A circular conducting loop of radius R carrie | vedclass

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Question

$\mathrm{d} \mathrm{F}_{\mathrm{res}}=2(\mathrm{dF}) \sin 	heta$

$=2 \mathrm{I}(\mathrm{d} \ell) \mathrm{B} \sin 	heta$

$=2 \mathrm{I}(\mathrm

Vedclass · Accepted Answer

$\mu_0 I^2$

Vedclass · Answer

$\mathrm{d} \mathrm{F}_{\mathrm{res}}=2(\mathrm{dF}) \sin 	heta$

$=2 \mathrm{I}(\mathrm{d} \ell) \mathrm{B} \sin 	heta$

$=2 \mathrm{I}(\mathrm{R} \mathrm{d} 	heta)\left(\frac{\mu_{0} \mathrm{I}}{2 \pi r}ight) \sin 	heta$

$=\frac{\mu_{0} I^{2} R}{\pi} \frac{\sin 	heta d 	heta}{(r)}$

$=\frac{\mu_{0} I^{2} R}{\pi} \frac{\sin 	heta}{(R \sin 	heta)} d 	heta=\frac{\mu_{0} I^{2}}{\pi}(d 	heta)$

$\mathrm{F}_{\mathrm{res}}=\int \mathrm{dF}_{\mathrm{res}}=\frac{\mu_{0} \mathrm{I}^{2}}{\pi} \int_{0}^{\pi} \mathrm{d} 	heta=\mu_{0} \mathrm{I}^{2}$

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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