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Write neat please. Show step by step please. Read instructions carefully. Box in answer please. Thank you
Q TO TWO infinitely long solenoids (seen in cross-section) thread ach the figure. The magnitude of B inside each is the same
Chapter 28 for you. torrono FB = q vxB FB = q vB sin o r =mv qB VB Rm
K=mv2 = open 1_2 21 21 vwqB EF =qE + qvxB VE m_rB, B k 1 mv2_qBR qE FB = |LXB Tmax = 1 AB Tmax = 1 A x B = uxB = IA U=-.B EH
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Answer #1

In the loop on the left, the induced emf is

e dos = A2 = 7(0.10m)2100T/s = TV dB =

And it attempt to produce a counterclockwise current in this loop.

In the loop on the right, the induced emf is,

e = dos = A2 = (0.15m)2100T/s = 2.25TV dB

and it attempt to produce a clockwise current.

Now, let us assume, that the I_1 flows down through the 6 \ohm resistor, I_2 flows down through the 5 \ohm resistor and I_3 flows up through the 3 ohm resistor.

From Kirchhoff point and junction rule give us,

Point rule : 13 = 11+12

Loop Rule:.

On the left loop,

611 +313 = 1

And on the right loop,

51, +313 = 2.257

Solving these three equation gives us,

11 = 0.062A, L2 = 0.86 Aandl3 = 0.923A

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