Question

1.The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole human bodies has a field strength of 8.19 T, and the current in the solenoid is 2.10 × 102 A. What is the number of turns per meter of length of the solenoid?

You have a wire of length L = 1.0 m for making a square coil of a dc motor. The current in the coil is I = 2.0 A, and the magnetic field of the motor has a magnitude of B = 0.14 T. Find the maximum torque exerted on the coil when the wire is used to make (a) a single-turn square coil and (b) a two-turn square coil.

Answer 1

Question 1

Given: B = 8.19 T, I = 2.1 x 10² A

As we know the number of turns for s solenoid can by calculated by using the magnetic field produced by the solenoid,

8.19 = 4 x 3.14 x 10^-7 x n x 2.1 x 10²

8.19 = 26.376 x 10^-5 x n

n = 31050 turns.

Questions 2

A) a single turn square coil,

For a single turn square coil, the area is given by

A = side x side =( 1/4) x (1/4) = 0.25 x 0.25 = 0.0625m2

Hence torques is find out by using

T =1x 0.0625 x 2 x 0.14 x 1

T = 0.0175 N-m

B) a two turn square coil,

A = (0.25/2) x (0.25/2)

A = 0.015625

Therefore,

T = 2 x 0.015625 x 2 x 0.14 x 1

T = 0.00825