A solenoid of length 0.4 m and having 500 turns of wire carries a current of 3 A. A thin coil having 10 turns of wire and of radius 0.01 m carries a current of 0.4 A. Calculate the torque required to hold the coil in the middle of the solenoid with its axis perpendicular to the axis of the solenoid.

The torque acting on a current loop of turn N and current i in a magnetic field of induction B due to a solenoid is given by
τ = (N i A) B sin θ
where,
i is the current flowing in N turns of the loop with face area A and making an angle with the field.
N i A can be regarded as the magnetic dipole moment M.
The magnetic dipole moment M of the loop lies along the axis of the loop.
Magnetic field B due to a solenoid is given by

B = μ_{0} n i_{0}

where,
n is the number of turns per unit length of the solenoid and,
i₀ is the current in the solenoid.
The direction of B is along its axis.
Therefore, when the small coil is held with its axis perpendicular to the axis of the solenoid, the torque becomes

τ = μ_{0} {ni}_{0} N i A

Here,

N = 10, n = \frac{500}{0.4} = 1250, i_{0} = 3 A, i = 0.4 A

A = {πr}^{2} = π {(0.01)}^{2} = π \times 10^{- 4} m^{2} μ_{0} = 4 π \times 10^{- 7}

Therefore,

τ = 4 π \times 10^{- 7} \times 1250 \times 3 \times 10 \times 0.4 \times π \times 10^{- 4}

= 5.922 \times 10^{- 6} N m

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