Q. The magnetic field normal to the plane of a wire of $n$ turns and radius r which carries a current $i$ is measured on the axis of the coil at a small distance $h$ from the centre of the coil. This is smaller than the magnetic field at the centre by the fraction
A) $(2/3)r^{2}/h^{2}$
B) $(3/2)r^{2}/h^{2}$
C) $(2/3)h^{2}/r^{2}$
D) $(3/2)h^{2}/r^{2}$
Explanation
(d) $B_{1} = \frac{\mu_{1}}{4\pi}\frac{2\pi ni}{r}\ \ \ \ \ \ \ and\ B_{2} = \frac{\mu_{0}}{4\pi}\ \frac{2\pi nir^{2}}{\left( r^{2} + h^{2} \right)^{3/2}}$
So, $\frac{B_{2}}{B_{1}} = \left( 1 + \frac{h^{2}}{r^{2}} \right)^{- 3/2}$
Fractional decrease in the magnetic field will be
$= \ \frac{B_{1} - B_{2}}{B_{1}} = \left( 1 - \frac{B_{2}}{B_{1}} \right)\ $
$= \left\lbrack 1 - \left( 1 + \frac{h^{2}}{r^{2}} \right)^{- 3/2} \right\rbrack$
$= 1 - \left( 1 - \frac{3}{2}\ \frac{h^{2}}{r^{2}} \right) = \frac{3}{2}\ \frac{h^{2}}{r^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.