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 $BC$
$B_{B C}=\frac{\mu _{0}}{4 \pi }\frac{10}{3 \times 10^{- 2}}\times \frac{\pi }{4}$ (outwards)
Simplifying field due to $DA$
$B_{S P}=\frac{\mu _{0}}{4 \pi }\frac{10}{5 \times 10^{- 2}}\times \frac{\pi }{4}$ (inwards)
Net field $B=\frac{\left(\mu \right)_{o}}{4 \pi }\frac{10 \pi }{4 \times \left(10\right)^{- 2}}\left(\frac{1}{3} - \frac{1}{5}\right)$

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