Question

Two identical loops P and Q each of radius 5 cm are lying in perpendicular planes such that they have a common centre as shown in the figure. Find the magnitude and direction of the net magnetic field at the common centre of the two coils, if they carry currents equal to 3 A and 4 A respectively.

Answer 1

$B_1=\frac{{\mu }_0}{2R}.I_1{B}_1=\frac{{\mu }_0}{2}\times \frac{3}{5\times {10}^{-2}}$
Similarly, $B_2=\frac{{\mu }_0}{2}\times \frac{4}{5\times {10}^{-2}}$
Resulatant magnetic field is ,
$B=\sqrt{B_1^2+B_2^2}B=\sqrt{{(\frac{{\mu }_0}{2}\times \frac{3}{5\times {10}^{-2}})}^2+{(\frac{{\mu }_0}{2}\times \frac{4}{5\times {10}^{-2}})}^2}B=\frac{{\mu }_0}{2\times 5\times {10}^{-2}}\times 5B=\frac{{\mu }_0}{2}\times 100=50{\mu }_0=50\times 4\pi \times {10}^{-7}=62.83\times {10}^{-6}Ttan\theta =\frac{B_1}{B_2}tan\theta =\frac{3}{4}\theta =ta{n}^{-1}\left(\frac{3}{4}\right)$
Thus direction of magnetic field makes an angle $\theta =ta{n}^{-1}\left(\frac{3}{4}\right)$ with the vertical.