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medium difficulty10 min

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 θ, where N is number of turns, I is current, A is area, B is magnetic field, and θ is the angle between the magnetic moment and magnetic field.

In one line: Loop torque turns the coil; spring torque measures the current.

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Understand it

  • 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.

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.

Formula and rule

Torque on coil: tau = N I A B sin θ.

Magnetic moment m = N I A, so vector form tau = m × 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 m2, current in ampere.

See it worked out

Follow the example line by line, then use the same method on new questions.

Worked example

A galvanometer has N = 100 turns, A = 2.0 × 10−4 m2, B = 0.50 T, and torsional constant k = 1.0 × 10−6 N m rad−1.

  1. 1Current sensitivity phi/I = NAB/k = (100 × 2.0 × 10−4 × 0.50)/(1.0 × 10−6) = 1.0 × 104 rad A−1.
  2. 2A current of 1 microampere gives deflection 0.01 rad.

Method you can reuse

  1. 1For torque, mark forces on opposite arms of the loop.
  2. 2Show they form a couple.
  3. 3Write tau = force x perpendicular distance and obtain tau = NIAB sin θ.
  4. 4For galvanometer, write deflecting torque and restoring torque, equate them, and derive proportionality between current and deflection.

Sketch it in your answer

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.

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

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

Think first, then show answer
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.

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.

Don't mix this up

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

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.

Write it in the exam

Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.

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.
  1. 1

    Enough for 1 mark

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

  2. 2

    Add for 2 marks

    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.

  3. 3

    Add for 3 marks

    The two opposite sides of a rectangular current loop experience equal and opposite magnetic forces that form a couple.

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