(1196) Problem 9: A loop of wire with radius r= 0.13 m is in a magnetic field wi

Question

(1196) Problem 9: A loop of wire with radius r= 0.13 m is in a magnetic field with magnitude B as shown in the graph. B changes from B,-0.12T to B:-7.5 T in St 6.5 s at a constant rate. Randomized Variables r= 0.13 m B-0.12 T B-75 T At= 6.5 s theexpertta.com @ 20% Part(a) Express the magnetic flux going through a loop of radius r assuming a constant magnetic field B @ 20% Part (b) Express the change in the magnetic flux going through this loop, , in terms ofB.. B2 and r @ 2096 Part (c) Calculate the numerical value of in T-m. as 20% Part (d) Express the magnitude of the average induced electric field, E, induced in the loop in terms of14, and . Grade Su 806 9296 | ( |) | 7 | 8 | 9 | HOME At Attempts remaining: 15 (490 per attempt) detailed view 123- 496 496 HACKSPACEDCLEAR I give up Hints: 06 deduction per hint. Hints remaining:- Feedback: 1 for a 06 deduction No meaningful feedback available for the current submission. Submission History Hints Feedback Totals 406 406 86 This is induced emf, but the question is about the induced electric field. EMF is the work which the electric field does while moving charge along the closed loop 4 006 096 46 Totals 006 86 20% Part (e) Calculate the numerical value of E in N/O

Explanation / Answer

we know that

integral(E*dl) = emf = e = (d/dt)(phi)

E*L = (d/dt)(phi)

L = 2*pi*r

then

E = [(d(phi))/dt]*(1/(2*pi*r))

b) d(phi)/dt = A*(dB/dt) = pi*r^2(7.5-0.12)/6.5 = (3.142*0.13^2*(7.5-0.12))/6.5 = 0.060288 Wb

2*pi*r = 2*3.142*0.13 =0.82 m

then

E = [(d(phi))/dt]*(1/(2*pi*r)) = 0.060288*(1/0.82) = 0.0735 N/C

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