A closely wound solenoid of 1000 turns and area of cross section $1.2 cm^2$ carries a current of 5 A. It is suspended horizontally in uniform magnetic field of $5×10^{-2} T$ making an angle of 30° with the axis of solenoid. The torque acting on the solenoid is:

$1.5×10^{-2} N m$

The correct answer is Option (1) → $1.5×10^{-2} N m$

Torque ($τ$) acting on a solenoid in a uniform magnetic field is given by the formula -

$τ=NIAB\sinθ$

where,

N, Number of turns = 1000

I, Current = 5 A

A, Area of cross-section = $1.2×10^{-4}m^2$

B, Magnetic field strength = $5×10^{-2}T$

θ, Angle between the magnetic field and the axis of the solenoid = 30°

$τ=NIAB\sinθ$

$=1000×5×1.2×10^{-4}×5×10^{-2}×\sin 30°$

$=15×10^{-3}$

$=1.5×10^{-2}Nm$