Question

Two concentric current loops lie in the same plane. The smaller loop has a radius of 3.0 cm and a current of 12 A. The bigger loop has a current of 20 A. The magnetic field at the center of the loops is found to be zero.
What is the radius of the bigger loop?

Answer 1

Concepts and reason

The concept used to solve this problem is Biot-Savart’s law.

At first, use the concept of Biot-savart’s law to derive the expression for the magnetic field at the center of the circular coil carrying current and then, use this expression to determine the radius of the bigger loop.

Fundamentals

Biot-Savart’s law:

This law is used to find the magnetic field at a point due to a current carrying element.

According to this law, the magnitude of the magnetic field induction at a point $B$ , which is at a distance $r$ from the central axis of the wire is:

• Directly proportional to the current $I$ in the element, $dB \propto I$

• Directly proportional to the small element $dl$ of the length of the conductor $l$ , $dB \propto dl$

• Directly proportional to the sine of the angle $\theta$ between the line joining the element $dl$ and the point $B$ , $dB \propto \sin \theta$

• Inversely proportional to the square of the distance $r$ from the current carrying element $dl$ to the point $B$ , $dB \propto \frac{1}{{{r^2}}}$

Here, $dB$ is the magnitude of the magnetic field at point $B$ .

The following diagram explains the concept of Biot-savart’s law.

Thus, $dB \propto \frac{{Idl\sin \theta }}{{{r^2}}}$

Replace the proportionality sign with a proportionality constant.

$dB = \frac{{{\mu _0}}}{{4\pi }}\frac{{Idl\sin \theta }}{{{r^2}}}$

Here, $\frac{{{\mu _0}}}{{4\pi }}$ is the proportionality constant and ${\mu _0}$ is the permeability of the free space.

According to Biot-Savart’s law, the magnetic field at a point $B$ due to a current carrying element is given as:

$dB = \frac{{{\mu _0}}}{{4\pi }}\frac{{Idl\sin \theta }}{{{r^2}}}$ ...... (1)

The magnetic field at the center of the circular coil carrying current can be calculated with the help of the following diagram:

The angle between $\overrightarrow {dl}$ and $\overrightarrow r$ is ${90^ \circ }$ .

Substitute ${90^ \circ }$ for $\theta$ in the above equation,

$dB = \frac{{{\mu _0}}}{{4\pi }}\frac{{Idl\sin {{90}^ \circ }}}{{{r^2}}}$

$dB = \frac{{{\mu _0}}}{{4\pi }}\frac{{Idl}}{{{r^2}}}$

The total magnetic field at a point $O$ in the circular coil can be obtained by integrating the above equation.

$B = \frac{{{\mu _0}}}{{4\pi }}\frac{I}{{{r^2}}}\int {dl}$

The total length of the circular coil is $2\pi r$ .

Substitute $2\pi r$ for $\int {dl}$ in the above equation.

$B = \frac{{{\mu _0}I}}{{2r}}$

This is the required magnetic field at the center of the circular coil carrying current.

The magnetic field at the center of the circular coil carrying current is given by:

$B = \frac{{{\mu _0}I}}{{2r}}$

The magnetic field of the two concentric loops is the same.

Thus, ${B_1} = {B_2}$

Here, ${B_1}$ is the magnetic field of the smaller loop and ${B_2}$ is the magnetic field of the bigger loop.

Substitute $\frac{{{\mu _0}{I_1}}}{{2{r_1}}}$ for ${B_1}$ and $\frac{{{\mu _0}{I_2}}}{{2{r_2}}}$ for ${B_2}$ .

$\begin{array}{l}\\\frac{{{\mu _0}{I_1}}}{{2{r_1}}} = \frac{{{\mu _0}{I_2}}}{{2{r_2}}}\\\\\frac{{{I_1}}}{{{r_1}}} = \frac{{{I_2}}}{{{r_2}}}\\\end{array}$

Here, ${I_1}$ is the current flowing through the smaller loop, ${I_2}$ is the current flowing through the larger loop, ${r_1}$ is the radius of the smaller loop and ${r_2}$ is the radius of the larger loop.

Rearrange the above equation for ${r_2}$ .

${r_2} = \frac{{{I_2} \cdot {r_1}}}{{{I_1}}}$

Substitute $12{\rm{ A}}$ for ${I_1}$ , $3{\rm{ cm}}$ for ${r_1}$ and $20{\rm{ A}}$ for ${r_2}$ .

$\begin{array}{c}\\{r_2} = \frac{{\left( {20{\rm{ A}}} \right) \cdot \left( {3{\rm{ cm}}} \right)}}{{\left( {12{\rm{ A}}} \right)}}\\\\ = 5{\rm{ cm}}\\\end{array}$

This is the required radius of the bigger loop.

Ans:

The radius of the bigger loop is $5{\bf{ cm}}$ .