Question

A rectangular loop of conductor of length $a$ and breadth $b$ carrying current $I$ is shown in the figure. The magnetic field at the centre $O$ of the loop is

• A. $\frac{{\mu }_{0}I(a+b)}{2\pi \sqrt{a^2+b^2}}$
• B. $\frac{{\mu }_{0}Iab}{2\pi \sqrt{a^2+b^2}}$
• C. $\frac{{\mu }_{0}I(a+b)}{\pi \sqrt{a^2+b^2}}$
• D. $\frac{2{\mu }_{0}I\sqrt{a^2+b^2}}{\pi ab}$

Answer 1

Answer: (D)

The total magnetic field at $O$ due to current $I$ through rectangular loop is $B$.
Let $B_1,B_2,B_3$, and $B_4$ be the magnetic fields at $O$ due to current in arms $1,2,3$ and 4 respectively.
$\therefore B=B_1+B_2+B_3+B_4$
where $\left(B_1\right)=\left(B_3\right)$
$B_1=B_3=\frac{{\mu }_{0}I}{4\pi (a/2)}\left[\sin \left(90^{\circ }-\alpha \right)+\sin \left(90^{\circ }-\alpha \right)\right]$
$=\frac{{\mu }_{0}I}{4\pi (a/2)}(\cos \alpha +\cos \alpha )=\frac{{\mu }_{0}I\cos \alpha }{\pi a}$
$\left(B_2\right)=\left(B_4\right)=\frac{{\mu }_{0}I}{4\pi (b/2)}\left[\sin \left(90^{\circ }-\beta \right)+\sin \left(90^{\circ }-\beta \right)\right]$
$=\frac{{\mu }_{0}I}{4\pi (b/2)}(\cos \beta +\cos \beta )=\frac{{\mu }_{0}I\cos \beta }{\pi b}$
As $B_1,B_2,B_3$, and $B_4$ are acting in the same direction i.e. perpendicular to the plane of rectangular loop downwards.
$\therefore B=B_1+B_2+B_3+B_4$
or $B=\frac{2{\mu }_{0}I}{\pi }\left[\frac{\cos \alpha }{a}+\frac{\cos \beta }{b}\right]$
Putting $\cos \alpha =\frac{b}{\sqrt{a^2+b^2}}$ and $\cos \beta =\frac{a}{\sqrt{a^2+b^2}}$
we get, $B=\frac{2{\mu }_{0}I\sqrt{a^2+b^2}}{\pi ab}$