JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Magnetic Effects of Current

A galvanometer coil has 500 turns and each turn has an average area of

3 \times 1 0^{- 4} m^{2}

. 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 ______.

Answer: 20

Solution

The torque on a current-carrying coil is

τ = N B I A sin θ

, where

θ

is the angle between the coil's normal and the field. Keeping the coil's plane parallel to the field puts its normal at right angles to the field, so

θ = 9 0^{\circ}

and

sin θ = 1

. Then

B = \frac{τ}{N I A} = \frac{1.5}{500 \times 0.5 \times 3 \times 1 0 ^{- 4}} = \frac{1.5}{0.075} = 20 T

JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Integer answer

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