In two separate experiments,the neutral point | vedclass

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(D) The magnetic field $B$ at a distance $d$ on the equatorial line (broadside-on position) of a short bar magnet is given by $B = \frac{\mu_0}{4\pi}

Vedclass · Accepted Answer

$1:8$

Vedclass · Answer

(D) The magnetic field $B$ at a distance $d$ on the equatorial line (broadside-on position) of a short bar magnet is given by $B = \frac{\mu_0}{4\pi} \frac{M}{d^3}$.At the neutral point,the magnetic field of the magnet is equal and opposite to the horizontal component of the Earth's magnetic field $(B_H)$.Thus,$\frac{\mu_0}{4\pi} \frac{M_1}{r^3} = B_H$ and $\frac{\mu_0}{4\pi} \frac{M_2}{(2r)^3} = B_H$.Equating the two expressions: $\frac{M_1}{r^3} = \frac{M_2}{8r^3}$.Therefore,the ratio of the magnetic moments is $\frac{M_1}{M_2} = \frac{r^3}{8r^3} = \frac{1}{8}$.

In two separate experiments,the neutral points due to two small magnets are at a distance of $r$ and $2r$ in the broadside-on position. The ratio of their magnetic moments will be:

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