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Question

(A) The magnetic field at point $P$ is the vector sum of the magnetic fields produced by the two straight wire segments and the semicircular arc.$1$.

Vedclass · Accepted Answer

$\frac{\mu_{0} I }{4 \pi r }(2+\pi)$

Vedclass · Answer

(A) The magnetic field at point $P$ is the vector sum of the magnetic fields produced by the two straight wire segments and the semicircular arc.$1$. For each semi-infinite straight wire,the magnetic field at distance $r$ is $B_{	ext{straight}} = \frac{\mu_{0} I}{4 \pi r}$. Since both wires carry current in directions that produce magnetic fields in the same direction at point $P$,the total field due to both straight wires is $B_{1} = 2 	imes \frac{\mu_{0} I}{4 \pi r} = \frac{\mu_{0} I}{2 \pi r}$.$2$. For the semicircular arc of radius $r$,the magnetic field at its center is $B_{	ext{arc}} = \frac{1}{2} 	imes \frac{\mu_{0} I}{2 r} = \frac{\mu_{0} I}{4 r}$.$3$. The total magnetic field is $B = B_{1} + B_{	ext{arc}} = \frac{\mu_{0} I}{2 \pi r} + \frac{\mu_{0} I}{4 r}$.$4$. Factoring out $\frac{\mu_{0} I}{4 \pi r}$,we get $B = \frac{\mu_{0} I}{4 \pi r} (2 + \pi)$.

$A$ hairpin-like shape as shown in the figure is made by bending a long current-carrying wire. What is the magnitude of the magnetic field at point $P$,which lies at the center of the semicircle?

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