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 10^{2} 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 10^{2}

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

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