Question

Two identical coils P and Q each of radius R are lying in perpendicular planes such that they have a common centre. Find the magnitude and direction of the magnetic field at the common centre of the two coils, if they carry currents equal to I and \[\sqrt{3}\] I respectively.

Answer 1

Magnetic field at the centre of the coils due to coil P, having current I is 

\[B_P = \frac{\mu_0 I}{2R}\]

And magnetic field due to coil Q having current 

\[\sqrt{3}I\] is \[B_Q = \frac{\mu_0 \sqrt{3}I}{2R}\] 

Since both coils are inclined to each other at an angle of 90°, the magnitude of their resultant magnetic field at the common centre will be

\[B = \sqrt{{B_P}^2 + {B_Q}^2} = \frac{\mu_0 I}{2R}\sqrt{1 + 3} = \frac{\mu_0 I}{R}\]

The directions of B_P and B_Q are as indicated in the figure. The direction of the resultant field is at an angle θ given by

\[\theta = \tan^{- 1} \left( \frac{B_P}{B_Q} \right) = \tan^{- 1} \left( \frac{1}{\sqrt{3}} \right) = 30°\]

Hence, the direction of the magnetic field will be at an angle 30° to the plane of loop P.