Question

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?

Answer 1

Answer:

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

The average  induced emf     $\epsilon =0.11232 V$

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  $D = 19cm = \frac{19}{100} = 0.19 m$

The area of the loop is mathematically represented as

        $A = \pi [\frac{D}{2} ]^2$

             $= 3.142 \frac{0.19}{2}$

             $= 0.02834 m^2$

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

           $\epsilon = \Delta \o v$

              $= 0.1404 * 0.80$

             $\epsilon =0.11232 V$