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 2.20-mm-diameter superconducting wire.

Required:
What current is needed?

Answer 1

Answer:

The current needed is 2625.72 A.

Explanation:

Given;

magnetic field strength, B = 1.5 T

length of the solenoid, L = 1.8 m

diameter of the solenoid, d = 75 cm

The number of turns is given by;

$N = \frac{Length \ of solenoid }{diameter \ of wire}\\\\N = \frac{1.8}{2.2*10^{-3}} = 818.18 \ turns$

The magnetic field is given by;

$B = \frac{\mu_o NI}{l}$

where;

I is the current needed

$I = \frac{Bl}{\mu_o N} \\\\I = \frac{(1.5)(1.8)}{(4\pi*10^{-7})(818.18)} \\\\I = 2625.72 \ A$

Therefore, the current needed is 2625.72 A.