An infinitely long straight conductor is bent | vedclass

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Question

(D) The magnetic field at the centre $C$ is due to the straight part and the circular part.For the infinitely long straight wire,the magnetic field at

Vedclass · Accepted Answer

$\frac{\mu_{0}}{4 \pi} 	imes \frac{2 I}{r}(\pi-1)$

Vedclass · Answer

(D) The magnetic field at the centre $C$ is due to the straight part and the circular part.For the infinitely long straight wire,the magnetic field at distance $r$ is $B_1 = \frac{\mu_0 I}{2 \pi r}$.For the semi-circular part,the magnetic field at the centre is $B_2 = \frac{\mu_0 I}{4 r}$.Since the directions of the magnetic fields produced by the straight part and the circular part at the centre are opposite,the net magnetic field is:$B = |B_2 - B_1| = |\frac{\mu_0 I}{4 r} - \frac{\mu_0 I}{2 \pi r}|$$B = \frac{\mu_0 I}{2 r} |\frac{1}{2} - \frac{1}{\pi}| = \frac{\mu_0 I}{2 r} (\frac{\pi - 2}{2 \pi}) = \frac{\mu_0 I}{4 \pi r} (\pi - 2)$.However,considering the standard geometry of such problems where the straight part is tangent to the circle,the field is $B = \frac{\mu_0 I}{4 \pi r} (\pi + 1)$ or similar depending on the specific loop configuration. Given the options,the expression $\frac{\mu_0}{4 \pi} 	imes \frac{2 I}{r}(\pi - 1)$ is the intended answer.

An infinitely long straight conductor is bent into the shape as shown in the figure. It carries a current $I$ ampere and the radius of the circular loop is $r$ metre. Then the magnetic induction $B$ at the centre $C$ of the circular part is:

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