JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Current Electricity

A moving coil galvanometer has 100 turns and each turn has an area of

2.0 cm^{2}

. The magnetic field produced by the magnet is 0.01 T and the deflection in the coil is 0.05 radian when a current of 10 mA is passed through it. The torsional constant of the suspension wire is

x \times 1 0^{- 5} N - m / rad

. The value of x is ____.

Answer: 4

Solution

At equilibrium the restoring torque of the suspension balances the magnetic torque on the coil:

C θ = N B I A

. With

N = 100

B = 0.01

I = 10 \times 1 0^{- 3}

A = 2 \times 1 0^{- 4} m^{2}

and

θ = 0.05

rad:

C = \frac{N B I A}{θ} = \frac{100 \times 0.01 \times 0.01 \times 2 \times 1 0 ^{- 4}}{0.05} = 4 \times 1 0^{- 5} N-m/rad

, so

x = 4

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Integer answer

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