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Q.

The torque required to hold a small circular coil of 10 turns, $2 \times 10^{- 4} m^{2}$ area and carrying 0.5 A current in the middle of a long solenoid of $10^{3}$ turns/m and carrying 3 A current, with its axis perpendicular to the axis of the solenoid, is

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a

$2 π \times 10^{- 7} N m$

b

$12 π \times 10^{- 7} N m$

c

$6 π \times 10^{- 7} N m$

d

$4 π \times 10^{- 7} N m$

answer is A.

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Detailed Solution

Magnetic dipole moment of current loop is
$M = N i A = 10 \times 0.5 \times 2 \times 10^{- 4} = 10^{- 3} A m^{2}$
Magnetic field due to the solenoid carrying current i e . . ,
$B = μ_{0} n i = 4 π 10^{- 7} \times 10^{- 3} \times 3 = 12 π \times 10^{- 4} T$
$∴$ Torque,
, $\begin{array}{l} τ = M B \sin θ \\ = 10^{- 3} \times 12 π \times 10^{- 4} \times \sin 90^{0} \\ = 12 π \times 10^{- 7} N m \end{array}$

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The torque required to hold a small circular coil of 10 turns, 2×10−4m2 area and carrying 0.5 A current in the middle of a long solenoid of 103 turns/m and carrying 3 A current, with its axis perpendicular to the axis of the solenoid, is