The magnetic field at the centre C of the cir | vedclass

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(A) The total magnetic field at point $C$ is the vector sum of the magnetic fields due to the three segments: the straight wire $AB$, the quarter-circ

Vedclass · Accepted Answer

$\frac{{\mu _0}i}{{4\pi R}}\left[ {\frac{1}{{\sqrt 2 }} + \frac{\pi }{2} + 1} \right]$

Vedclass · Answer

(A) The total magnetic field at point $C$ is the vector sum of the magnetic fields due to the three segments: the straight wire $AB$, the quarter-circular arc $BC$, and the semi-infinite straight wire starting from $C$.$1$. Magnetic field due to the straight wire $AB$ at distance $R$ from the wire:$B_1 = \frac{\mu_0 i}{4\pi R} (\sin 45^\circ + \sin 0^\circ) = \frac{\mu_0 i}{4\pi R} \left(\frac{1}{\sqrt{2}}ight) \otimes$$2$. Magnetic field due to the quarter-circular arc $BC$ of radius $R$:$B_2 = \frac{\mu_0 i}{4\pi R} 	imes \frac{\pi}{2} = \frac{\mu_0 i}{8R} = \frac{\mu_0 i}{4\pi R} 	imes \frac{\pi}{2} \otimes$$3$. Magnetic field due to the semi-infinite straight wire starting from $C$:$B_3 = \frac{\mu_0 i}{4\pi R} (\sin 90^\circ + \sin 0^\circ) = \frac{\mu_0 i}{4\pi R} \otimes$Since all fields are directed into the plane $(\otimes)$, the net magnetic field is:$B_{	ext{net}} = B_1 + B_2 + B_3 = \frac{\mu_0 i}{4\pi R} \left[ \frac{1}{\sqrt{2}} + \frac{\pi}{2} + 1 ight] \otimes$

The magnetic field at the centre $C$ of the circular arc is:

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