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(B) For a short bar magnet of magnetic moment $M$,the magnetic intensity at an axial point (end-on position) at distance $d$ is given by $H_{axial} =

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$H_1:H_2 = 1:2$

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(B) For a short bar magnet of magnetic moment $M$,the magnetic intensity at an axial point (end-on position) at distance $d$ is given by $H_{axial} = \frac{2M}{4\pi d^3}$.For the same magnet,the magnetic intensity at an equatorial point (broad side-on position) at the same distance $d$ is given by $H_{equatorial} = \frac{M}{4\pi d^3}$.Given $OP_1 = OP_2 = d = 10 \ m$,where $P_1$ is at the equatorial position and $P_2$ is at the axial position.Therefore,$H_1 = H_{equatorial} = \frac{M}{4\pi d^3}$ and $H_2 = H_{axial} = \frac{2M}{4\pi d^3}$.The ratio is $\frac{H_1}{H_2} = \frac{M/4\pi d^3}{2M/4\pi d^3} = \frac{1}{2}$.Thus,the ratio $H_1:H_2 = 1:2$.

$A$ bar magnet having centre $O$ has a length of $4 \ cm$. Point $P_1$ is in the broad side-on (equatorial) position and $P_2$ is in the end side-on (axial) position with $OP_1 = OP_2 = 10 \ m$. The ratio of magnetic intensities $H$ at $P_1$ and $P_2$ is:

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