Question

The magnetic induction at O due to a current in conductor shaped as shown in figure is:

• A. $\frac{{\mu }_{0}i}{4\pi }[\frac{3\pi }{2a}+\frac{\sqrt{2}}{b}]$
• B. $\frac{{\mu }_{0}i}{4\pi }[\frac{3\pi }{4a}-\frac{\sqrt{2}}{b}]$
• C. $\frac{{\mu }_{0}i}{2\pi }[\frac{3\pi }{4a}-\frac{1}{\sqrt{2}b}]$
• D. $\frac{{\mu }_{0}i}{2\pi }[\frac{1}{a}+\frac{1}{b}]$

Answer 1

The correct option is A $\frac{{\mu }_{0}i}{4\pi }[\frac{3\pi }{2a}+\frac{\sqrt{2}}{b}]$

Magnetic Induction due to $HID$ is $\frac{3}{4}$ time the magnetic induction due to complete circular loop

$B_1=\frac{3}{4}\frac{{\mu }_{0}I}{2a}$ downward direction.

Magnetic induction at $0$ due to line $EF$ and $FG$ will be same and equal to $B_2$.

$B_2=2\left\{\frac{{\mu }_{0}I}{4\pi b}\left[\cos \frac{\pi }{4}-\cos 90°\right]\right\}=\frac{{\mu }_{0}I}{4\pi b}\frac{2}{\sqrt{2}}=\frac{{\mu }_{0}I}{4\pi }\frac{\sqrt{2}}{b}$

Now magnetic induction due to hole system is $B$.

$B=B_1+B_2=\frac{3{\mu }_{0}I}{4\times 2a}+\frac{{\mu }_{0}I}{4\pi }\frac{\sqrt{2}}{b}=\frac{{\mu }_{0}I}{4\pi }[\frac{3\pi }{2a}+\frac{\sqrt{2}}{b}]$