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A current I flows in an infinitely long wire with cross section in the form of a semicircular ring of radius $\mathrm{R}$. The magnitude of the magnetic induction along its axis is
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Verified Answer
The correct answer is:
$\frac{\mu_0 l}{\pi^2 R}$
$\frac{\mu_0 l}{\pi^2 R}$
$\mathrm{d} \overrightarrow{\mathrm{B}}=\frac{\mu_0 \mathrm{di}}{2 \pi \mathrm{R}}[-\cos \theta \hat{\mathbf{i}}-\sin \theta \hat{\mathbf{j}}]$
$\mathrm{di}=\frac{\mathrm{T}}{\pi \mathrm{R}} \mathrm{Rd} \theta$
$=\frac{\mathrm{I}}{\pi} \mathrm{d} \theta$
$\mathrm{d} \overrightarrow{\mathrm{B}}=\frac{\mu_0 \mathrm{I}}{2 \pi^2 \mathrm{R}}(-\cos \theta \hat{\mathrm{i}}-\sin \theta \hat{\mathrm{j}})$
$\vec{B}=-\frac{\mu_0 I}{\pi^2 R} \hat{j}$
$\mathrm{di}=\frac{\mathrm{T}}{\pi \mathrm{R}} \mathrm{Rd} \theta$
$=\frac{\mathrm{I}}{\pi} \mathrm{d} \theta$
$\mathrm{d} \overrightarrow{\mathrm{B}}=\frac{\mu_0 \mathrm{I}}{2 \pi^2 \mathrm{R}}(-\cos \theta \hat{\mathrm{i}}-\sin \theta \hat{\mathrm{j}})$
$\vec{B}=-\frac{\mu_0 I}{\pi^2 R} \hat{j}$
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