Question

A coil having $ N $ turns carry a current as shown in the figure. The magnetic field intensity at point $ P $ is:

• A. $\frac{\mu_{0} N i R^{2}}{2\left(R^{2}+x^{2}\right)^{3 / 2}}$
• B. $ \frac{{{\mu }_{0}}Ni}{2\,R} $
• C. $ \frac{{{\mu }_{0}}Ni{{R}^{2}}}{{{\left( R+x \right)}^{2}}} $
• D. Zero

Answer 1

Answer: (A)

The resultant magnetic field at $P$ is found only along the axis.
Magnetic field being a vector quantity can be resolved into two components, one of magnitude $\delta B \sin \phi$ along the axis of the coil and $\delta B \cos \phi$ at right angles. The resultant field at $P$ is divided along the axis and its magnitude is given by
$B=\int d B \sin \phi$
For current $i$ flowing in the coil of radius $r$.
$B=\frac{\mu_{0}}{4 \pi} \frac{i}{r^{2}} \int d l \sin \phi$
From figure, $\sin \phi=\frac{R}{r}$

$B=\frac{\mu_{0}}{4 \pi} \frac{i R}{r^{3}} d l$
Also, $\int d l=2 \pi R$ and $r=\left(R^{2}+x^{2}\right)^{1 / 2}$
$\therefore B=\frac{\mu_{0}}{4 \pi} \frac{2 \pi i R^{2}}{\left(R^{2}+x^{2}\right)^{3 / 2}}=\frac{\mu_{0} i R^{2}}{2\left(R^{2}+x^{2}\right)^{3 / 2}}$
Since, coil has $N$ turns, therefore
$B=\frac{\mu_{0} N i R^{2}}{2\left(R^{2}+x^{2}\right)^{3 / 2}}$