For Q1 please just summarize the observation in general. For Q5, Example 30.9 is attached

1. Summarize the observations of Faraday regarding electromagnetic induction as described at the beginning of the chapter. 2. What is an eddy current? 3. We talked about electric flux in the past when we talked about Gauss's law and now we are talking about magnetic flux. Describe the concept of flux in general without making reference to electric or magnetic fields. Use any analogies that you find useful. 4. Write down and describe Lenz's law and Faraday's law. How are they related? 5. Reproduce Example 30.9: "Current induced in an MRI machine" showing all your work.EXAMPLE 30.9 I Current induced by an MRI machine The body is a conductor. so rapid magnetic eld changes in an MRI machine can induce currents in the body. To estimate the size of these currents. and any biological hazard they might impose. consider the \"loop\" of muscle tissue shown in FIGURE 30.29. This might be muscle circling the bone of your arm or thigh. Although muscle is not a great conductorits resistivity is 1.5 mwe can consider it to be a conducting loop with a rather high resistance. Suppose the magnetic eld along the axis of the loop drops from FIGURE 30.29 Edge View ofa loop of muscle tissue in a magnetic field. E 1.0 cm 1.6 T to 0 T in 0.30 s, which is about the largest possible rate of change for an MRI solenoid. What current will be induced? MODEL Model the muscle as a conducting loop. Assume that B decreases linearly with time. SOLVE The magnetic field is parallel to the axis of the loop, with B = 0". so the magnetic flux through the loop is (13m = AB = 1'23. The ux changes with time because 8 changes. According to Faraday's law, the magnitude of the induced emf is (134'... d: dB dr = m2 8: The rate at which the magnetic eld changes is ==-1-60T d: A: 0.303 =-5.3 T15 (13/11! is negative because the eld is decreasing. but all we need for Faraday's iaw is the absolute value. Thus dB 5= 2 173' d! = (0.040 m)1(5.3 Tls) = 0.027 v To nd the current, we need to know the resistance of the loop. Re call. from Chapter 27. that a conductor with resistivity p, length L. and cross-section area A has resistance R = leA. The length is the circumference of the loop, calculated to be L = 0.25 m, and we can use the 1.0 cm diameter of the \"wire" to nd A = 7.9 X l0"j m2. With these values, we can compute R = 4700 Q. As a result. the induced current is ASSESS This is a very small current. Powerthe rate of energy dissipation in the muscleis P= 11R = (5.7 x10\" A)2(4700 ) = 1.5 x10'7w The current is far too small to notice. and the tiny energy dissipation will certainly not heat the tissue