Question

An element $dl=dxi=(wheredx=1cm)$ is placed at origin and carries a current $10A$. What is the magnetic field on y- axis at a dist. of $0.5m$?

• A. $2\times {10}^{-8}kT$
• B. $4\times {10}^{-8}kT$
• C. $-2\times {10}^{-8}kT$
• D. $-4\times {10}^{-8}kT$

Answer 1

The correct option is B $4\times {10}^{-8}kT$

$dB=\frac{{\mu }_0}{4\pi }\frac{\pm dL\sin \theta }{r^2}$

$=\frac{4\pi \times {10}^{-7}}{4\pi }\times \frac{10\times 1\times {10}^{-2}\times \sin 90}{0.5\times 0.5}T$

( The axis of element and y-axis are perpendicular to each other )

$=4\times {10}^{-8}T$

$\Rightarrow \overrightarrow{dB}=\frac{\mu oI}{4\pi }\times \frac{\overrightarrow{dL}\times \overrightarrow{r}}{r^3}$

$\Rightarrow \overrightarrow{dI}\times \overrightarrow{r}=dx\hat{i}\times y\hat{j}$

$=ydx(\hat{i}\times \hat{j})$

$=ydx\hat{u}$

Therefore the direction of $\overrightarrow{dB}$ is in the +z direction.

Hence, the answer is $4\times {10}^{-8}T.$