Question

A $100\mathrm{A}$current circulates around a $2.0\mathrm{mm}$diameter superconducting ring.

a. What is the ring’s magnetic dipole moment?

b. What is the on-axis magnetic field strength $5.0\mathrm{cm}$from the ring?

Short answer

a. The ring magnetic moment is $3.14\times {10}^{-4}\mathrm{A}.{\mathrm{m}}^2.$

b. the magnetic field strength is$0.5\mathrm{\mu T}.$

Step-by-step solution

Part (a) Step 1: Given information

We have given,

current = $100\mathrm{A}$

Diameter = $2\mathrm{mm}$

We have to find the magnetic dipole moment.

Simplify

The magnetic dipole moment of ring is

$$\begin{gathered} \mathrm{\mu }=\mathrm{A}\mathrm{I} \\ \mathrm{\mu }={\mathrm{\pi r}}^2\mathrm{I} \\ \mathrm{\mu }=3.14\times {10}^{-6}\times 100 \\ \mathrm{\mu }=3.14\times {10}^{-4}\mathrm{A}.{\mathrm{m}}^2 \end{gathered}$$

Part (b) Step 1: Given information

We have given,

Distance = $5\mathrm{cm}$

We have to find the magnetic field strength.

Simplify

Magnetic field due to dipole is

$$\begin{gathered} \mathrm{B}=\frac{{\mathrm{\mu }}_0\overrightarrow{\mathrm{\mu }}}{2{\mathrm{\pi R}}^3} \\ \mathrm{B}=\frac{(4\mathrm{\pi }\times {10}^{-7})(3.14\times {10}^{-4})}{2\mathrm{\pi }{(5\times {10}^{-2})}^3} \\ \mathrm{B}=0.5\mathrm{\mu T} \end{gathered}$$