A galvanometer coil has 500 turns and each turn has an average area of 3 \times 10–4 m2 . If a torque of 1.5 Nm is required to keep this coil parallel to a magnetic field when a current of 0.5 A is flowing through it, the strength of the field (in T) is \underline\hspace2cm .

Magnetics PYQ — JEE Main 2020 · 3 Sep · Shift 2 · Q42 with Solution

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Correct answer: 20

For a coil of $N$ turns, area $A$ , carrying current $I$ in a magnetic field $B$ , the torque is

\tau = N I A B \sin\theta

where $\theta$ is the angle between the normal to the coil and the magnetic field.

If the plane of the coil is parallel to the magnetic field, then the normal to the coil is perpendicular to the field.

Hence,

\theta = 90^\circ \Rightarrow \sin\theta = 1

So the torque becomes

\tau = N I A B

Given:

N = 500, \quad A = 3 \times 10^{-4}\,\text{m}^2, \quad I = 0.5\,\text{A}, \quad \tau = 1.5\,\text{N m}

Thus,

1.5 = 500 \times 0.5 \times 3 \times 10^{-4} \times B

500 \times 0.5 = 250

250 \times 3 \times 10^{-4} = 750 \times 10^{-4} = 0.075

So,

1.5 = 0.075 B

Therefore,

B = \frac{1.5}{0.075} = 20

The magnetic field strength is

\boxed{20\,\text{T}}