Answer:
I = 2387.32 A
Explanation:
Given that,
Number of turns in the loop, N = 400
The radius of the circular coil, r = 0.66 m
The magnetic field inside the MRI, B = 1.2 T
We need to find the current in the loop. The magnetic field inside the solenoid is given by :
[tex]B=\mu_o NI\\\\I=\dfrac{B}{\mu_o N}\\\\I=\dfrac{1.2}{4\pi \times 10^{-7}\times 400}\\\\I=2387.32\ A[/tex]
So, the required current is equal to 2387.32 A.
