Q74 / 75 · JEE Main 2026 (04 April Shift 2) · Physics
A circular coil of radius $2$ cm and $125$ turns carries a current of $1$ A. The coil is placed in a uniform magnetic field of magnitude $0.4$ T. The axis of the coil makes an angle of $30°$ with the direction of the magnetic field. The torque acting on the coil is $\alpha \times 10^{-4}$ N.m. The value of $\alpha$ is ______.
($\pi=3.14$)
($\pi=3.14$)
Correct answer: 314
Solution
The magnetic moment of the coil is given by $M = NIA$.
The torque acting on the coil is $\tau = MB \sin\theta = NIAB \sin\theta$.
Substituting the given values:
$\tau = 125 \times 1 \times (\pi \times (0.02)^2) \times 0.4 \times \sin 30^\circ$
$\tau = 125 \times 1 \times 4\pi \times 10^{-4} \times 0.4 \times 0.5$
$\tau = 100\pi \times 10^{-4}$
$\tau = 100 \times 3.14 \times 10^{-4} = 314 \times 10^{-4}$ N.m
Comparing with $\alpha \times 10^{-4}$ N.m, we get $\alpha = 314$.
Answer: $314$
The torque acting on the coil is $\tau = MB \sin\theta = NIAB \sin\theta$.
Substituting the given values:
$\tau = 125 \times 1 \times (\pi \times (0.02)^2) \times 0.4 \times \sin 30^\circ$
$\tau = 125 \times 1 \times 4\pi \times 10^{-4} \times 0.4 \times 0.5$
$\tau = 100\pi \times 10^{-4}$
$\tau = 100 \times 3.14 \times 10^{-4} = 314 \times 10^{-4}$ N.m
Comparing with $\alpha \times 10^{-4}$ N.m, we get $\alpha = 314$.
Answer: $314$
Hence, the correct answer is 314.