A rectangular coil of 40 turns, each of area 25 cm2, is placed in a uniform magnetic field of 0.4 T. The axis of the coil makes an angle of 30° with the field. The current in the coil that produces a torque of 0.02 N m on it is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1 A
A coil of \( N \) turns in a uniform field feels a torque \( \tau=NIAB\sin\theta \). Here \( \theta \) is the angle between the field and the normal (axis) of the coil.
Given:
\( N=40 \)
\( A=25\ \mathrm{cm^{2}}=2.5\times10^{-3}\ \mathrm{m^{2}} \)
\( B=0.4\ \mathrm{T} \)
\( \theta=30^\circ \)
\( \tau=0.02\ \mathrm{N\,m} \)
\( \begin{aligned} I &= \frac{\tau}{NAB\sin\theta} \\ &= \frac{0.02}{40\times2.5\times10^{-3}\times0.4\times0.5} \\ &= \frac{0.02}{0.02} \\ &= 1\ \mathrm{A} \end{aligned} \)
Therefore, a current of 1 A is needed in the coil.
NCERT (2026-27):
Torque on a rectangular current loop in a uniform magnetic field – “… We take the angle between the field and the normal to the coil to be angle θ … The magnitude of the torque on the loop is,”
\( \tau=IAB\sin\theta \)\( \qquad (4.21) \)
“If the loop has N closely wound turns, the expression for torque, … still holds, with”
\( m=NIA \)\( \qquad (4.24) \)