A magnet is in stable equilibrium in a uniform magnetic field. It is deflected by $60^{o}$ and the workdone is equal to $W$ . In deflecting the magnet further by $30^{o}$ , Work done is:

\frac{W \sqrt{3}}{2}

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\frac{W}{4}

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\frac{W}{2}

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W

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Solution

The correct option is D $W$
$U = m . B$
$U_{1} = - m B$
$U_{2} = - m B cos 60^{o}$
$= \frac{- m B}{2}$
$U_{3} = - m B cos 90^{o}$
$= 0$
$W = U_{2} - U_{1}$
$= \frac{m B}{2}$
$W^{1} = U_{3} - U_{2}$
$= \frac{m B}{2}$
$W^{1} = W$

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A magnet is in stable equilibrium in a uniform magnetic field. It is deflected by 60o and the workdone is equal to W. In deflecting the magnet further by 30o, Work done is:

The correct option is D WU=m.BU1=−mBU2=−mBcos60o=−mB2U3=−mBcos90o=0W=U2−U1=mB2W1=U3−U2=mB2W1=W

A magnet is in stable equilibrium in a uniform magnetic field. It is deflected by $60^{o}$ and the workdone is equal to $W$ . In deflecting the magnet further by $30^{o}$ , Work done is:

The correct option is D $W$
$U = m . B$
$U_{1} = - m B$
$U_{2} = - m B cos 60^{o}$
$= \frac{- m B}{2}$
$U_{3} = - m B cos 90^{o}$
$= 0$
$W = U_{2} - U_{1}$
$= \frac{m B}{2}$
$W^{1} = U_{3} - U_{2}$
$= \frac{m B}{2}$
$W^{1} = W$