First slide

Magnetic field due to a current carrying circular ring

Question

Two circular coils X and Y have equal number of turns and carry equal currents in the same sense and subtend same solid angle at point O. The smaller coil X is midway between O and Y. If we represent the magnetic induction due to bigger coil Y at O as B_Y and that due to smaller coil
X at O as B_X, then

Moderate

A

$\frac{B_{Y}}{B_{X}} = 1$
B

$\frac{B_{Y}}{B_{X}} = 2$
C

$\frac{B_{Y}}{B_{X}} = \frac{1}{2}$
D

$\frac{B_{Y}}{B_{X}} = \frac{1}{4}$

Solution

As two coils subtend the same solid angle at O, hence Area of coil Y = 4 x Area of coil X
$[Solid angle = \frac{area}{{(⊥ distance)}^{2}}]$
i. e., Radius of coil Y = 2 x Radius of coil X
$\begin{array}{l} ∴ B_{y} = \frac{μ_{0}}{4 π} \times \frac{2 π I {(2 r)}^{2}}{{[{(2 r)}^{2} + (d^{2})]}^{3 / 2}} \\ B_{x} = \frac{μ_{0}}{4 π} \times \frac{2 π {(r)}^{2}}{{[{(r)}^{2} + {(\frac{d}{2})}^{2}]}^{3 / 2}} \end{array}$
$\begin{array}{l} ∴ \frac{B_{y}}{B_{x}} = \frac{4}{{(4 r^{2} + d^{2})}^{3 / 2}} \times {[\frac{4 r^{2} + d^{2}}{4}]}^{3 / 2} \\ = \frac{4}{{(4)}^{3 / 2}} = \frac{4}{8} = \frac{1}{2} \end{array}$

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