Question

A rectangular loop of metallic wire is of length $a$ and breadth $b$ and carries a current $i$ . The magnetic field at the centre of the loop is

• A. $\frac{{\mu }_{0}i}{4\pi }\frac{8\sqrt{a^2+b^2}}{ab}$
• B. $\frac{{\mu }_{0}i}{4\pi }\frac{4\sqrt{a^2+b^2}}{ab}$
• C. $\frac{{\mu }_{0}i}{4\pi }\frac{2\sqrt{a^2+b^2}}{ab}$
• D. $\frac{{\mu }_{0}i}{4\pi }\frac{\sqrt{a^2+b^2}}{ab}$

Answer 1

Answer: (A)

From figure it is clear that

$B_{AB}=B_{CD}$ $=\frac{{\mu }_{0}i}{4\pi \left(\frac{b}{2}\right)}\left(\sin {\varphi }_1+\sin {\varphi }_2\right)=\frac{{\mu }_0}{4\pi }\cdot \frac{4i}{b}\cdot \frac{a}{\sqrt{a^2+b^2}}$ $B_{BC}=B_{DA}=\frac{{\mu }_0}{4\pi }\cdot \frac{4i}{a}\cdot \frac{b}{\sqrt{a^2+b^2}}$
$$\begin{gathered} \therefore B=B_{AB}+B_{BC}+B_{CD}+B_{DA} \\ B=2\left(\frac{{\mu }_0}{4\pi }\cdot \frac{4i}{b}\cdot \frac{a}{\sqrt{a^2+b^2}}\right)+2\left(\frac{{\mu }_0}{4\pi }\cdot \frac{4i}{a}\cdot \frac{b}{\sqrt{a^2+b^2}}\right) \end{gathered}$$
$B=\frac{{\mu }_0}{4\pi }\cdot \frac{2\times 4i}{\sqrt{a^2+b^2}}\left[\frac{a}{b}+\frac{b}{a}\right]$
$B=\frac{{\mu }_0}{4\pi }\cdot \frac{8i\sqrt{a^2+b^2}}{ab}$