The magnetic field at the centre O due to the | vedclass

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Question

(C) The total magnetic field $B$ at the centre $O$ is the vector sum of the magnetic fields due to the circular arc and the two straight wire segments

Vedclass · Accepted Answer

$\frac{\mu_0 I}{4 R}\left[\frac{3}{2}+\frac{1}{\pi}\right] \otimes$

Vedclass · Answer

(C) The total magnetic field $B$ at the centre $O$ is the vector sum of the magnetic fields due to the circular arc and the two straight wire segments.$1$. For the circular arc of radius $R$ subtending an angle $	heta = \frac{3\pi}{2}$ at the centre,the magnetic field is $B_1 = \frac{\mu_0 I}{4\pi R} \cdot 	heta = \frac{\mu_0 I}{4\pi R} \cdot \frac{3\pi}{2} = \frac{3\mu_0 I}{8R}$. Using the right-hand rule,the direction is into the page $(\otimes)$.$2$. For the two semi-infinite straight wires,each contributes a magnetic field at distance $R$ from the wire. The field due to one semi-infinite wire is $B_2 = \frac{\mu_0 I}{4\pi R}$. Since there are two such wires,the total field from them is $B_{straight} = 2 	imes \frac{\mu_0 I}{4\pi R} = \frac{\mu_0 I}{2\pi R}$. Both wires produce a field into the page $(\otimes)$.$3$. Adding these,$B = B_1 + B_{straight} = \frac{3\mu_0 I}{8R} + \frac{\mu_0 I}{2\pi R} = \frac{\mu_0 I}{4R} \left[ \frac{3}{2} + \frac{1}{\pi} ight]$.The direction is into the page $(\otimes)$.

The magnetic field at the centre $O$ due to the given structure is:

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