Question

A clinical MRI (magnetic resonance imaging) superconducting magnet can be approximated as a solenoid with a diameter of \(1.00 \mathrm{~m}\) a length of \(1.50 \mathrm{~m},\) and a uniform magnetic field of \(3.00 \mathrm{~T}\). Determine (a) the energy density of the magnetic field and (b) the total energy in the solenoid.

Short answer

Answer: The energy density of the magnetic field inside the solenoid is approximately 10789.45 J/m³, and the total energy in the solenoid is approximately 12713.3 J.

Step-by-step solution

Find the energy density of the magnetic field

To find the energy density of the magnetic field, we can use the formula \(u = \frac{1}{2} \cdot \frac{B^2}{\mu_0}\). We are given the magnetic field, B = 3.00 T, and the permeability of free space, \(\mu_0 = 4\pi \times 10^{-7} \mathrm{Tm/A}\). Plug these values into the formula to find the energy density: \(u = \frac{1}{2} \cdot \frac{(3.00 \mathrm{T})^2}{4\pi \times 10^{-7} \mathrm{Tm/A}}\)

Calculate the Energy Density

Now, we can calculate the energy density with the given values: \(u = \frac{1}{2} \cdot \frac{9 \mathrm{T}^2}{4\pi \times 10^{-7} \mathrm{Tm/A}}\) \(u \approx 10789.45 \mathrm{J/m^3}\) So, the energy density of the magnetic field inside the solenoid is approximately \(10789.45 \mathrm{J/m^3}\).

Calculate the volume of the solenoid

We are given the diameter of the solenoid as 1m. So, the radius is half of the diameter, which is 0.5m. The length of the solenoid is given as 1.50m. Now we can calculate the volume of the solenoid using the formula for the volume of a cylinder, \(Vol = \pi r^2 h\), where r is the radius and h is the height (or length) of the solenoid: \(Vol = \pi (0.5\mathrm{m})^2 (1.50\mathrm{m})\) \(Vol \approx 1.1781\mathrm{m^3}\) The volume of the solenoid is approximately \(1.1781\mathrm{m^3}\).

Calculate the total energy in the solenoid

To find the total energy in the solenoid, we can use the formula \(W = u \times Vol\) where u is the energy density we found in step 2 and Vol is the volume of the solenoid calculated in step 3: \(W = 10789.45 \mathrm{J/m^3} \times 1.1781 \mathrm{m^3}\) \(W \approx 12713.3\mathrm{J}\) The total energy in the solenoid is approximately \(12713.3\mathrm{J}\).