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 $100N$ ? (Take $\sqrt{5}=2.24$ ).

Answer 1

Answer: 224

Magnetic field induction due to vertical loop at centre $O$ is, ${\vec{B}}_1=\frac{{\mu }_{0}2I}{2R}\hat{i}$ . Magnetic field due to horizontal loop at centre $O$ is, ${\vec{B}}_2=\frac{{\mu }_{0}4I}{2R}\hat{j}$
Now, $B_{net}=\sqrt{B_1^2+B_2^2}$
$\Rightarrow B_{net}=\sqrt{{\left(\frac{{\left(\mu \right)}_{0}2I}{2R}\right)}^2+{\left(\frac{{\left(\mu \right)}_{0}4I}{2R}\right)}^2}$
$=\frac{{\mu }_{0}I}{2R}\sqrt{(2{)}^2+(4{)}^2}$
$=\frac{\sqrt{20}}{2}\frac{{\mu }_{0}I}{R}$
$\therefore B_{net}=\frac{\sqrt{5}{\mu }_{0}I}{R}=2.24\frac{{\mu }_{0}I}{R}$