COMEDK · Physics · 21. Magnetic Effects of Current
A bar magnet is suspended in a uniform magnetic field of field strength 0.36 T . If the magnetic moment of the bar magnet is \(4 \mathrm{Am}^2\) then the work done in rotating the magnet from its most stable position to its most unstable position is:
- A 18 J
- B 0.72 J
- C 2.88 J
- D 0.045 J
Answer & Solution
Correct Answer
(C) 2.88 J
Step-by-step Solution
Detailed explanation
The potential energy \(U\) of a magnetic dipole with magnetic moment \(\vec{M}\) in a uniform magnetic field \(\vec{B}\) is given by \(U = -\vec{M} \cdot \vec{B} = -MB \cos \theta\), where \(\theta\) is the angle between \(\vec{M}\) and \(\vec{B}\). The most stable position…
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