A long wire carrying current I comes in along…

A long wire carrying current I comes in along a straight line, goes round a semicircle of radius R and goes out along a second straight line. The two…

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A long wire carrying current

I

comes in along a straight line, goes round a semicircle of radius

R

and goes out along a second straight line. The two straight segments are semi-infinite, parallel to each other and perpendicular to the diameter of the semicircle; each begins at an end of the diameter and extends away from the semicircle. The magnetic field at the centre of the semicircle is

A. $\dfrac{\mu_0 I}{4R}\left(1-\dfrac{2}{\pi}\right)$
B. $\dfrac{\mu_0 I}{4R}$
C. $\dfrac{\mu_0 I}{4R}\left(1+\dfrac{2}{\pi}\right)$
D. $\dfrac{\mu_0 I}{2R}\left(1+\dfrac{1}{\pi}\right)$

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Answer: (C) $\dfrac{\mu_0 I}{4R}\left(1+\dfrac{2}{\pi}\right)$

The semicircle gives

B_1 = \dfrac{\mu_0 I}{4R}

. Each straight semi-infinite wire ends at the foot of the perpendicular from the centre (distance

R

), so each gives

\dfrac{\mu_0 I}{4\pi R}

. Applying the right-hand rule to the three parts, the current goes up the left wire, clockwise round the arc, and down the right wire, so all three fields point into the plane and add:

B = \dfrac{\mu_0 I}{4R} + 2\cdot\dfrac{\mu_0 I}{4\pi R} = \dfrac{\mu_0 I}{4R}\left(1+\dfrac{2}{\pi}\right)

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