Two magnets are suspended by a given wire one by one in a horizontal uniform magnetic field. In order to deflect the first magnet through $45^{0},$ the wire has to be twisted through $540^{0}$ whereas with the second magnet, the wire requires a twist of $360^{0}$ for the same deflection. Then the magnetic moments of the first and second magnets are in the ratio $n : 14 .$ Find the value of $n$

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answer is 22.

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Detailed Solution

If C is torque per unit angular twist of the wire, then for a twist $ϕ, τ = C ϕ = M H \sin θ$

In the $1^{s t}$ case, $ϕ_{1} = 540^{0} - 45^{0} = 495^{0}, θ_{1} = 45^{0}$

In the $2^{n d}$ case, $ϕ_{2} = 360^{0} - 45^{0} = 315^{0},$

$θ_{2} = 45^{0}$

$∴ C (495^{0}) = M_{1} H \sin 45^{0} ....... (i)$ And $C (315^{0}) = M_{2} H \sin 45^{0} .... (i i)$

Dividing (i) by (ii), we get $\frac{M_{1}}{M_{2}} = \frac{495}{315} = \frac{11}{7} = \frac{22}{14}$

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