Question

A circular loop of wire 50 mmin radius carries a current of 100 A. (a) Find the magnetic field strength. (b) Find the energy density at the center of the loop

Short answer

$$\begin{gathered} a)B=1.3\times {10}^{-3}\mathrm{T} \\ \mathrm{b})u_B=0.63\frac{J}{m^3} \end{gathered}$$

Step-by-step solution

Given

i) Radius of the circular loop $R=50\mathrm{mm}=50\times {10}^{-3}\mathrm{m}$

ii) Current in the circular loop$i=100A$

Understanding the concept

We can use the formula of magnetic field at the center of the current loop to find it. Using this magnetic field into the formula of the energy density at that point, we can find the energy density at the center of the loop.

Formula:

$$\begin{gathered} B={\mathrm{\mu }}_{0}i/2R \\ {u}_B=B^2/2{\mathrm{\mu }}_0 \end{gathered}$$

(a) Calculate the magnetic field strength

The magnetic field at the center of loop is given by

$$\begin{gathered} B=\frac{{\mathrm{\mu }}_{0}i}{2R} \\ B=\frac{\left(4\times {10}^{-7}{\displaystyle \frac{\mathrm{H}}{\mathrm{m}}}\right)\left(100A\right)}{2\left(50\times {10}^{-3}\mathrm{m}\right)} \\ B=1.3\times {10}^{-3}\mathrm{T} \end{gathered}$$

Therefore, the magnetic field strength is $1.3\times {10}^{-3}\mathrm{T}$.

(b) Calculate the energy density at the center of the loop

The energy density at the centre of the loop is given by

$$\begin{gathered} u_B=\frac{B^2}{2{\mathrm{\mu }}_0} \\ {\mathrm{u}}_{\mathrm{B}}=\frac{{\left(1.3\times {10}^{-3}\mathrm{T}\right)}^2}{2\left(4\mathrm{\pi }\times {10}^{-7}{\displaystyle \frac{\mathrm{H}}{\mathrm{m}}}\right)} \\ {\mathrm{u}}_{\mathrm{B}}=0.63\frac{\mathrm{J}}{{\mathrm{m}}^3} \end{gathered}$$

Therefore, energy density at the center of the loop is role="math" localid="1661424939827" $0.63\mathrm{J}/{\mathrm{m}}^3$.