Question

The magnitude of the magnetic field at $O$ (centre of the circular part) due to the current-carrying coil as shown is:

• A. $\frac{\left(\mu \right)_{0} i}{4 \pi }\left(\frac{3 \pi }{a} + \frac{\sqrt{2}}{b}\right)$
• B. $\frac{\left(\mu \right)_{0} i}{2 \pi }\left(\frac{3 \pi }{2 a} + \frac{\sqrt{2}}{b}\right)$
• C. $\frac{\left(\mu \right)_{0} i}{2 \pi }\left(\frac{\pi }{3 a} + \frac{3}{\sqrt{2} b}\right)$
• D. $\frac{\left(\mu \right)_{0} i}{4 \pi }\left(\frac{3 \pi }{2 a} + \frac{\sqrt{2}}{b}\right)$

Answer 1

Answer: (D)

Magnetic field due to the circular segment $=\frac{3}{4}.\frac{\mu _{0} i}{2 a}$
Due to one straight wire segment $=\frac{\left(\mu \right)_{0} i}{4 \pi b}\left(\right. \, sin \, 45^\circ + \, sin \, 0^\circ \left.\right)=\frac{\left(\mu \right)_{0} i}{4 \sqrt{2} \pi b}$
Net field
$=\frac{3 \left(\mu \right)_{0} i}{8 a}+2 \, \times \frac{\left(\mu \right)_{0} i}{4 \sqrt{2} \pi b}=\frac{\left(\mu \right)_{0} i}{4 \pi }\left(\frac{3 \pi }{2 a} + \frac{\sqrt{2}}{b}\right)$