A bar magnet of 10 ,cm length is kept with it | vedclass

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(D) The length of the magnet $2l = 10 \,cm$, so $l = 5 \,cm = 0.05 \,m$.The distance from each pole to the neutral point $P$ is $d = 15 \,cm = 0.15 \,

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$1.35$

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(D) The length of the magnet $2l = 10 \,cm$, so $l = 5 \,cm = 0.05 \,m$.The distance from each pole to the neutral point $P$ is $d = 15 \,cm = 0.15 \,m$.The distance of the neutral point from the center of the magnet $O$ is $r = \sqrt{d^2 - l^2} = \sqrt{15^2 - 5^2} = \sqrt{200} \,cm = 10\sqrt{2} \,cm = 0.1414 \,m$.The magnetic field at the neutral point due to the magnet is equal to the horizontal component of Earth's magnetic field $B_H = 0.4 \,Gauss = 0.4 	imes 10^{-4} \,T$.The magnetic field at a point on the equatorial line of a bar magnet is given by $B = \frac{\mu_0}{4\pi} \frac{M}{(r^2 + l^2)^{3/2}}$, where $M$ is the magnetic moment.Here, $r^2 + l^2 = d^2 = (0.15)^2 = 0.0225 \,m^2$.So, $0.4 	imes 10^{-4} = 10^{-7} 	imes \frac{M}{(0.0225)^{3/2}}$.$M = \frac{0.4 	imes 10^{-4} 	imes (0.0225)^{3/2}}{10^{-7}} = 0.4 	imes 10^3 	imes (0.15)^3 = 400 	imes 0.003375 = 1.35 \,A-m^2$.The pole strength $m$ is given by $M = m 	imes (2l)$.$m = \frac{M}{2l} = \frac{1.35 \,A-m^2}{0.10 \,m} = 13.5 \,A-m$.

$A$ bar magnet of $10 \,cm$ length is kept with its north $(N)$-pole pointing north. $A$ neutral point is formed at a distance of $15 \,cm$ from each pole. Given the horizontal component of Earth's magnetic field is $0.4 \,Gauss$, the pole strength of the magnet is: (in $\,A-m$)

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