Question

$A$ long straight wire in the horizontal plane carries a current of $75 \,A$ in north to south direction, magnitude and direction of field $B$ at a point $3 \,m$ east of the wire is

• A. $4 \times 10^{-6}\, T$, vertical up
• B. $5 \times 10^{-6}\, T$, vertical down
• C. $5 \times 10^{-6}\, T$, vertical up
• D. $4 \times 10^{-6}\, T$, vertical down

Answer 1

Answer: (C)

From Ampere circuital law
$\oint \, \vec{B} \cdot\overrightarrow{dl}=\mu_{0}I_{enc}$
$B\times2\pi R=\mu_{0}I_{enc}$
$B=\frac{\mu_{0}I_{enc}}{2\pi R}=2\times10^{-7}\times\frac{75}{3}$
$=5\times10^{-6}\,T$
The direction of field at the given point will be vertical up determined by the screw rule or right hand rule