8) A magnetic field passes through a stationary wire loop, and its magnitude cha

Question

8) A magnetic field passes through a stationary wire loop, and its magnitude changes in time according to the graph in the drawing. The direction of the field remains constant, however. There are three equal time intervals indicated in the graph:

0—3.0 s, 3.0—6.0 s, and 6.0—9.0 s. The loop consists of 49 turns of wire and has an area of 0.10 m2. The magnetic field is oriented parallel to the normal to the loop. For purposes of this problem, this means that = 0° in Equation 22.2.

(a) For each interval, determine the induced emf.

(b) The wire has a resistance of 0.50 . Determine the induced current for the first and third intervals.
I1 =
I3 =

9) 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 m2 and a normal that is parallel to a magnetic field of 3.5 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.

10) The magnetic flux that passes through one turn of a 18-turn coil of wire changes to 2.0 Wb from 8.0 Wb in a time of 0.048 s. The average induced current in the coil is 275 A. What is the resistance of the wire?

Explanation / Answer

Only one question can be posted as per chegg guidelines

8.

a)

For 0<t<3 s

Induced emf

E=-NAdB/dt =-49*0.1*(0.4-0)/(3-0)

E=-0.6533 Volts

3<t<6 s

Induced emf

E=-NAdB/dt =-49*0.1*(0.4-0.4)/(3-0)

E=0 Volts

6<t<9 s

Induced emf

E=-NAdB/dt =-49*0.1*(0.2-0.4)/(9-6)

E=0.3267 Volts

b)

Induced current

FOr 0<t<3 s

I1=-0.6533/0.5 =-1.3067 A

For 6<t<9 s

I3=0.3267/0.5 =0.6533 A

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