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Question 3, chap 131, sect 1. part 1 of 1 10 points Question 9, chap 131, sect 4 part 3 of 3 10 points What is the direction

Questions 3,4,7,9,12...and thanks for the help...

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Answer #1

3. Area of the loop, A = \pir^2 = [3.14*(0.628)^2]m^2 = 1.24 m^2

Flux linked with the coil with 2 number of turns \phi1 = NBA = 2BA

& flux linked with one turn, \phi2 = NBA = BA

So, induced emf, e = -(change in flux)/time interval

Or, e = -(\phi2 - \phi1)/t = 2BA - BA/t = BA/t = (0.325*1.24/0.0419) V

=   9.62 V

7. Motional emf induced in the bar, e = BLv , where 'B' is the magnetic field, 'v' is the velocity of the bar and 'L' is the length of the bar.

So, e = (3*7*6) = 126 volt  

Hence, magnitude of current in the resistor is given by, I = e/R = (126/9) = 14 ampere

8. External force required, F = ILB = (14*7*3) N =   294 N

12. When the switch is closed for long time, the left to right flowing current produces a magnetic field in the location of the loop, which, by right hand thumb rule, acts perpendicular to the plane of the loop and coming out of the plane. When switch is suddenly opened current in the wire and hence the magnetic field decreases from maximum to zero. So, according to Lenz's law, the induced current through the resistor will flow from right to left so as to give an induced magnetic field coming out of the plane of the loop to support the decreasing field. Option 2.

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