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:
Answer & explanation
Correct answer: option 1
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°
$τ=NIAB\sinθ$
$=1000×5×1.2×10^{-4}×5×10^{-2}×\sin 30°$
$=15×10^{-3}$
$=1.5×10^{-2}Nm$