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

Answer 1

Answer: (A)

From figure it is clear that

$B_{A B}=B_{C D} $ $=\frac{\mu_0 i}{4 \pi\left(\frac{b}{2}\right)}\left(\sin \phi_1+\sin \phi_2\right)=\frac{\mu_0}{4 \pi} \cdot \frac{4 i}{b} \cdot \frac{a}{\sqrt{a^2+b^2}} $ $ B_{B C}=B_{D A}=\frac{\mu_0}{4 \pi} \cdot \frac{4 i}{a} \cdot \frac{b}{\sqrt{a^2+b^2}} $
$ \therefore B=B_{A B}+B_{B C}+B_{C D}+B_{D A} \\ B=2\left(\frac{\mu_0}{4 \pi} \cdot \frac{4 i}{b} \cdot \frac{a}{\sqrt{a^2+b^2}}\right)+2\left(\frac{\mu_0}{4 \pi} \cdot \frac{4 i}{a} \cdot \frac{b}{\sqrt{a^2+b^2}}\right) $
$ B=\frac{\mu_0}{4 \pi} \cdot \frac{2 \times 4 i}{\sqrt{a^2+b^2}}\left[\frac{a}{b}+\frac{b}{a}\right] $
$ B=\frac{\mu_0}{4 \pi} \cdot \frac{8 i \sqrt{a^2+b^2}}{a b}$