Question

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

• A. $\frac{{\mu }_{0}Ni{R}^2}{2{\left(R^2+x^2\right)}^{3/2}}$
• B. $\frac{{\mu }_{0}Ni}{2R}$
• 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 $\delta B\sin \varphi$ along the axis of the coil and $\delta B\cos \varphi$ at right angles. The resultant field at $P$ is divided along the axis and its magnitude is given by
$B=\int dB\sin \varphi$
For current $i$ flowing in the coil of radius $r$.
$B=\frac{{\mu }_0}{4\pi }\frac{i}{r^2}\int dl\sin \varphi$
From figure, $\sin \varphi =\frac{R}{r}$

$B=\frac{{\mu }_0}{4\pi }\frac{iR}{r^3}dl$
Also, $\int dl=2\pi R$ 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
$B=\frac{{\mu }_{0}Ni{R}^2}{2{\left(R^2+x^2\right)}^{3/2}}$