The magnetic moment of a bar magnet is 0.5 . It is suspended in a uniform magnetic field of T. The work done in rotating it from its most stable to most unstable position is

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Solution

Q.
The magnetic moment of a bar magnet is 0.5 $A m^{2}$ . It is suspended in a uniform magnetic field of $8 \times 10^{- 2}$ T. The work done in rotating it from its most stable to most unstable position is [2024]

1 $16 \times 10^{- 2} J$

2 zero

3 $8 \times 10^{- 2} J$

4 $4 \times 10^{- 2} J$

Ans.
(3)

$At stable equilibrium, U = - M B \cos 0 ° = - M B$

$At unstable equilibrium, U = - M B \cos 180 ° = + M B$

$∆ U = 2 M B = 2 \times 0.5 \times 8 \times 10^{- 2}$

$Work done, W = Δ U = 2 M B$

$W = 8 \times 10^{- 2} J$

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