Question

Two co-axial rings of same radius $R = 10$ cm are placed parallel to the y-z plane, such that x-axis is the axis of the rings. Ring 1 carries a current of 2 Ampere and ring 2 carries a current of 1 Ampere as shown in the figure. The separation between the rings is $d = 50$ cm. Find the magnitude of $\int_{-\infty}^{+\infty} \frac{\overrightarrow{B}.\overrightarrow{dx}}{\mu_0}$, where $\overrightarrow{B}$ is the net magnetic field at any point on the axis. (Current in both the rings are in opposite sense)

Answer 1

1

Answer 2

The problem requires evaluating an integral of the magnetic field along the axis of two co-axial current-carrying rings. The magnetic field on the axis of a single current loop of radius $R$ at a distance $x$ from its center is $B_x = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$. For two rings with currents $I_1$ and $I_2$ in opposite directions, the net magnetic field on the axis is $B_x = \frac{\mu_0 I_1 R^2}{2(R^2 + (x-x_1)^2)^{3/2}} - \frac{\mu_0 I_2 R^2}{2(R^2 + (x-x_2)^2)^{3/2}}$. The integral to be calculated is $\int_{-\infty}^{+\infty} \frac{B_x dx}{\mu_0}$. This integral simplifies to $\frac{I_1 R^2}{2} \int_{-\infty}^{+\infty} \frac{dx}{(R^2 + (x-x_1)^2)^{3/2}} - \frac{I_2 R^2}{2} \int_{-\infty}^{+\infty} \frac{dx}{(R^2 + (x-x_2)^2)^{3/2}}$. The definite integral $\int_{-\infty}^{+\infty} \frac{du}{(R^2 + u^2)^{3/2}}$ evaluates to $\frac{2}{R^2}$. Substituting this result, the entire expression simplifies to $I_1 - I_2$. Given $I_1 = 2$ A and $I_2 = 1$ A, the value is $2 - 1 = 1$. The magnitude is $1$.