Two identical circular wires P and Q each of radius R and carrying current I are kept in perpendicular planes such that they have a common center as shown in figure. Find the magnitude and direction of the net magnetic field at the common center of the two coils
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{\mu_0 I}{\sqrt{2}R},\ \beta = 45^\circ \)
The magnetic field at the centre of a circular loop acts along its axis, perpendicular to the plane of the loop.
\( \displaystyle B = \frac{\mu_0 I}{2R} \)
Both loops have the same radius and current, so each gives the same field \( B \) at the common centre. The loops lie in perpendicular planes, so their axes, and the two fields, are at right angles. The net field is their vector sum:
\( \displaystyle B_{\mathrm{net}} = \sqrt{B^{2} + B^{2}} = \sqrt{2}\,\frac{\mu_0 I}{2R} = \frac{\mu_0 I}{\sqrt{2}\,R} \)
Two equal perpendicular fields give a resultant midway between them:
\( \displaystyle \tan\beta = \frac{B}{B} = 1 \;\Rightarrow\; \beta = 45^\circ \)
So the net field makes \( 45^\circ \) with the field of each coil, that is, with the axis of each loop.