First slide

Magnetic field due to a current carrying circular ring

Question

A circular current carrying coil has a radius .R. The distance from the centre of the coil, on the axis where the magnetic induction will be 1/8th to its value at the centre of the coil is

Easy

A

$R / \sqrt{3}$
B

$R \sqrt{3}$
C

$2 R \sqrt{3}$
D

$(2 / \sqrt{3}) R$

Solution

The magnetic field on the axis of a current i carrying coil of turns n, radius R and at a distance x from the centre of coil is $B = \frac{μ_{0}}{4 π} \times \frac{2 {πiR}^{2}}{{(R^{2} + x^{2})}^{3 / 2}}$
At centre $B_{centre} = \frac{μ_{0}}{4 π} \times \frac{2 πi}{R} (∵ x = 0)$
In the given problem
$\frac{μ_{0}}{4 π} \times \frac{2 {πiR}^{2}}{{(R^{2} + x^{2})}^{3 / 2}} = \frac{1}{8} [\frac{μ_{0}}{4 π} \times \frac{2 πi}{R}]$
This gives ${(R^{2} + x^{2})}^{3 / 2} = 8 R^{3}$
Solving, we get $x = \sqrt{3} \cdot R$

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