TS EAMCET · Physics · Magnetic Effects of Current
A bar magnet of magnetic moment \(M\) and moment of inertia \(I\) is freely suspended such that the magnetic axial line is in the direction of magnetic meridian. If the magnet is displaced by a very small angle \((\theta)\), the angular acceleration is (Magnetic induction of earth's horizontal field \(=B_H\) )
- A \(\frac{M B_H \theta}{I}\)
- B \(\frac{I B_H \theta}{M}\)
- C \(\frac{M \theta}{I B_H}\)
- D \(\frac{I \theta}{M B_H}\)
Answer & Solution
Correct Answer
(A) \(\frac{M B_H \theta}{I}\)
Step-by-step Solution
Detailed explanation
When magnet is displaced by a very small angle \(\theta\), then restoring couple acting on the magnet is \(\tau=-M B_H \sin \theta\) Negative sign shows the restoring nature of torque. Now since \(\tau=I \alpha\) and \(\sin \theta \approx \theta\) for small angular displacement…
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