on the speed. A faster electron travels a proportionately larger circle, so the extra distance exactly offsets the extra speed and the time per revolution is unchanged. This speed-independence is what makes the cyclotron possible.

    ✦ f = 1.7 x 10^7 Hz, about 17 MHz; it does NOT depend on the speed, because a faster electron moves in a proportionately larger circle

13. 134 marksNCERT Exercises, Chapter 4

    (a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 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 were replaced by a planar coil of some irregular shape enclosing the same area?

    Hint. Compute the torque on the coil; the counter torque must be equal and opposite. For (b), ask which property of the coil the formula actually uses.

    (a) The area of the circular coil is:

    $A=\pi r^2=\pi (0.08{)}^2=2.01\times 10^{-2}{\text{m}}^2$

    The torque on the coil is $\tau =NIAB\sin \theta$, with $\theta =60^{\circ }$ measured from the normal as stated:

    $\tau =30\times 6.0\times 2.01\times 10^{-2}\times 1.0\times \sin 60^{\circ }$

    $\tau =30\times 6.0\times 2.01\times 10^{-2}\times 0.866\approx 3.1\text{N m}$

    To prevent the coil turning, the counter torque must equal this in magnitude and oppose it in direction.

    (b) No, the answer would not change. The torque formula involves only the area enclosed by the coil, not its shape.

    Any planar coil enclosing the same area, whatever its outline, experiences the same torque under otherwise identical conditions.

    The shape is irrelevant because the torque arises from the magnetic moment NIA, which depends only on the enclosed area.

    ✦ (a) Counter torque = 3.1 N m (b) No — the torque depends only on the enclosed area, not on the shape of the coil

Solutions written by the tuition.in editorial team and checked against leph104.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.