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(D) The magnetic field at the centre $O$ due to a circular arc of radius $r$ subtending an angle $	heta$ at the centre is given by $B = \frac{\mu_0 I

Vedclass · Accepted Answer

$\frac{3\mu_0 I}{8a} + \frac{\mu_0 I}{8b} \otimes$

Vedclass · Answer

(D) The magnetic field at the centre $O$ due to a circular arc of radius $r$ subtending an angle $	heta$ at the centre is given by $B = \frac{\mu_0 I 	heta}{4\pi r}$.For the inner arc of radius $a$, the angle subtended is $270^\circ$ or $\frac{3\pi}{2}$ radians. The direction of the magnetic field is inwards $(\otimes)$ by the right-hand rule.$B_1 = \frac{\mu_0 I (3\pi/2)}{4\pi a} = \frac{3\mu_0 I}{8a}$ (inwards).For the outer arc of radius $b$, the angle subtended is $90^\circ$ or $\frac{\pi}{2}$ radians. The direction of the magnetic field is also inwards $(\otimes)$ by the right-hand rule.$B_2 = \frac{\mu_0 I (\pi/2)}{4\pi b} = \frac{\mu_0 I}{8b}$ (inwards).Since both fields are in the same direction, the total magnetic field is:$B = B_1 + B_2 = \frac{3\mu_0 I}{8a} + \frac{\mu_0 I}{8b} \otimes$.

The magnetic induction at the centre $O$ is:

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