The work done in moving a dipole from its most stable to most unstable position in a $0.09$ $T$ uniform magnetic field is?(dipole moment of this dipole $= 0.5$ $A$ $m^{2}$ ).

0.07 J

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0.08 J

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0.09 J

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0.1 J

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Solution

The correct option is C $0.09 J$
Since the most stable position is at $θ = 0$ and the most unstable position is at $θ = 180^{o}$ , then the work done is given by
$W = \int_{θ = 0^{o}}^{θ = 180^{o}} τ θ d θ = \int_{0^{o}}^{180^{o}} M B sin θ d θ = - M B [cos θ]_{0}^{180^{o}}$

$= - M B [cos 180^{o} - cos 0^{o}] = - M B [- 1 - 1]$
$= - M B [- 2] = 2$ MB
Here, $M = 0.5 A m^{2}$ and $B = 0.09 T$
$W = 2 \times 0.50 \times 0.09 = 0.09$ J

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The work done in moving a dipole from its most stable to most unstable position in a 0.09T uniform magnetic field is?(dipole moment of this dipole =0.5Am2).

The work done in moving a dipole from its most stable to most unstable position in a $0.09$ $T$ uniform magnetic field is?(dipole moment of this dipole $= 0.5$ $A$ $m^{2}$ ).