A bar magnet of length 14 , cm is placed in t | vedclass

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(C) The length of the magnet $2l = 14 \, cm$,so $l = 7 \, cm = 0.07 \, m$.The distance of the neutral point from the center is $d = 18 \, cm = 0.18 \,

Vedclass · Accepted Answer

$2.880 \, J \, T^{-1}$

Vedclass · Answer

(C) The length of the magnet $2l = 14 \, cm$,so $l = 7 \, cm = 0.07 \, m$.The distance of the neutral point from the center is $d = 18 \, cm = 0.18 \, m$.The distance $r$ from each pole to the neutral point is $r = \sqrt{d^2 + l^2} = \sqrt{18^2 + 7^2} = \sqrt{324 + 49} = \sqrt{373} \, cm$.The magnetic field due to one pole at the neutral point is $B_0 = \frac{\mu_0}{4\pi} \frac{m}{r^2}$.At the neutral point,the horizontal component of the Earth's magnetic field $B_H$ is balanced by the resultant magnetic field of the two poles: $B_H = 2 B_0 \sin 	heta$,where $\sin 	heta = \frac{l}{r}$.Substituting the values: $B_H = 2 \left( \frac{\mu_0}{4\pi} \frac{m}{r^2} ight) \left( \frac{l}{r} ight) = \frac{\mu_0}{4\pi} \frac{2ml}{r^3} = \frac{\mu_0}{4\pi} \frac{M}{r^3}$,where $M = m(2l)$ is the magnetic moment.$0.4 	imes 10^{-4} = 10^{-7} 	imes \frac{M}{(r 	imes 10^{-2})^3} = 10^{-7} 	imes \frac{M}{(373 	imes 10^{-4})^{3/2}}$.$M = \frac{0.4 	imes 10^{-4} 	imes (373 	imes 10^{-4})^{3/2}}{10^{-7}} = 0.4 	imes 10^3 	imes (373)^{3/2} 	imes 10^{-6} = 0.4 	imes 10^{-3} 	imes 7203.82 \approx 2.88 \, J \, T^{-1}$.

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