Question

A loop $ABCD$ has current $I=10A$ , as shown in the figure. $AD$ and $BC$ are circular arcs with centre at $O$ , for both. The magnetic field at point $O$ is

• A. $10^{-4}T$
• B. $10^{-5}T$
• C. $1.5\times 10^{-5}T$
• D. $2\times 10^{-5}T$

Answer 1

Answer: (B)

Magnetic field due to $AB$ and $CD$ are zero at $O$
Magnetic field due to $BC$
$B_{BC}=\frac{{\mu }_0}{4\pi }\frac{10}{3\times 10^{-2}}\times \frac{\pi }{4}$ (outwards)
Simplifying field due to $DA$
$B_{SP}=\frac{{\mu }_0}{4\pi }\frac{10}{5\times 10^{-2}}\times \frac{\pi }{4}$ (inwards)
Net field $B=\frac{{\left(\mu \right)}_o}{4\pi }\frac{10\pi }{4\times {\left(10\right)}^{-2}}\left(\frac{1}{3}-\frac{1}{5}\right)$

$=1.0\times 10^{-5}T$