Answer to Question #168423 in Atomic and Nuclear Physics for Elizabethv

Question

The line integral of the magnetic field on a closed path surrounding a wire has the value 10.0μT·m. What is the current in the wire?

Answer 1

According to Ampere's law,

$\oint B.dl=\mu _0 I$

The current passing through the wire is

$I= \frac{\oint B.dl}{ \mu _0} = \frac{10.0 \times 10^{-6}T·m}{4 \pi \times 10^{-7} H/m}=7.96 A$