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Work Energy Theorem Application

The work done...

Question

The work done in turning a magnet of magnetic moment $M$ by an angle of $90^{o}$ from the meridian is $n$ times the corresponding work done to turn it through an angle of $60^{o}$ . Where $n$ is given by

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Solution

The correct option is A $2$
The work done in turning a magnet of magnetic moment M by an angle of 90⁰ from the meridian is

$w_{1} = M B (cos θ_{1} - cos θ_{2})$

$w_{1} = M B (cos 0 - cos 90)$

$w_{1} = M B . . . . . . . (1)$

Similarly, The work done in turning a magnet of magnetic moment M by an angle of 60⁰ from the meridian is

$w_{2} = M B (cos 0 - cos 60)$
$w_{2} = \frac{M B}{2} . . . . . . . . (2)$

It is given that
$w_{1} = n w_{2}$

$M B = n \frac{M B}{2}$

$n = 2$

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The work done in turning a magnet of magnetic moment M by an angle of 90o from the meridian is n times the corresponding work done to turn it through an angle of 60o. Where n is given by

The work done in turning a magnet of magnetic moment $M$ by an angle of $90^{o}$ from the meridian is $n$ times the corresponding work done to turn it through an angle of $60^{o}$ . Where $n$ is given by