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(C) The magnetic field $B$ at a point on the equatorial line (perpendicular to the axis) of a short bar magnet at a distance $r$ from its centre is gi

Vedclass · Accepted Answer

$27:1$

Vedclass · Answer

(C) The magnetic field $B$ at a point on the equatorial line (perpendicular to the axis) of a short bar magnet at a distance $r$ from its centre is given by $B = \frac{\mu_0}{4\pi} \frac{M}{r^3}$,where $M$ is the magnetic dipole moment.Since the points are at large distances,we can use the formula for a short dipole.Thus,$B \propto \frac{1}{r^3}$.Given distances are $r_A = X$ and $r_B = 3X$.The ratio of the magnetic fields is $\frac{B_A}{B_B} = \left( \frac{r_B}{r_A} \right)^3$.Substituting the values,$\frac{B_A}{B_B} = \left( \frac{3X}{X} \right)^3 = (3)^3 = 27$.Therefore,the ratio is $27:1$.

Points $A$ and $B$ are situated perpendicular to the axis of a $2 \ cm$ long 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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