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(D) The magnetic field $B$ produced by an infinitely long wire at a distance $r$ is $B = \frac{\mu_0 I}{2\pi r}$.For a small current loop with magneti

Vedclass · Accepted Answer

$F \propto \left( \frac{a}{d} \right)^2$

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(D) The magnetic field $B$ produced by an infinitely long wire at a distance $r$ is $B = \frac{\mu_0 I}{2\pi r}$.For a small current loop with magnetic moment $M = I_L \pi a^2$ (where $I_L$ is the loop current) placed in a non-uniform magnetic field,the force is given by $F = 
abla (M \cdot B)$.Since the magnetic field $B$ from the wire varies as $1/r$,the gradient of the field varies as $1/r^2$.Thus,the force $F$ on the loop due to the wire (or vice-versa) is proportional to $M 	imes (	ext{gradient of } B)$.$F \propto M 	imes \frac{1}{d^2} \propto (I_L a^2) 	imes \frac{1}{d^2}$.Given the options,we look for the dependence on $a$ and $d$. The force is proportional to $a^2/d^2$,which is equivalent to $(a/d)^2$.

An infinitely long current-carrying wire and a small current-carrying loop are in the plane of the paper as shown. The radius of the loop is $a$ and the distance of its centre from the wire is $d$ $(d >> a)$. If the loop applies a force $F$ on the wire,then:

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