Question

Current $i=2.5\, A$ flows along the circular coil whose equation is given by $x^{2}+y^{2}=9 (x$ and $y$ are in $cm )$. If the magnetic field is given by $\pi x \times 10^{-9} T$ at point $(0,0,4)\, cm$, then find the value of $x$.

Answer 1

Magnetic field on the axis of circular loop,
$B=\left(\frac{\mu_{0}}{4 \pi}\right) \times \frac{2 \pi I R^{2}}{\left(R^{2}+z^{2}\right)^{3 / 2}}$
where, $R=$ Radius of loop $=3 \times 10^{-2} m$
$\Rightarrow B =10^{-7} \times \frac{2 \pi \times 2.5 \times 3^{2} \times 10^{-4}}{125 \times 10^{-6}} \hat{k} $
$=\left(\frac{9 \pi}{25} \times 10^{-5} T \right) \hat{k}=\left(36 \pi \times 10^{-7} T \right) \hat{k}$