Two concentric circular loops of wire lie on a tabletop, one inside the other. The inner wire has a diameter of 20.0 cm and carries a clockwise current of 12.0 A, as viewed from above, and the outer wire has a diameter of 30.0 cm. What must be the magnitude and direction (as viewed from above) of the current in the outer wire so that the net magnetic field due to this combination of wires is zero at the common centre of the wires?

When current I is passing through the conductor, magnetic field will be produced around that conductor. If conductor is shaped as circular loop with radius r, magnitude of the magnetic field in the center of that loop is equal to $\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{I}}{\mathbf{2}\mathbf{r}}\mathbf{\text{where}}{\mathit{\mu }}_{\mathbf{0}}\mathbf{=}\mathbf{4}\mathit{\pi }\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{H}\mathbf{/}\mathbf{m}$is permeability of free space.

Magnetic field is produced by the current carried by the inner loop and magnetic field is produced by the current carried by the outer loop. The magnitude of magnetic field produced by inner loop be $B_1$and the magnitude of magnetic field produced by outer loop be$B_2$. The magnitude of magnetic field in the center of inner loop with radius $R_1$that carries current$l_1$by substituting $r=R_1$and $l=l_1$we get,$B_1=\frac{{\mu }_0{l}_1}{2R_1}$ and the magnitude of magnetic field in the center of inner loop with radius$R_2$ that carries current $l_2$by substituting$r=R_2$ and $l=l_2$we get$B_2=\frac{{\mu }_0{l}_2}{2R_2}$

In order to have zero magnetic field in the center of concentric loops, magnitudes of magnetic fields$B_1$ and$B_2$must be equal in the center of concentric loops, but with opposite directions:$B_1=B_2$

$\frac{{\mu }_0{l}_1}{2R_1}=\frac{{\mu }_0{l}_2}{2R_2}$

From equation above, multiply both sides of equation by$\frac{2{\mathrm{R}}_2}{{\mu }_0}$ we get,

$\begin{array}{c}\frac{{\mu }_0{\mathrm{l}}_1}{2{\mathrm{R}}_1}\times \frac{2{\mathrm{R}}_2}{{\mu }_0}=\frac{{\mu }_0{\mathrm{I}}_2}{2{\mathrm{R}}_2}\times \frac{2{\mathrm{R}}_2}{{\mu }_0}\\ \frac{{\mathrm{I}}_1}{{\mathrm{R}}_1}\times {\mathrm{R}}_2={\mathrm{I}}_2\end{array}$

Substitute the values we get,

role="math" localid="1668249949373" $\begin{array}{r}{I}_2=\frac{12\mathrm{A}\times 15\mathrm{cm}}{10\mathrm{cm}}\\ =18\mathrm{A}\end{array}$

Therefore, the magnitude of current flowing through the outer loop is 18 A

To determine direction of magnetic field produced use right hand rule by the current-carrying loop. If we direct curled fingers of our right hand in the direction of current, our extended thumb will point in the direction of magnetic field. Since current through the inner loop is flowing in the clockwise direction, that means that the magnetic field is directed downwards from our perspective. This means that outer loop must produce magnetic field directed upwards, which means that current flowing through the outer loop must flow in the counter-clockwise direction.

Therefore, the current is flowing in clockwise direction.