A current I flows in an infinitely long wire | vedclass

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(A) Consider a cross-section of the wire as a semicircular arc of radius $R$. The total current $I$ is distributed over the arc length $\pi R$.The cur

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

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(A) Consider a cross-section of the wire as a semicircular arc of radius $R$. The total current $I$ is distributed over the arc length $\pi R$.The current per unit arc length is $\lambda = \frac{I}{\pi R}$.Consider a small element of arc length $dl = R d	heta$ at an angle $	heta$ from the axis. The current in this element is $di = \lambda dl = \frac{I}{\pi R} (R d	heta) = \frac{I}{\pi} d	heta$.The magnetic field $dB$ at the center $O$ due to this infinite wire element is given by the formula for an infinite wire: $dB = \frac{\mu_0 di}{2\pi R} = \frac{\mu_0}{2\pi R} \left( \frac{I}{\pi} d	heta ight) = \frac{\mu_0 I}{2\pi^2 R} d	heta$.Due to symmetry,the components of the magnetic field perpendicular to the axis cancel out. The net magnetic field is the integral of the components along the axis: $B = \int_{-\pi/2}^{\pi/2} dB \cos	heta = 2 \int_{0}^{\pi/2} \frac{\mu_0 I}{2\pi^2 R} \cos	heta d	heta$.$B = \frac{\mu_0 I}{\pi^2 R} \int_{0}^{\pi/2} \cos	heta d	heta = \frac{\mu_0 I}{\pi^2 R} [\sin	heta]_{0}^{\pi/2} = \frac{\mu_0 I}{\pi^2 R}$.

$A$ 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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