Two small identical circular coils marked 1, 2 carry equal currents and are placed with their geometric axes perpendicular to each other as shown in the figure. Derive an expression for the resultant magnetic field at 0.

Solution

Given, two identical circular loops which are placed perpendicular to each other carry equal currents.

Now,

Magnetic field at O due to loop 1.

B_{1} = \frac{μ_{0} {iR}^{2}}{2 {(x^{2} + R^{2})}^{3 / 2}}

, acting towards left.

Magnetic field at O due to loop 2.

B_{2} = \frac{μ_{0} {iR}^{2}}{2 {(x^{2} + R^{2})}^{3 / 2}}

,acting vertically upwards .

where, R is the radius of each loop.

Therefore,
Resultant magnetic field at O will be

B = \sqrt{B_{1}^{2} + B_{2}^{2}} = \sqrt{2 B_{1}}

(∵ B_{1} = B_{2})

= \frac{μ_{0}}{\sqrt{2}} \frac{i R^{2}}{{(x^{2} + R^{2})}^{3 / 2}}

This resultant field acts at an angle of 45° with the axis of loop 1.

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