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: Â Analogy between Electrostatics & Magnetostatics |
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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: Â Analogy between Electrostatics & Magnetostatics |
Â 69%
From NCERT
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NEET 2023 - Target Batch - Aryan Raj Singh
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