Question

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

If $\Delta x=1\, cm$, 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,
$|d B |=\frac{\mu_{0}}{4 \pi} \frac{I d l \sin \theta}{r^{2}}$

i.e., $|d B |=\frac{10^{-7} \times 10 \times 10^{-2}}{25 \times 10^{-2}}$
$=4 \times 10^{-8} T$
As, $ d l \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.