A rectangular loop of wire with sides 0.19 and 0.31 m lies in a plane perpendicu
ID: 1546988 • Letter: A
Question
A rectangular loop of wire with sides 0.19 and 0.31 m lies in a plane perpendicular to a constant magnetic field (see part a of the drawing). The magnetic field has a magnitude of 0.58 T and is directed parallel to the normal of the loop's surface. In a time of 0.20 s, one half of the loop is then folded back onto the other half, as indicated in part b of the drawing. Determine the magnitude of the average emf induced in the loop. ______________ V Magnetic resonance imaging (MRI) is a medical technique for producing pictures of the interior of the body. The patient is placed within a strong magnetic field. One safety concern is what would happen to the positively and negatively charged particles in the body fluids if an equipment failure caused the magnetic field to be shut off suddenly. An induced emf could cause these particles to flow, producing an electric current within the body. Suppose the largest surface of the body through which flux passes has an area of 0.042 m^2 and a normal that is parallel to a magnetic field of 3.0 T. Determine the smallest time period during which the field can be allowed to vanish if the magnitude of the average induced emf is to be kept less than 0.010 V. _______________ sExplanation / Answer
16#The area of the loop A = 0.19*0.31 m = 0.0589 m^2
The magnetic field B = 0.58 T
the time interval t = 0.20 s
when the loop folded then area A = 0
therefore the induced emf
e = - df/dt = - BdA/dt
= - (0.58)(0 - 0.0589)/0.20
= 0.17081 V (Answer)
17#
Area (A) of body = 0.042m2
Change in the magnetic field (dB) = 3.0T
Time duration (dt) = ?
Average induced emf () = 0.010V
We know the formula for the average induced emf is
= d / dt
But we know the formula for the magnetic flux is = BA
Then = (d/dt)(BA)
= A(dB / dt)
dt = (A)(dB)/
=(0.042)(3.0T)/0.010
=12.6 s(Answer)