Physics - Magnetism Question with Solution | TestHub

The magnetic field at the center of a circular coil of radius $R$, having $N$ turns and carrying current $I$ is given by:

$B=\frac{{\mu }_{0}NI}{2R}$

Since both coils are identical and carry the same current, the magnitude of the magnetic field produced by each coil at their common center is the same:

$B_1=B_2=\frac{{\mu }_{0}NI}{2R}$

From the figure, the two coils are placed perpendicular to each other. Therefore, the magnetic fields produced by them at the center are also perpendicular to each other (i.e., the angle between ${\vec{B}}_1$ and ${\vec{B}}_2$ is $90^{\circ }$).

The resultant magnetic field $B_{net}$ at the center is the vector sum of ${\vec{B}}_1$ and ${\vec{B}}_2$:

$B_{net}=\sqrt{B_1^2+B_2^2}$

Substituting the values of $B_1$ and $B_2$:

$B_{net}=\sqrt{{\left(\frac{{\mu }_{0}NI}{2R}\right)}^2+{\left(\frac{{\mu }_{0}NI}{2R}\right)}^2}$

$B_{net}=\sqrt{2}\left(\frac{{\mu }_{0}NI}{2R}\right)$

$B_{net}=\frac{{\mu }_{0}NI}{\sqrt{2}R}$

Answer: $\frac{{\mu }_0\mathrm{NI}}{\sqrt{2}\mathrm{R}}$