A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in the equilibrium state. The energy required to rotate it by 60o is W. Now the torque required to keep the magnet in this new position is:

1.  $\frac{W}{\sqrt{3}}$

2.  $\sqrt{3}W$

3.  $\frac{\sqrt{3}W}{2}$

4.  $\frac{2W}{\sqrt{3}}$

Subtopic:  Analogy between Electrostatics & Magnetostatics |
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A magnetic needle suspended parallel to a magnetic field requires $\sqrt{3}$ J of work to turn it through $60°$. The torque needed to maintain the needle in this position will be:

1.  3 J

2.

3.  $\frac{3}{2}J$

4.

Subtopic:  Analogy between Electrostatics & Magnetostatics |
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A short bar magnet of magnet moment 0.4 ${\mathrm{JT}}^{-1}$ is placed in a uniform magnetic field of 0.16 $\mathrm{T}$. The magnet is in stable equilibrium when the potential energy is :
1. 0.064 J
2. – 0.064 J
3. Zero
4. – 0.082 J

Subtopic:  Analogy between Electrostatics & Magnetostatics |
To view explanation, please take trial in the course below.
NEET 2023 - Target Batch - Aryan Raj Singh
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NEET 2023 - Target Batch - Aryan Raj Singh
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