Question

A current loop $ABCD$ is held fixed on the plane of the paper as shown in the figure. The arcs $BC$ (radius = $b$) and $DA$ (radius = $a$) of the loop are joined by two straight wires $AB$ and $CD$. A steady current $I$ is flowing in the loop. Angle made by $AB$ and $CD$ at the origin $O$ is ${30}^o$. Another straight thin wire with steady current $I_1$ flowing out of the plane of the paper is kept at the origin. The magnitude of the magnetic field due to the loop $ABCD$ at the origin $\left(O\right)$ is:

• A. $zero$
• B. $\frac{{\mu }_{0}I(b-a)}{24ab}$
• C. $\frac{{\mu }_{0}I}{4\pi }\left[\frac{b-a}{ab}\right]$
• D. $\frac{{\mu }_{0}I}{4\pi }\left[2\right(b-a)+\frac{\pi }{3}(a+b\left)\right]$.

Answer 1

Answer: (B)

$\frac{{\mu }_{0}I(b-a)}{24ab}$

The net magnetic at the origin O can be given as the sum of the field due to each arm of the loop.

So, the field due to all loop is:

$B=B_{AB}+B_{BC}+B_{CD}+B_{DA}$

$B=0+\frac{{\mu }_{0}I}{4\pi a}\times \frac{\pi }{6}+0-\frac{{\mu }_{0}I}{4\pi b}\times \frac{\pi }{6}$

$B=\frac{{\mu }_{0}I}{24ab}(b-a)$