A bar magnet of magnetic moment \(1.5~\text{J/T}\) lies aligned with the direction of a uniform magnetic field of \(0.22~\text{T}\). What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment normal to the field direction?
1. \(0.66\) J
2. \(0.33\) J
3. \(0\)
4. \(0.44\) J

Subtopic:  Analogy between Electrostatics & Magnetostatics |
 78%
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A short bar magnet has a magnetic moment of \(0.48~\text{JT}^{-1}\). The direction and magnitude of the magnetic field produced by the magnet at a distance of \(10~\text{cm}\) from the centre of the magnet on the equatorial lines (normal bisector) of the magnet are:

1. \(0.38~\mathrm{G}\) along the \(\text{N-S}\) direction.
2. \(0.48~\mathrm{G}\) along the \(\text{N-S}\) direction.
3. \(0.38~\mathrm{G}\) along the \(\text{S-N}\) direction.
4. \(0.48~\mathrm{G}\) along the \(\text{S-N}\) direction.
Subtopic:  Analogy between Electrostatics & Magnetostatics |
 69%
From NCERT
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NEET 2023 - Target Batch - Aryan Raj Singh
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NEET 2023 - Target Batch - Aryan Raj Singh