Question

A circular loop of radius \(\mathrm{R}\) carrying current I lies in \(\mathrm{X}-\mathrm{Y}\) plane with its centre at origin. The total magnetic flux through \(\mathrm{X}-\mathrm{Y}\) plane is...... (a) Directly proportional to I (b) Directly proportional to \(\mathrm{R}\) (c) Directly proportional to \(\mathrm{R}^{2}\) (d) Zero

Short answer

The total magnetic flux through the XY plane for a circular loop of radius R carrying current I is zero.

Step-by-step solution

Expression for Magnetic Field of a Circular Loop

We know that the magnetic field due to a circular loop carrying current I at a perpendicular distance x from the center of the loop is given by the expression: \[B(x) = \frac{\mu_0 I R^2}{2 \pi (R^2 + x^2)^{3/2}}\] Where, \(B(x)\) = Magnetic field at a perpendicular distance x from the center of the loop \(\mu_0\) = Permeability of free space I = Current in the loop R = Radius of the circular loop

Calculate the Magnetic Flux

The magnetic flux (Φ) through the XY plane is the integral of the magnetic field over the XY plane area: \[\Phi = \int\int_{XY} B(x) \, dA\] Considering the symmetry of the problem, the magnetic flux through each small elemental area (dA) will be: \(d\Phi = B(x) \, dA\) Now integrating the flux, for the entire XY plane (from x = -∞ to x = +∞): \[\Phi = \int\limits_{-∞}^{+∞} B(x) \, dA = \int\limits_{-∞}^{+∞} \frac{\mu_0 I R^2}{2 \pi (R^2 + x^2)^{3/2}} \, dA\]

Determine the Magnetic Flux for the entire XY plane

We notice that the integrand contains the term \((R^2 + x^2)^{3/2}\) in the denominator, which means that as x goes from -∞ to +∞, the value of this term in the denominator will approach infinity, making the entire integrand approach zero. Therefore, the magnetic flux through the XY plane will be zero: \[\Phi = 0\] Hence, the correct answer is: (d) Zero