A magnetic dipole is under the influence of t | vedclass

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(D) Let the magnitude of the first magnetic field be $B_{1} = 1.2 	imes 10^{-2} \; T$.Let the magnitude of the second magnetic field be $B_{2}$.The a

Vedclass · Accepted Answer

$4.39 	imes 10^{-3}\; T$

Vedclass · Answer

(D) Let the magnitude of the first magnetic field be $B_{1} = 1.2 	imes 10^{-2} \; T$.Let the magnitude of the second magnetic field be $B_{2}$.The angle between the two fields is $	heta = 60^{\circ}$.At stable equilibrium,the dipole makes an angle $	heta_{1} = 15^{\circ}$ with the first field $B_{1}$.The angle between the dipole and the second field $B_{2}$ is $	heta_{2} = 	heta - 	heta_{1} = 60^{\circ} - 15^{\circ} = 45^{\circ}$.For the dipole to be in rotational equilibrium,the net torque acting on it must be zero.Therefore,the torque due to $B_{1}$ must balance the torque due to $B_{2}$:$M B_{1} \sin 	heta_{1} = M B_{2} \sin 	heta_{2}$where $M$ is the magnetic moment of the dipole.Canceling $M$ from both sides,we get:$B_{2} = \frac{B_{1} \sin 	heta_{1}}{\sin 	heta_{2}}$Substituting the values:$B_{2} = \frac{1.2 	imes 10^{-2} 	imes \sin 15^{\circ}}{\sin 45^{\circ}}$Using $\sin 15^{\circ} \approx 0.2588$ and $\sin 45^{\circ} \approx 0.7071$:$B_{2} = \frac{1.2 	imes 10^{-2} 	imes 0.2588}{0.7071} \approx 4.39 	imes 10^{-3} \; T$.

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