Question

A 108 A current circulates around a 2.40-mm-diameter superconducting ring.


1. What is the ring's magnetic dipole moment?

2. What is the on-axis magnetic field strength 4.70 cm from the ring?

Answer 1

Concepts and reason

The concepts required to solve this problem is the magnetic moment and the magnetic field.

Initially, calculate the magnetic moment using the current and the area and the area can be calculated by the diameter of the ring. Then, calculate the magnetic field using the expression of the magnetic field at a point on the axis of a current carrying circular loop.

Fundamentals

The expression of the magnetic moment is,

$\mu = IA$

Here, $\mu$ is the magnetic moment, I is the current, and A is the area.

The expression of the magnetic field at a point on the axis of a current carrying circular loop is,

$B = \frac{{{\mu _0}}}{{4\pi }}\frac{{2\pi I{R^2}}}{{{{\left( {{x^2} + {R^2}} \right)}^{3/2}}}}$

Here, B is the magnetic field, ${\mu _0}$ is the permeability of free space, R is the radius of the ring, x is the distance of point from the center of the loop, and I is the current.

(1)

The area of the circular loop is,

$A = \pi {\left( {\frac{d}{2}} \right)^2}$

The expression of the magnetic moment is,

$\mu = IA$

Substitute $\pi {\left( {\frac{d}{2}} \right)^2}$ for A.

$\mu = I\pi {\left( {\frac{d}{2}} \right)^2}$

Substitute 108 A for I and 2.40 mm for d in equation $\mu = I\pi {\left( {\frac{d}{2}} \right)^2}$.

$\begin{array}{c}\\\mu = \left( {108{\rm{ A}}} \right)\pi {\left( {\frac{{2.40{\rm{ mm}}}}{2}\left( {\frac{{{{10}^{ - 3}}{\rm{ m}}}}{{1{\rm{ mm}}}}} \right)} \right)^2}\\\\ = 4.89 \times {10^{ - 4}}{\rm{ A}} \cdot {{\rm{m}}^2}\\\end{array}$

(2)

The radius R is,

$R = \frac{d}{2}$

Substitute 2.40 mm for d in equation $R = \frac{d}{2}$.

$\begin{array}{c}\\R = \left( {\frac{{2.40{\rm{ mm}}}}{2}} \right)\left( {\frac{{{{10}^{ - 3}}{\rm{ m}}}}{{1{\rm{ mm}}}}} \right)\\\\ = 1.20 \times {10^{ - 3}}{\rm{ m}}\\\end{array}$

The magnetic field at a point on the axis of a current carrying circular loop is,

$B = \frac{{{\mu _0}}}{{4\pi }}\frac{{2\pi I{R^2}}}{{{{\left( {{x^2} + {R^2}} \right)}^{3/2}}}}$

Substitute $1.20 \times {10^{ - 3}}{\rm{ m}}$ for R, $4\pi \times {10^{ - 7}}{\rm{ T}} \cdot {\rm{m/A}}$ for ${\mu _0}$, and 4.70 cm for x.

$\begin{array}{c}\\B = \frac{{4\pi \times {{10}^{ - 7}}{\rm{ T}} \cdot {\rm{m/A}}}}{{4\pi }}\frac{{2\pi \left( {108{\rm{ A}}} \right){{\left( {1.20 \times {{10}^{ - 3}}{\rm{ m}}} \right)}^2}}}{{{{\left( {{{\left( {4.70{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)}^2} + {{\left( {1.20 \times {{10}^{ - 3}}{\rm{ m}}} \right)}^2}} \right)}^{3/2}}}}\\\\ = 9.40 \times {10^{ - 7}}{\rm{ T}}\\\end{array}$

Ans: Part 1

The magnetic moment of the ring is $4.89 \times {10^{ - 4}}{\rm{ A}} \cdot {{\rm{m}}^2}$.

Part 2

The magnetic field is $9.40 \times {10^{ - 7}}{\rm{ T}}$.