Question

An element $\Delta I=\Delta x\hat{i}$ is placed at the origin and carries a current $I=10A$.

If $\Delta x=1cm$, magnetic field at point $P$ is

• A. $4\times 10^{-8}\hat{k}T$
• B. $4\times 10^{-8}\hat{i}T$
• C. $4\times 10^{-8}\hat{j}T$
• D. $-4\times 10^{-8}\hat{j}T$

Answer 1

Answer: (A)

The magnitude of magnetic field,
$|dB|=\frac{{\mu }_0}{4\pi }\frac{Idl\sin \theta }{r^2}$

i.e., $|dB|=\frac{10^{-7}\times 10\times 10^{-2}}{25\times 10^{-2}}$
$=4\times 10^{-8}T$
As, $dl\times r=\Delta x\hat{i}\times y\hat{j}$
$=y\Delta x(\hat{i}\times \hat{j})=y\Delta x\hat{k}$
So, the direction of the field is in the $+Z$-direction.