Question

Two similar coils of radius $R$ are lying concentrically with their planes at right angles to each other. The currents flowing in them are $2I$ and $4I$ respectively. Let the resultant magnetic field induction at the centre be found to be $N$ times $\frac{\mu _{0} I}{R}$ . What is the value of $100\,N$ ? (Take $\sqrt{5}=2.24$ ).

Answer 1

Magnetic field induction due to vertical loop at centre $O$ is, $\vec{B}_{1}=\frac{\mu _{0} 2 I}{2 R}\hat{i}$ . Magnetic field due to horizontal loop at centre $O$ is, $\vec{B}_{2}=\frac{\mu _{0} 4 I}{2 R}\hat{j}$
Now, $B_{n e t}=\sqrt{B_{1}^{2} + B_{2}^{2}}$
$\Rightarrow B_{n e t}=\sqrt{\left(\frac{\left(\mu \right)_{0} 2 I}{2 R}\right)^{2} + \left(\frac{\left(\mu \right)_{0} 4 I}{2 R}\right)^{2}}$
$=\frac{\mu_0 I}{2 R} \sqrt{(2)^2+(4)^2}$
$=\frac{\sqrt{20}}{2}\frac{\mu _{0} I}{R}$
$\therefore B_{n e t}=\frac{\sqrt{5} \mu _{0} I}{R}=2.24\frac{\mu _{0} I}{R}$