Question

The magnetic induction at the center O (fig.) is

• A. $\frac{{\mu }_{0}I}{2a}+\frac{{\mu }_{0}I}{2b}⨂$
• B. $\frac{3{\mu }_{0}I}{8a}+\frac{{\mu }_{0}I}{8b}⨂$
• C. $\frac{3{\mu }_{0}I}{8a}-\frac{{\mu }_{0}I}{8b}⨂$
• D. $\frac{3{\mu }_{0}I}{8a}+\frac{{\mu }_{0}I}{8b}⨀$

Answer 1

The correct option is B $\frac{3{\mu }_{0}I}{8a}+\frac{{\mu }_{0}I}{8b}⨂$

Given: Three-quarter of the circular loop with radius $a$ and the quarter portion of a circular loop of radius $b$ carrying current $I$.

Solution :

We know that the magnetic field at the center due to the circular loop is given by

$B=\frac{{\mu }_{0}I}{2r}$, where r is the radius of loop.

Let's call the magnetic field due to three-quarter of the circular loop of radius a $B_1$.

Let's call the magnetic field due to one-quarter of the circular loop of radius b $B_2$.

So, $B_1=\frac{3}{4}\cdot \frac{{\mu }_{0}I}{2a}$ downwards

Also, $B_2=\frac{1}{4}\cdot \frac{{\mu }_{0}I}{2b}$ downwards

So the net magnetic field B is given by

$B=\frac{3{\mu }_{0}I}{8a}+\frac{{\mu }_{0}I}{8b}$ $⨂$

The Correct Opt : [B}