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 1.90-mm-diameter superconducting wire.

What current is needed?

Answer 1

Answer:$I=2.27\times 10^3 A$

Explanation:

Given

Magnetic Field Strength $B=1.5 T$

Length of solenoid $L=1.8 m$

Diameter of Solenoid $D=75 cm$

diameter of wire $d=1.90 mm$

No of turns $N=\frac{Length\ of\ solenoid}{diameter\ of\ wire}$

$N=\frac{1.8}{1.90\times 10^{-3}}=947.36 turns$

Magnetic Field is given by

$B=\frac{\mu _0NI}{L}$

where $\mu _0=magnetic\ permeability\ of\ free\ space=4\pi \times 10^{-7}$

$I =current$

Thus $I=\frac{BL}{\mu _0N}$

$I=\frac{1.5\times 1.8}{4\pi \times 10^{-7}\times 947.36}$

$I=2270 A$

$I=2.27\times 10^3 A$

Answer 2

The current needed or flowing around the solenoid is 2270 Ampere.

Given the data in the question;

• Solenoid magnetic flux density; $B = 1.5T$

• Length of the solenoid; $l = 1.8m$

• Diameter of the solenoid; $D = 75cm = 0.75m$

• Diameter of wire; $d = 1.90mm = 0.0019m$

Current is needed; $I = \ ?$

Solenoid

A solenoid is simply is a type of electromagnet which generates a controlled magnetic field, it is created by a helical coil of wire whose diameter is less than its length.

Solenoid magnetic flux density is expressed as;

$B = u_0\frac{NI}{l}$

Where B is the solenoid magnetic flux density, $u_0$ is the magnetic constant ($u_0= 4\pi *10^{-7}H/m$, I is the current, $l$ is the length of the solenoid and N is the number of turns $N = \frac{length}{diameter}$.

First we determine the number of turns

$N = \frac{length}{diameter} \\\\N = \frac{l}d} \\\\N = \frac{1.8m}{0.0019m}\\ \\N = 947.368turns\\\\N = 947turns$

Now, from the expression for Solenoid magnetic flux density, we make current I the subject of the formular.

$B = u_0\frac{NI}{l}\\ \\Bl = u_0NI\\\\I = \frac{Bl}{u_0N}$

We substitute our values into the equation

$I = \frac{Bl}{u_0N} \\\\I = \frac{(1.5T) * (1.8m)}{(4\pi *10^{-7}H/m)(947turns)} \\\\I = 2270A$

Therefore, the current needed or flowing around the solenoid is 2270 Ampere.

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