A current carrying circular ring, when placed in a uniform magnetic field of 0.25 T with one diameter perpendicular to the field experiences a torque of 0.3 N-m, when the ring is rotated about this diameter through an angle of $90^{o}$ , it experiences a torques of 0.4 N-m. Then magnetic dipole moment of the ring is

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$2 A - m^{2}$

$4 A - m^{2}$

$2 .5 A - m^{2}$

$1 .25 A - m^{2}$

answer is A.

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Detailed Solution

$τ_{1} = MBsinθ$
Where $θ$ is the angle between $\vec{M} and \vec{B} .$
When the ring is rotated, new angle between $\vec{M} and \vec{B}$ is $(90^{0} - θ)$ .
$∴ τ_{2} = MBsin (90^{0} - θ) = MBcosθ$

$∴ τ_{1}^{2} + τ_{2}^{2} = {(MB)}^{2} \Rightarrow M = \frac{\sqrt{τ_{1}^{2} + τ_{2}^{2}}}{B} = \frac{\sqrt{{(0 .3)}^{2} + {(0 .4)}^{2}}}{0 .25} Amp - m^{2}$