A circular conducting loop of radius R carrie | vedclass

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(B) Consider a small element $d\ell = R d	heta$ on the loop at an angle $	heta$ from the straight wire.The magnetic field $B$ due to the straight wi

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$\mu_0 I^2$

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(B) Consider a small element $d\ell = R d	heta$ on the loop at an angle $	heta$ from the straight wire.The magnetic field $B$ due to the straight wire at a distance $r = R \sin 	heta$ is $B = \frac{\mu_0 I}{2\pi r} = \frac{\mu_0 I}{2\pi R \sin 	heta}$.The force on the element $d\ell$ is $dF = I (d\ell) B = I (R d	heta) \left( \frac{\mu_0 I}{2\pi R \sin 	heta} ight) = \frac{\mu_0 I^2}{2\pi \sin 	heta} d	heta$.Due to symmetry,the components of force perpendicular to the wire cancel out,and the components parallel to the wire add up.The net force on the element is $dF_{res} = 2(dF) \sin 	heta = 2 \left( \frac{\mu_0 I^2}{2\pi \sin 	heta} d	heta ight) \sin 	heta = \frac{\mu_0 I^2}{\pi} d	heta$.Integrating from $	heta = 0$ to $	heta = \pi$,the total force is $F_{res} = \int_0^{\pi} \frac{\mu_0 I^2}{\pi} d	heta = \frac{\mu_0 I^2}{\pi} [	heta]_0^{\pi} = \mu_0 I^2$.

$A$ circular conducting loop of radius $R$ carries a current $I.$ Another straight infinite conductor carrying current $I$ passes through the diameter of this loop as shown in the figure. The magnitude of force exerted by the straight conductor on the loop is

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