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

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

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

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

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

Solution

Field at the centre, $B_{1} = \frac{μ_{0}}{4 π} \times \frac{2 πin}{r} = \frac{μ_{0}}{2} . \frac{ni}{r}$
Field at a distance h from the centre,
$\begin{array}{l} B_{2} = \frac{μ_{0}}{4 π} . \frac{2 {πnir}^{2}}{{(r^{2} + h^{2})}^{3 / 2}} = \frac{μ_{0}}{2} . \frac{{nir}^{2}}{r^{3} {(1 + \frac{h^{2}}{r^{2}})}^{3 / 2}} \\ = B_{1} {(1 + \frac{h^{2}}{r^{2}})}^{- 3 / 2} = B_{1} (1 - \frac{3}{2} . \frac{h^{2}}{r^{2}}) \end{array}$
(By binomial theorem)
Hence B₂ is less than B₁ by a fraction $= \frac{3}{2} \frac{h^{2}}{r^{2}}$

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