A current I flows in an infinitely long wire | vedclass

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(A) Consider an infinitely long wire with a semicircular cross-section of radius $R$. The total current $I$ is distributed over the semicircular arc o

Vedclass · Accepted Answer

$\frac{\mu_0 I}{\pi^2 R}$

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(A) Consider an infinitely long wire with a semicircular cross-section of radius $R$. The total current $I$ is distributed over the semicircular arc of length $\pi R$.The linear current density is $\lambda = \frac{I}{\pi R}$.Consider a small element of arc length $d\ell = R d	heta$ at an angle $	heta$ from the axis of symmetry. The current in this element is $di = \lambda d\ell = \frac{I}{\pi R} (R d	heta) = \frac{I}{\pi} d	heta$.This element acts as an infinitely long wire carrying current $di$. The magnetic field $dB$ at the center $O$ due to this element is $dB = \frac{\mu_0 di}{2\pi R} = \frac{\mu_0}{2\pi R} \left( \frac{I}{\pi} d	heta ight) = \frac{\mu_0 I}{2\pi^2 R} d	heta$.By symmetry,the components of the magnetic field perpendicular to the axis of symmetry cancel out,while the components parallel to the axis add up. The component of $dB$ along the axis is $dB_x = dB \cos 	heta$.Integrating from $	heta = -\pi/2$ to $\pi/2$,or $2 \int_{0}^{\pi/2} dB \cos 	heta$:$B = 2 \int_{0}^{\pi/2} \frac{\mu_0 I}{2\pi^2 R} \cos 	heta d	heta = \frac{\mu_0 I}{\pi^2 R} [\sin 	heta]_{0}^{\pi/2} = \frac{\mu_0 I}{\pi^2 R} (1 - 0) = \frac{\mu_0 I}{\pi^2 R}$.

$A$ current $I$ flows in an infinitely long wire with a 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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