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Question: The torque required to hold a small circular coil of 10 turns, area 1 mm2 and carrying a current of ...

The torque required to hold a small circular coil of 10 turns, area 1 mm2 and carrying a current of (2144)\left( \frac { 21 } { 44 } \right) A in middle of a long solenoid of 103 turns/m carrying a current of 2.5 A, with its axis perpendicular to the axis of the solenoid is

A

1.5 x 10-6 N m

B

1.5 x l0-8Nm

C

1.5 x 106 N m

D

1.5 x l08Nm

Answer

1.5 x l0-8Nm

Explanation

Solution

: Here, For small circular coil, Number of turns,

Area, A = 1 mm2=1×10−6 m2\mathrm { mm } ^ { 2 } = 1 \times 10 ^ { - 6 } \mathrm {~m} ^ { 2 }

Current,

For a long solenoid

Number of turns per metre, n=103\mathrm { n } = 10 ^ { 3 } per m

Current, I2=2.5 A\mathrm { I } _ { 2 } = 2.5 \mathrm {~A}

Magnetic field due to a long solenoid on its axis is

B=μ0nI2\mathrm { B } = \mu _ { 0 } \mathrm { nI } _ { 2 } …(i)

Magnetic moment of a circular coil is,

M=NAI1\mathrm { M } = \mathrm { NAI } _ { 1 } …(ii)

Torque,

τ=mBsin⁡θ=MB\tau = \mathrm { mB } \sin \theta = \mathrm { MB } (∵θ=90∘\because \theta = 90 ^ { \circ }(Given)

τ=(NAI1)(μ0nI2)\tau = \left( \mathrm { NAI } _ { 1 } \right) \left( \mu _ { 0 } \mathrm { nI } _ { 2 } \right) (Using (i) and (ii))

τ=10×1×10−6×2144×4×227×10−7×103×2.5\tau = 10 \times 1 \times 10 ^ { - 6 } \times \frac { 21 } { 44 } \times 4 \times \frac { 22 } { 7 } \times 10 ^ { - 7 } \times 10 ^ { 3 } \times 2.5

=1.5×10−8Nm= 1.5 \times 10 ^ { - 8 } \mathrm { Nm }