Question

$A$ long straight wire in the horizontal plane carries a current of $75A$ in north to south direction, magnitude and direction of field $B$ at a point $3m$ 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
$\oint \vec{B}\cdot \overrightarrow{dl}={\mu }_0{I}_{enc}$
$B\times 2\pi R={\mu }_0{I}_{enc}$
$B=\frac{{\mu }_0{I}_{enc}}{2\pi R}=2\times 10^{-7}\times \frac{75}{3}$
$=5\times 10^{-6}T$
The direction of field at the given point will be vertical up determined by the screw rule or right hand rule