Question

A long straight wire carries a current of $10A$ directed along the negative y-axis as shown in the figure. A uniform magnetic field $B_0$ of magnitude ${10}^{-6}T$ is directed parallel to the x-axis. What is the resultant magnetic field at the following points,
(a) $x=0,z=2m$
(b) $x=2m,z=0$
(c) $x=0,z=-0.5m$

Answer 1

Magnetic field B due to current $i=\frac{{\mu }_{o}i}{2\pi d}$

d= distance from line to point.

(a) x=0, Z=2m, d=2m.

$B_1=\frac{{\mu }_{o}i}{2\pi \times 2}$ along -x direction.

$B_1=\frac{4\pi \times {10}^{-7}\times 10}{4\pi }={10}^{-6}\overrightarrow{l}{B}_0={10}^{-6}\overrightarrow{l}$

Resultant field $B=B_1+B_0=0$

(b) $Q(x=2m,z=0)B_1=\frac{4\pi \times {10}^{-7}\times 10}{4\pi }\left(\hat{k}\right)B_0={10}^{-6}\hat{k}{B}_0={10}^{-6}\hat{l}\overrightarrow{B}=\overrightarrow{B_1}+\overrightarrow{B_0}={10}^{-6}(\hat{l}+\hat{k})\left|\overrightarrow{B}\right|=\sqrt{2}\times {10}^{-6}T$

(c) $R=(x=0,z=-0.5m)B_1=\frac{{\mu }_{o}i}{2\pi d}(d=o.5m)B_1=\frac{4\pi \times {10}^{-7}\times 10}{2\pi \times 0.5}={4\times 10}^{-6}\hat{l}{B}_0={10}^{-6}\hat{l}\overrightarrow{B}=\overrightarrow{B_0}+\overrightarrow{B_1}=({10}^{-6}+{4\times 10}^{-6})\hat{l}={5\times 10}^{-6}\hat{l}$