A current I flows in an infinitely long wire | vedclass

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Question

If we cu a thin pipe along its length then we get a conductor as given in quation.

consider a piece of small length $d\ell$ then current through th

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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