A current I flows in an infinitely long wire | vedclass

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Question

If we cut a thin pipe along its length then we get a conductor as given in question.

consider a piece of small length $dl$ then current through thi

Vedclass · Accepted Answer

$\frac{{{\mu _0}I}}{{{\pi ^2}R}}$

Vedclass · Answer

If we cut a thin pipe along its length then we get a conductor as given in question.

consider a piece of small length $dl$ then current through this part is

$\mathrm{i}=\frac{\mathrm{I}}{\pi \mathrm{R}} \mathrm{d} \ell=\frac{\mathrm{I}}{\pi \mathrm{R}} \cdot \mathrm{R} \mathrm{d} 	heta=\frac{\mathrm{I}}{\pi} \mathrm{d} 	heta$

and magnetic field at $O$ due to this part of width $dl$ is $d B=\frac{\mu_{0} i}{2 \pi R}=\frac{\mu_{0}}{2 \pi R} \cdot \frac{I}{\pi} d 	heta=\frac{\mu_{0} I}{2 \pi^{2} R} d 	heta$

similarly take element $dl$ on opposite side

so magnetic field at $O$ due to these two elements $=2 \,d B\, \cos 	heta$

$	herefore B=2 \int_{0}^{\pi / 2} d B \cos 	heta=2 \int_{0}^{\pi / 2} \frac{\mu_{0} I}{2 \pi^{2} R} \cos 	heta d 	heta$

$=\frac{\mu_{0} I}{\pi^{2} R}(\sin 	heta)_{0}^{\pi / 2} \Rightarrow B=\frac{\mu_{0} I}{\pi^{2} R}$

A current $I$ flows in an infinitely long wire with cross-section in the form of a semicircular ring of radius $R$ . The magnitude of the magnetic induction along its axis is

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