AP EAMCET · PHYSICS · Magnetic Effects of Current
A magnetic dipole of moment \(2.5 \mathrm{Am}^2\) is free to rotate about a vertical axis passing through its centre. It is released from East-West direction. Its kinetic energy at the moment, it takes North-South position is \(\left(B_H=3 \times 10^{-5} \mathrm{~T}\right)\)
- A \(50 \mu \mathrm{J}\)
- B \(100 \mu \mathrm{J}\)
- C \(175 \mu \mathrm{J}\)
- D \(75 \mu \mathrm{J}\)
Answer & Solution
Correct Answer
(D) \(75 \mu \mathrm{J}\)
Step-by-step Solution
Detailed explanation
When magnetic dipole is released from \(\mathrm{E}-\mathrm{W}\) a torque acts on it. So, in the displacement from E-W to N-S work is done by the torque.…
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