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 $60^{o}$ from the meridian, where $n$ is given by :

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Toppr Admin · Accepted Answer

Work done in 1st case- W1-MB-cos00-x2212-cos900-Work done in 2nd case- W2-MB-cos00-x2212-cos600-W1-2W2therefore- n-2-

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 $60^{o}$ from the meridian, where $n$ is given by :

$2$
$\frac{1}{2}$
$\frac{1}{4}$
$1$

Work done in 1st case, $W_{1} = M B (c o s 0^{0} - c o s 90^{0})$

Work done in 2nd case, $W_{2} = M B (c o s 0^{0} - c o s 60^{0})$

$W_{1} = 2 W_{2}$

therefore, $n = 2.$

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 from the meridian, where n is given by :212141

Work done in 1st case, W1=MB(cos00−cos900)Work done in 2nd case, W2=MB(cos00−cos600)W1=2W2therefore, n=2.

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 $60^{o}$ from the meridian, where $n$ is given by :

$2$
$\frac{1}{2}$
$\frac{1}{4}$
$1$

Work done in 1st case, $W_{1} = M B (c o s 0^{0} - c o s 90^{0})$

Work done in 2nd case, $W_{2} = M B (c o s 0^{0} - c o s 60^{0})$

$W_{1} = 2 W_{2}$

therefore, $n = 2.$