Previous year questions: Moving Charges and Magnetism

Ten questions follow, taken from official CBSE papers in the collected corpus. The label on each question gives the paper and the marks. Work the question first, then compare your direction conventions with the answer.

Q1 (CBSE Board 2023, 5 marks)

State the Biot-Savart law. Apply it to a circular current loop and find the magnetic field at a point on its axis, stating the direction of the field. An electron revolves around a nucleus in a circular orbit of radius 10^-10 m with a speed of 10^7 m s^-1. Find the current constituted by the revolving electron.

Answer: The field of a current element is (mu0 / 4pi) I dl × r-hat / r^2. For a loop of radius R with the point at distance x on the axis, each element's field has an axial component and a perpendicular component, and the perpendicular components cancel by symmetry. Integrating the axial components gives B = mu0 N I R^2 / (2 (R^2 + x^2)^(3/2)), directed along the axis by the right hand rule. For the electron, current equals charge divided by time period: I = e / T, where T = 2 pi r / v = 2 pi × 10^-10 / 10^7 = 6.28 × 10^-17 s. Therefore I = 1.6 × 10^-19 / 6.28 × 10^-17 ≈ 2.5 × 10^-3 A, about 2.5 mA.

Q2 (CBSE Board 2023, 5 marks)

Write the principle of a moving coil galvanometer and explain its working. Why can a galvanometer not be used as such to measure the current in a circuit? Why is the magnetic field made radial in a moving coil galvanometer, and how is this achieved?

Answer: A moving coil galvanometer works on the torque on a current carrying coil in a magnetic field. A coil of N turns and area A suspended in a radial field B carries current I, so it feels a torque N I A B. The spring supplies a restoring torque k phi. At equilibrium N I A B = k phi, so the deflection is proportional to the current and the scale is linear. A galvanometer cannot measure large currents directly because its coil is delicate and full scale deflection occurs for a small current; a larger current could damage the coil. The radial field keeps the plane of the coil parallel to B at every deflection, so the torque is independent of the angle and the scale stays uniform. The radial field is produced by concave shaped pole pieces and a soft iron cylinder at the centre.

Q3 (CBSE Board 2024, 1 mark)

A galvanometer of resistance G is converted into an ammeter of range 0 to I A. If the current through the galvanometer is 0.1 percent of I A, what is the resistance of the ammeter?

Answer: The shunt carries the remaining current, so the current through the galvanometer is Ig = 0.001 I. The shunt satisfies Ig G = (I - Ig) S, which gives S = Ig G / (I - Ig) = 0.001 I G / 0.999 I = G / 999. The ammeter resistance is the parallel combination of G and G/999, which is approximately G/1000. The current through the meter is 0.1 percent of the range, so nearly all of it passes through the shunt.

Q4 (CBSE Board 2024, 3 marks)

Two long, straight, parallel conductors carry steady currents in opposite directions. Explain the nature of the force of interaction between them. Draw the field lines and find the force per unit length.

Answer: Each wire lies in the magnetic field of the other. The field of wire 1 at wire 2 is mu0 I1 / (2 pi d), perpendicular to both the wire and the separation. The force on a length L of wire 2 is I2 L B, so the force per unit length is mu0 I1 I2 / (2 pi d). For currents in opposite directions the force pushes the wires apart, so it is repulsive. For parallel currents the same expression gives attraction. The field lines around each wire are circles, and between the wires the two fields point in the same direction when the currents are opposite, which raises the field there.

Q5 (CBSE Board 2025, 3 marks)

Write the vector form of the Biot-Savart law. Two insulated long straight wires, each carrying 2.0 A, are kept along the x and y axes. Find the magnitude and direction of the resultant magnetic field at the point (4 m, 5 m).

Answer: The vector form is dB = (mu0 / 4pi) I dl × r-hat / r^2. The point is 5 m from the x-axis wire and 4 m from the y-axis wire. The field of the x-axis wire circles it; at the point its direction is along positive z with magnitude mu0 I / (2 pi × 5) = 2 × 10^-7 × 2 / 5 = 8 × 10^-8 T. The field of the y-axis wire at the point points along negative z with magnitude 2 × 10^-7 × 2 / 4 = 1 × 10^-7 T. The net field is 2 × 10^-8 T along negative z.

Q6 (CBSE Board 2025, 3 marks)

A charged particle enters a uniform magnetic field. Describe the path followed by the particle and draw its shape. How will the magnetic field inside a long solenoid be affected when the radius of the turns is increased, and when the length as well as the total number of turns are doubled?

Answer: A charge entering perpendicular to the field follows a circular path, because the magnetic force supplies the centripetal force. If it enters at an angle, the path is a helix with axis along the field. Inside a long solenoid B = mu0 n I, where n is turns per unit length. Increasing the radius of the turns does not change n, so the field is unchanged. Doubling both the length and the total number of turns keeps the turns per unit length the same, so the field again stays the same.

Q7 (CBSE Board 2026, 4 marks)

A moving coil galvanometer is used to detect small currents. Explain its working principle and the role of the radial magnetic field. A galvanometer of resistance 20 ohm gives full scale deflection at 5 mA. Convert it into an ammeter of range 10 A and find the required shunt.

Answer: The coil feels a magnetic torque N I A B and the spring a restoring torque k phi; at balance phi = N I A B / k, so the deflection measures the current. The radial field keeps the torque independent of deflection. For the ammeter, a shunt S in parallel carries the excess current. The condition is Ig G = (I - Ig) S, so S = 5 × 10^-3 × 20 / (10 - 5 × 10^-3) ≈ 1.0 × 10^-2 ohm. The shunt is connected in parallel with the galvanometer, and the ammeter resistance is then the parallel combination, about 0.01 ohm.

Q8 (CBSE Board 2026, 1 mark)

A galvanometer of resistance 15 ohm shows full scale deflection at 3 mA. What series resistance converts it into a voltmeter of range 0 to 12 V?

Answer: The series resistance must drop the remaining voltage while the full scale current flows. V = Ig (G + R), so R = V / Ig - G = 12 / 0.003 - 15 = 4000 - 15 = 3985 ohm. The voltmeter of this range therefore has 3985 ohm in series with the 15 ohm coil, and its total resistance is 4000 ohm.

Q9 (CBSE Additional PQ 2023-24, 1 mark)

A positively charged particle is projected with uniform velocity along the axis of a current carrying long solenoid. Which statement is true: the particle is accelerated along the axis, follows a circular path, moves with the same velocity along the axis, or follows a parabolic path?

Answer: The particle moves with the same velocity along the axis. The field inside a long solenoid is parallel to the axis, so the particle's velocity is parallel to the magnetic field. The cross product v × B is zero, the magnetic force is zero, and the particle continues in a straight line at constant velocity.

Q10 (CBSE Additional PQ 2022-23, 1 mark)

A deuteron and an alpha particle move with the same kinetic energy in identical magnetic fields. What is the ratio of the radii of their circular paths?

Answer: The radius of the circular path is r = m v / (q B), and with the same kinetic energy K the momentum is p = sqrt(2 m K). So r = sqrt(2 m K) / (q B). For the deuteron, mass 2 u and charge e; for the alpha particle, mass 4 u and charge 2 e. The ratio r_d / r_alpha is sqrt(2)/1 divided by sqrt(4)/2, which is sqrt(2). The deuteron path is larger by a factor of root two.