MRI (magnetic resonance imaging) is a medical technique that produces detailed “pictures” of the interior of the body. The patient is placed into a solenoid that is 40 cm in diameter and 1.0 m long. A 100 A current creates a 5.0 T magnetic field inside the solenoid. To carry such a large current, the solenoid wires are cooled with liquid helium until they become superconducting (no electric resistance).

a. How much magnetic energy is stored in the solenoid? Assume that the magnetic field is uniform within the solenoid and quickly drops to zero outside the solenoid.

b. How many turns of wire does the solenoid have?

We have given that, the solenoid has 40 cm diameter and 1.0 m length. A 100 A current creates a 5.0 T magnetic field inside the solenoid.

To calculate the total energy of solenoid, we have to determine volume and energy density.

The formula to calculate volume is :

$V_{sol}=\pi {r^2}_{sol}l$

Here, $V_{sol}$= volume of solenoid

$r_{sol}$= radius of solenoid

$l$= length of solenoid

Substituting the values in the formula :

localid="1651030697382" $V=\pi \times {(0.20m)}^2\times (1.0m)$

$V=0.126m^3$

Now we will calculate the the value of energy density of solenoid.

The formula of energy density is :

${\mu }_B=\frac{1}{2{\mu }_0}{B}^2$

Here, ${\mu }_B$= Energy density

${\mu }_{0=}$Permeability constant

$B$= Magnetic field

Substituting the values in the formula :

localid="1651030882723" ${\mu }_B=\frac{{(5.0T)}^2}{2(4\pi \times {10}^{-7}H/m)}$

localid="1651031066068" ${\mu }_B=9.95\times {10}^{6}J/m^3$

Now we will calculate the value of stored energy.

The formula for the stored energy is :

$U_{tot}={\mu }_B{V}_{sol}$

localid="1651031003274" $U_{tot}=(9.95\times {10}^{6}J/m^{-3})(0.120m^3)$

$U_{tot}=1.25\times {10}^{6}J$

The stored energy of solenoid is$1.25\times {10}^{6}J$.

We have given that, the solenoid has 40 cm diameter and 1.0 m length. A 100 A current creates a 5.0 T magnetic field inside the solenoid.

The formula for the number of turns of wire is :

$N=\frac{B{l}_{sol}}{{\mu }_{0}I}$

Here, $N$= Number of turns of wire

$l_{sol}$= Length of solenoid

${\mu }_0$= Permeability constant

$I$= Current

Substituting the values in the formula :

$N=\frac{(1.0m)(5.0T)}{(4\pi \times {10}^{-7}H/m)(100A)}$

$N=4.0\times {10}^4$

Total number of turns is$4.0\times {10}^4$.