JEE PYQ: Magnetic Effects Of Current Question 16
Question 16 - 2020 (04 Sep 2020 Shift 1)
A small bar magnet placed with its axis at $30°$ with an external field of $0.06$ T experiences a torque of $0.018$ Nm. The minimum work required to rotate it from its stable to unstable equilibrium position is:
(1) $6.4 \times 10^{-2}$ J
(2) $9.2 \times 10^{-3}$ J
(3) $7.2 \times 10^{-2}$ J
(4) $11.7 \times 10^{-3}$ J
Show Answer
Answer: (3) $7.2 \times 10^{-2}$ J
Solution
$\tau = MB\sin\theta$, $M = 0.6$ A m$^2$. $W = 2MB = 2 \times 0.6 \times 0.06 = 7.2 \times 10^{-2}$ J.
BETA
