Question

A closed curve encircles several conductors.The line integral $\int \overrightarrow{B}\cdot d\overrightarrow{l}$ around this curve is $3.83\times {10}^{-7}T-m$. What is the net current in the conductors?

• A. $0.1A$
• B. $0.2A$
• C. $0.3A$
• D. $0.4A$

Answer 1

The correct option is C $0.3A$

From Ampere's law, for a closed loop carrying current

$\int \overrightarrow{B}.d\overrightarrow{l}={\mu }_{0}i$

$i=\frac{\int \overrightarrow{B}.d\overrightarrow{l}}{{\mu }_0}$

$=\frac{3.83\times {10}^{-7}}{{\mu }_0}$
$=0.3A$