Two circular coils P and Q carry equal currents. They are placed in the same uniform magnetic field, with their planes parallel to the field. Coil P has 100 turns of radius 5 cm. Coil Q has 50 turns of radius 10 cm. The ratio of the torque on P to the torque on Q is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1 : 2
The torque on a coil of \( N \) turns is \( \tau=NIAB\sin\theta \), where \( \theta \) is the angle between the normal to the coil and the field. Both coils have their planes parallel to the field, so \( \theta=90^\circ \) for each. The current and the field are also the same, so \( \tau\propto NA\propto Nr^{2} \).
\( \begin{aligned} \frac{\tau_P}{\tau_Q} &= \frac{N_P\,r_P^{2}}{N_Q\,r_Q^{2}} \\ &= \frac{100\times5^{2}}{50\times10^{2}} \\ &= \frac{2500}{5000} \\ &= \frac{1}{2} \end{aligned} \)
Therefore, \( \tau_P:\tau_Q=1:2 \). Coil Q has half the turns, but its area is four times larger.
NCERT (2026-27):
Torque on a rectangular current loop in a uniform magnetic field – “… The magnitude of the torque on the loop is,”
\( \displaystyle \tau=IAB\sin\theta \)\( \qquad (4.21) \)
Torque on a rectangular current loop in a uniform magnetic field – “If the loop has N closely wound turns, the expression for torque, … still holds, with”
\( \displaystyle m=NIA \)\( \qquad (4.24) \)