CUET · PHYSICS · PYQ PAPER 2023
A short bar magnet placed with its axis at 30\(^\circ\) with an external field of 800G, experiences a torque of 0.016 Nm. Work done in moving it from its most stable to most unstable position is :
- A 5.321 J
- B 1.843 J
- C 0.89 J
- D 0.064 J
Answer & Solution
Correct Answer
(D) 0.064 J
Step-by-step Solution
Detailed explanation
\(\tau = mB\sin\theta\) \(0.016 = m \times (800 \times 10^{-4}) \times \sin(30^\circ)\) \(0.016 = m \times 0.08 \times 0.5\) \(m = \frac{0.016}{0.04} = 0.4 Am^2\) \(W = U_{unstable} - U_{stable} = (-mB\cos(180^\circ)) - (-mB\cos(0^\circ))\) \(W = mB - (-mB) = 2mB\)…
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