Question

Magnetic resonance imaging needs a magnetic field strength of 1.5 T. The solenoid is 1.8 m long and 75 cm in diameter. It is tightly wound with a single layer of 2.0-mm-diameter superconducting wire. What size current is needed?

Short answer

$I=2.4\times {10}^{3}A$

Step-by-step solution

Part 1 : Given information

The solenoid is wounded in number of turns $N$and due to the current in this wounded wire, there is a magnetic field produced in a uniform shape at the center of solenoid where equation ($29.17$) gives the magnetic field every where inside the infinite solenoid and along the central axis of an infinite solenoid the magnetic field is given by

$B=\frac{{\mu }_{0}NI}{l}$

Where $N$is the number of turns, $L$its length, $I$is the current in the solenoid and ${\mu }_0$is the free space permeability and equals

$4\mathrm{\pi }\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A}$

calculation

By solving for $I$we get

role="math" localid="1649234083484" $I=\frac{Bl}{{\mu }_{0}N}$

The number of turns equals the length of the solenoid by the diameter of the thin layer $N=\frac{1.8m}{2\times {10}^{-3}m}=900turns$

Now let us plug the values for ${\mu }_0$,$B,landN$into equation (2) to get the current in the solenoid

role="math" localid="1649234517945" $$\begin{gathered} I=\frac{Bl}{{\mu }_{0}N} \\ =\frac{(1.5T)(1.8m)}{(4\mathrm{\pi }\times {10}^{-7}\mathrm{T}.\mathrm{m}/\mathrm{A})(900\mathrm{turns})} \\ =2.4\times {10}^{3}A \end{gathered}$$