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Torque on a Current Loop and Moving Coil Galvanometer

A current loop placed in a magnetic field experiences a torque tau = N I A B sin theta, where N is number of turns, I is current, A is area, B is magnetic field, and theta is the angle between the magnetic moment and magnetic field.

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Student-friendly explanation

The two opposite sides of a rectangular current loop experience equal and opposite magnetic forces that form a couple. This couple rotates the loop. In a moving coil galvanometer, a current-carrying coil placed in a radial magnetic field experiences a deflecting torque proportional to current. A spring provides restoring torque, so the steady deflection becomes proportional to current.

How to write this in exams

  1. 1

    Start with the exact idea

    A current loop placed in a magnetic field experiences a torque tau = N I A B sin theta, where N is number of turns, I is current, A is area, B is magnetic field, and theta is the angle between the magnetic moment and magnetic field.

  2. 2

    Then show how to use it

    For torque, mark forces on opposite arms of the loop. Show they form a couple. Write tau = force x perpendicular distance and obtain tau = NIAB sin theta. For galvanometer, write deflecting torque and restoring torque, equate them, and derive proportionality between current and deflection.

  3. 3

    Add one concrete example

    A galvanometer can detect small currents. By connecting a low resistance shunt in parallel, it can be converted into an ammeter. By connecting a high resistance in series, it can be converted into a voltmeter.

  4. 4

    Avoid this incomplete answer

    Saying that an ammeter is made by connecting high resistance in series reverses the conversions. Ammeter needs low parallel shunt; voltmeter needs high series resistance.

Definition

A current loop placed in a magnetic field experiences a torque tau = N I A B sin theta, where N is number of turns, I is current, A is area, B is magnetic field, and theta is the angle between the magnetic moment and magnetic field.

Example

A galvanometer can detect small currents. By connecting a low resistance shunt in parallel, it can be converted into an ammeter. By connecting a high resistance in series, it can be converted into a voltmeter.

Rule to remember

Torque on coil: tau = N I A B sin theta. Magnetic moment m = N I A, so vector form tau = m x B. Galvanometer balance: NIAB = k phi, so I = (k/(NAB)) phi. Current sensitivity is phi/I = NAB/k. Ammeter conversion uses low shunt resistance in parallel; voltmeter conversion uses high resistance in series. SI units: torque in N m, B in tesla, A in m^2, current in ampere.

Memory hook

Loop torque turns the coil; spring torque measures the current.

Examples and method

Worked example

A galvanometer has N = 100 turns, A = 2.0 x 10^-4 m^2, B = 0.50 T, and torsional constant k = 1.0 x 10^-6 N m rad^-1. Current sensitivity phi/I = NAB/k = (100 x 2.0 x 10^-4 x 0.50)/(1.0 x 10^-6) = 1.0 x 10^4 rad A^-1. A current of 1 microampere gives deflection 0.01 rad.

Method to apply

For torque, mark forces on opposite arms of the loop. Show they form a couple. Write tau = force x perpendicular distance and obtain tau = NIAB sin theta. For galvanometer, write deflecting torque and restoring torque, equate them, and derive proportionality between current and deflection.

Diagram support

Draw rectangular coil, magnetic field, current direction, force on opposite arms, rotation axis, spring, pointer, radial pole pieces, and scale. For conversions, show shunt parallel to galvanometer and series resistance for voltmeter.

How CBSE asks it

Asked as torque derivation, principle and working of moving coil galvanometer, reason for radial field, current sensitivity, or conversion to ammeter and voltmeter.

Avoid common mistakes

Common confusion

Students often write tau = NIAB without explaining that this is the maximum torque or the radial-field galvanometer case where sin theta remains effectively 1.

Common wrong answer

Saying that an ammeter is made by connecting high resistance in series reverses the conversions. Ammeter needs low parallel shunt; voltmeter needs high series resistance.

Exam tip

For galvanometer answers, write both torques: deflecting torque NIAB and restoring torque k phi. Then equate them to show phi is proportional to I.

Quick check

Why is the scale of a moving coil galvanometer uniform in a radial magnetic field?

The scale is uniform because the radial magnetic field keeps the plane of the coil effectively parallel to the field direction needed for maximum torque, so deflecting torque is proportional to current. Since restoring torque is proportional to deflection, deflection is directly proportional to current.

Answer writing and exam use

1-mark answer

A current loop placed in a magnetic field experiences a torque tau = N I A B sin theta, where N is number of turns, I is current, A is area, B is magnetic field, and theta is the angle between the magnetic moment and magnetic field.

2-mark answer

A current loop placed in a magnetic field experiences a torque tau = N I A B sin theta, where N is number of turns, I is current, A is area, B is magnetic field, and theta is the angle between the magnetic moment and magnetic field. Torque on coil: tau = N I A B sin theta. Magnetic moment m = N I A, so vector form tau = m x B. Galvanometer balance: NIAB = k phi, so I = (k/(NAB)) phi. Current sensitivity is phi/I = NAB/k. Ammeter conversion uses low shunt resistance in parallel; voltmeter conversion uses high resistance in series. SI units: torque in N m, B in tesla, A in m^2, current in ampere. A galvanometer can detect small currents. By connecting a low resistance shunt in parallel, it can be converted into an ammeter. By connecting a high resistance in series, it can be converted into a voltmeter.

3-mark answer

The two opposite sides of a rectangular current loop experience equal and opposite magnetic forces that form a couple. This couple rotates the loop. In a moving coil galvanometer, a current-carrying coil placed in a radial magnetic field experiences a deflecting torque proportional to current. A spring provides restoring torque, so the steady deflection becomes proportional to current. Torque on coil: tau = N I A B sin theta. Magnetic moment m = N I A, so vector form tau = m x B. Galvanometer balance: NIAB = k phi, so I = (k/(NAB)) phi. Current sensitivity is phi/I = NAB/k. Ammeter conversion uses low shunt resistance in parallel; voltmeter conversion uses high resistance in series. SI units: torque in N m, B in tesla, A in m^2, current in ampere. A galvanometer has N = 100 turns, A = 2.0 x 10^-4 m^2, B = 0.50 T, and torsional constant k = 1.0 x 10^-6 N m rad^-1. Current sensitivity phi/I = NAB/k = (100 x 2.0 x 10^-4 x 0.50)/(1.0 x 10^-6) = 1.0 x 10^4 rad A^-1. A current of 1 microampere gives deflection 0.01 rad. Asked as torque derivation, principle and working of moving coil galvanometer, reason for radial field, current sensitivity, or conversion to ammeter and voltmeter. Saying that an ammeter is made by connecting high resistance in series reverses the conversions. Ammeter needs low parallel shunt; voltmeter needs high series resistance.
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