Q. An element $d\ \overrightarrow{I} = dx\widehat{i}\ (where\ dx = 1$ cm) is placed at the origin and carries a large current $i = 10$ A. What is the magnetic field on the y-axis at a distance f 0.5 m?
A) $2 \times 10^{- 8}\widehat{k}\ $T
B) $4 \times 10^{- 8}\widehat{k}\ $T
C) $- 2 \times 10^{- 8}\widehat{k}\ $T
D) $- 4 \times 10^{- 8}\widehat{k}\ $T
Explanation
(b) Here, $dl = dx = 1\ cm = 10^{- 2}m;$
$i = 10Am,\ \ r = 0.5\ m$
$d\overrightarrow{B}\ = \frac{\mu_{0}}{4\pi}\ \frac{(d\ \overrightarrow{I\ } \times \overrightarrow{r})}{r^{3}}$
$= \frac{\mu_{0}}{4\pi}\ \frac{i\ dl}{r^{2}}\left( \widehat{i} \times \widehat{j} \right) = \ \frac{\mu_{0}}{4\pi}\ \frac{i\ dl}{r^{2}}\ \widehat{k}$
$= \ \frac{10^{- 7} \times 10 \times 10^{- 2}\sin{90{^\circ}}}{(0.5)^{2}}\widehat{k}$
$= 4\ \times 10^{- 8}\ \widehat{k}\ $T.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.