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

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

Q. If the work done in turning a magnet of magnetic moment

M

by an angle of

90^{o}

from the magnetic meridian is n times the corresponding work done to turn it through an angle of

60^{o}

, then the value of

n

Q. The work done in tuning a magnet of magnetic moment M by angle of

90^{o}

from the meridian, is n times the corresponding work done to turn it through an angle of

60^{o}

. The value of n is given by

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The work done in turning a magnet of magnetic moment ‘M′ by an angle of 90∘ from the meridian is ‘n′ times the corresponding work done to turn it through an angle of 60∘, where ‘n′ is given by

The correct option is B 2W1=MB(cos0∘−cos90∘) =MB(1−0)=MB W2=MB(cos0∘−cos60∘) =MB(1−12)=MB2 ∴W1=2W2⇒n=2

The work done in turning a magnet of magnetic moment $‘ M^{'}$ by an angle of $90^{\circ}$ from the meridian is $‘ n^{'}$ times the corresponding work done to turn it through an angle of $60^{\circ}$ , where $‘ n^{'}$ is given by

The correct option is B $2$
$W_{1} = M B (cos 0^{\circ} - cos 90^{\circ})$
$= M B (1 - 0) = M B$
$W_{2} = M B (cos 0^{\circ} - cos 60^{\circ})$
$= M B (1 - \frac{1}{2}) = \frac{M B}{2}$
$∴ W_{1} = 2 W_{2} \Rightarrow n = 2$