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

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

$I=2270 A$

$I=2.27\times 10^3 A$