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(C) The magnetic field $B$ at a point on the equatorial line (perpendicular to the axis) of a small bar magnet of magnetic moment $M$ at a distance $r

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$27: 1$

Vedclass · Answer

(C) The magnetic field $B$ at a point on the equatorial line (perpendicular to the axis) of a small bar magnet of magnetic moment $M$ at a distance $r$ is given by $B = \frac{\mu_0}{4 \pi} \frac{M}{r^3}$.For point $A$ at distance $r_A = x$,the magnetic field is $B_A = \frac{\mu_0}{4 \pi} \frac{M}{x^3}$.For point $B$ at distance $r_B = 3x$,the magnetic field is $B_B = \frac{\mu_0}{4 \pi} \frac{M}{(3x)^3} = \frac{\mu_0}{4 \pi} \frac{M}{27x^3}$.The ratio of the magnetic fields at $A$ and $B$ is $\frac{B_A}{B_B} = \frac{\frac{\mu_0}{4 \pi} \frac{M}{x^3}}{\frac{\mu_0}{4 \pi} \frac{M}{27x^3}} = 27$.Therefore,the ratio is $27: 1$.

Points $A$ and $B$ are situated perpendicular to the axis of a small bar magnet at large distances $x$ and $3x$ from its centre on opposite sides. The ratio of the magnetic fields at $A$ and $B$ will be approximately equal to

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