Magnetic Field on the Axis of a Circular Current Loop: Introduction & Derivation

Ans. The field lines are constructed according to the right-hand thumb rule. The drawing is given below:

Ans. The formula for magnetic fields due to a current carrying loop is given by,

\(B =B_xi = \frac{\mu_0 IR^2}{2(x^2 +R^2)^\frac{3}{2}} i\)

Ans. The answer is as follows:

Consider a very small conducting element ‘dl’ of the loop. This is given in the above figure. The magnitude dB of the magnetic field due to dl is specified by the Biot-Savart law.

If we consider a conducting element dl of the loop, the magnetic field at point P can be given by Biot Savart Law:

\(dB = \frac{\mu_0}{4\pi} \frac{I |dl \times r|} {r^3}\)

r^₂ = x² + R²

Also, any element on the loop is perpendicular to the displacement vector on the X-axis. This is because they are in different planes. Hence, |dl x r| = rdl. So the formula given above becomes : 

\(dB = \frac{\mu_0}{4\pi} \frac{Idl} {(x^2 +R^2)}\)-----------------(1)

This calculated dB is shown in the figure as well. It has 2 components, dBx and dBy. The component perpendicular to the X axis cancel each other (Because the field is generated from a loop and the diametrically opposite element dl will cancel out the vertical component). The net contribution along x-direction can be obtained by integrating dBx = dB cos θ over the loop. 

\(cos \theta= \frac{R} {(x^2 +R^2)^{\frac{1}{2}}}\)-----------------(2)

From equations (1) and (2), 

\(dB_x = \frac{\mu_0 Idl}{4\pi} \frac{R} {(x^2 +R^2)^{\frac{3}{2}}}\)

Since dl is a small element of a circular loop, summation of dl leads to 2πR, the loop circumference. The magnetic field thus becomes:

\(B =B_xi = \frac{\mu_0 IR^2}{2(x^2 +R^2)^\frac{3}{2}} i\)

Special Case : As a special case, when x = 0, we get the magnetic field at the center of the current carrying loop. The formula is given below:

\(B= \frac{\mu_0 I}{2R} \hat{i}\)

Ans. The magnetic field due to a circular loop at an axial point is given by:

\(B =B_xi = \frac{\mu_0 IR^2}{2(x^2 +R^2)^\frac{3}{2}} i\)

For both the loops, the magnitude of force will be the same (since the parameters a and r are the same) but the direction will differ. The net force acting at this point will be the vector sum of the two individual forces. The direction of forces and the net force is given in the diagram below: 

The net force can be calculated as:

Ans. The Biot Savart law gives the mathematical relation to calculate the magnetic field generated due to a current-carrying conductor. Its formula is given below:

\({\displaystyle \mathbf {B} (\mathbf {r} )={\frac {\mu _{0}}{4\pi }}\int _{C}{\frac {I\,d{\boldsymbol {\ell }}\times \mathbf {r'} }{|\mathbf {r'} |^{3}}}}\)

The Figure below portrays a circular loop carrying a stable current I. The loop is located in the y-z plane with its Centre at the source O also radius R. The X-axis is the alignment of the circle. We compute the magnetic field at the point P on this axis. Let x denote the distance of P from the Centre O of the loop.

Consider a very small conducting element ‘dl’ of the loop. This is given in Fig. 4.1. The magnitude dB of the magnetic field due to dl is specified by the Biot-Savart law.

Currently, r² = x² + R² 

If we consider a conducting element dl of the loop, the magnetic field at point P can be given by Biot Savart Law:

As explained, r² = x² + R²

Also, any element on the loop is perpendicular to the displacement vector on the X-axis. This is because they are in different planes. Hence, |dl x r| = rdl. So the formula given above becomes : 

-----------------(1)

This calculated dB is shown in the figure as well. It has 2 components, dB_x and dB_y. The component perpendicular to the X axis cancel each other (Because the field is generated from a loop and the diametrically opposite element dl will cancel out the vertical component). The net contribution along x-direction can be obtained by integrating dB_x = dB cos θ over the loop. 

-----------------(2)

From equations (1) and (2), 

Since dl is a small element of a circular loop, summation of dl leads to 2πR, the loop circumference. The magnetic field thus becomes:

Special Case : As a special case, when x = 0, we get the magnetic field at the center of the current carrying loop. The formula is given below:

The magnetic field lines are illustrated in the figure below. It is very similar to a magnet.