Question

a long straight wire carries a current of $10A$ direct along the negative $y-axis$ as shown in the figure. $A$ uniform magnetic field $B_0$ of magnitude ${10}^{-6}$ $T$ is direct parallel to the $x-axis$. What is the resultant magnetic field at the following points?
$\left(a\right)x=0,z=2m;$
$\left(b\right)x=2m,z=0;$

Answer 1

$\left(a\right)x=0,z=2m;$
magnetic field due to wire is:
$=\frac{\mu oi}{2\pi \left(2\right)}$ at $-vex-axis$
$={10}^{-6}\left(-\hat{i}\right)$
And there is a magnetic field at $x$ also
$\therefore$ resultant is $0$.
$b)x=2,z=0$
due to long wire $=\frac{\mu o\hat{i}}{2\pi \left(2\right)}\left(\hat{k}\right)={10}^{-6}k$
$\therefore$ net resultant $={10}^{-6}\hat{1}+{10}^{-6}\hat{k}$