Magnetic moment of bar magnet is $M$ . The work done in turning the magnet by $90^{o}$ in direction of magnetic field $B$ will be

Zero

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\frac{1}{2}

M B

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3

M B

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M B

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Solution

The correct option is B $M B$
Potential energy of a magnetic moment at an angle $θ$ with the magnetic field = $- M B cos θ$
Thus work done in rotating from angle $θ_{1}$ to $θ_{2} = M B (c o s θ_{1} - c o s θ_{2})$
$= M B (c o s 0^{\circ} - c o s 90^{\circ})$
$= M B (1 - 0)$
$= M B$

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Magnetic moment of bar magnet is M. The work done in turning the magnet by 90o in direction of magnetic field B will be

The correct option is B MBPotential energy of a magnetic moment at an angle θ with the magnetic field = −MBcosθThus work done in rotating from angle θ1 to θ2=MB(cosθ1−cosθ2)=MB(cos0∘−cos90∘)=MB(1−0)=MB

Magnetic moment of bar magnet is $M$ . The work done in turning the magnet by $90^{o}$ in direction of magnetic field $B$ will be

The correct option is B $M B$
Potential energy of a magnetic moment at an angle $θ$ with the magnetic field = $- M B cos θ$
Thus work done in rotating from angle $θ_{1}$ to $θ_{2} = M B (c o s θ_{1} - c o s θ_{2})$
$= M B (c o s 0^{\circ} - c o s 90^{\circ})$
$= M B (1 - 0)$
$= M B$