Question

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

Short answer

The current needed is$2.4\times {10}^3\mathrm{A}.$

Step-by-step solution

Given information

We have given,

Magnetic field strength =$1.5\mathrm{T}$

Length = $1.8\mathrm{m}$

Diameter = $75\mathrm{cm}$

We have to find the current through the solenoid.

Simplify

We know the magnetic field which is produced by uniform shape solenoid at its center.

The magnetic field is

$\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{NI}}{\mathrm{L}}$

where N is the no. of turns and L is the length of the solenoid.

$\mathrm{N}=\frac{1.8\mathrm{m}}{0.002\mathrm{m}}=900\mathrm{turns}$

Then current will be,

$$\begin{gathered} \mathrm{i}=\frac{\mathrm{B}.\mathrm{L}}{{\mathrm{\mu }}_0\mathrm{N}} \\ \mathrm{I}=\frac{1.5\mathrm{T}\times 1.8\mathrm{m}}{4\mathrm{\pi }\times {10}^{-7}\times 900} \\ \mathrm{I}=2.4\times {10}^3\mathrm{A} \end{gathered}$$