41.6k views
3 votes
An MRI scanner is based on a solenoid magnet that produces a large magnetic field. The magnetic field doesn't stop at the solenoid's edge, however, but extends into the area around the magnet. Suppose a technician walks toward the scanner at 0.80 m/s from a region 1.0 m from the scanner where the magnetic field is negligible, into a region next to the scanner where the field is 6.0 T and points horizontally. As a result of this motion, what is the maximum magnitude of the change in flux through a loop defined by the outside of the technician's head? Assume the loop is vertical and has a circular cross section with a diameter of 19 cm.

What is the magnitude of the average induced emf around the outside of the technician's head during the time she's moving toward the scanner?

User Gammelgul
by
4.7k points

1 Answer

5 votes

Answer:

The maximum change in flux is
\Delta \o = 0.1404 \ Wb

The average induced emf
\epsilon =0.11232 V

Step-by-step explanation:

From the question we are told that

The speed of the technician is
v = 0.80 m/s

The distance from the scanner is
d = 1.0m

The initial magnetic field is
B_i = 0T

The final magnetic field is
B_f = 6.0T

The diameter of the loop is
D = 19cm = (19)/(100) = 0.19 m

The area of the loop is mathematically represented as


A = \pi [(D)/(2) ]^2


= 3.142 (0.19)/(2)


= 0.02834 m^2

At maximum the change in magnetic field is mathematically represented as


\Delta \o = (B_f - B_i)A

=>
\Delta \o = (6 -0)(0.02834)


\Delta \o = 0.1404 \ Wb

The average induced emf is mathematically represented as


\epsilon = \Delta \o v


= 0.1404 * 0.80


\epsilon =0.11232 V

User Gelisam
by
4.2k points