Question

A straight wire carrying a current of $12\, A$ is sent into a semicircular loop of radius $2.0\, cm$. What will be magnetic field at the centre of the semicircular loop?

• A. $B =1.9 \times 10^{-4} \,T$
• B. $B =1.9 \times 10^{4} \,T$
• C. $B =1.9 \times 10^{-4}\, T, \otimes$
• D. $B =1.9 \times 10^{4} \,T, \otimes$

Answer 1

Answer: (C)

$B=\frac{\mu_{0}}{4 \pi} \frac{\pi i}{R}=10^{-7} \times \frac{22 \times 12}{7 \times 2 \times 10^{-2}}$
$=1.9 \times 10^{-4} T$
Thus, $B$ is $1.9 \times 10^{-4}$ T normal to the plane of the paper going into it.