First slide

Magnetic field due to a current carrying circular ring

Question

The 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 field at the centre by the fraction

Difficult

A

$\frac{3}{2} \cdot \frac{h^{2}}{r^{2}}$
B

$\frac{2}{3} \cdot \frac{h^{2}}{r^{2}}$
C

$\frac{3}{2} \cdot \frac{r^{2}}{h^{2}}$
D

$\frac{2}{3} \cdot \frac{r^{2}}{h^{2}}$

Solution

The magnetic field on the axis of a current i carrying coil of turns n, radius r and at a distance i from the centre of the coil
$B = \frac{μ_{0}}{4 π} \times \frac{2 {πir}^{2}}{{(r^{2} + h^{2})}^{3 / 2}}$ …………………(1)
The field at the centre is given by
$B_{centre} = \frac{μ_{0}}{4 π} \times \frac{2 πi}{r}$ …………………….(2)
$(∵ At centre h = 0)$
$\frac{B}{B_{centre}} = \frac{r^{3}}{{(r^{2} + h^{2})}^{3 / 2}} = \frac{r^{3}}{r^{3} {[1 + \frac{h^{2}}{r^{2}}]}^{3 / 2}}$
$= \frac{1}{(1 + \frac{3}{2} \frac{h^{2}}{r^{2}})} or B (1 + \frac{3}{2} \frac{h^{2}}{r^{2}}) = B_{centre}$
$∴ (B_{centre} - B) / B = \frac{3}{2} \frac{h^{2}}{r^{2}}$

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