A short bar magnet placed in a horizontal pla | vedclass

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(B) At the null point on the axis,the magnetic field of the bar magnet $(B_{axis})$ is equal and opposite to the Earth's horizontal magnetic field $(B

Vedclass · Accepted Answer

$0.4$

Vedclass · Answer

(B) At the null point on the axis,the magnetic field of the bar magnet $(B_{axis})$ is equal and opposite to the Earth's horizontal magnetic field $(B)$.$B_{axis} = \frac{\mu_0}{4 \pi} \frac{2M}{d^3} = B$On the equatorial line (normal bisector) at the same distance $d$,the magnetic field of the bar magnet $(B_{eq})$ is given by:$B_{eq} = \frac{\mu_0}{4 \pi} \frac{M}{d^3} = \frac{B_{axis}}{2} = \frac{B}{2}$Since the angle of dip is $0^{\circ}$,the Earth's magnetic field is entirely horizontal. On the equatorial line,the magnetic field of the magnet is in the same direction as the Earth's magnetic field.Therefore,the total magnetic field $(B_{total})$ at this point is:$B_{total} = B_{eq} + B = \frac{B}{2} + B = \frac{3B}{2}$Given that $B_{total} = 0.6 \ G$,we have:$0.6 = \frac{3B}{2}$$B = \frac{0.6 	imes 2}{3} = 0.4 \ G$

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