Question

A straight current element is carrying current $i=100A$, as shown in the figure. The magnitude of magnetic field at point $P$ which is at perpendicular distance $(\sqrt{3}-1)m$ from the current element, if end $A$ and end $B$ of the element subtends angle $30^{\circ }$ and $60^{\circ }$ at point $P$, respectively, as shown, is $n\times 10^{-6}$ tesla. Find the value of $n$.

Answer 1

Answer: 5

$B=\frac{{\mu }_0.i}{4\pi d}\left(\sin {\theta }_1+\sin {\theta }_2\right)$
$=\frac{10^{-7}\times 100}{\sqrt{3}-1}\left[\frac{\sqrt{3}}{2}-\frac{1}{2}\right]$
$=5\times 10^{-6}T$