2. A wire carries current /1 to the right as shown For all situations below find

Question

2. A wire carries current /1 to the right as shown For all situations below find the direction of induced current: CW, CCW or 0 (a) 1(t) increases (b) /1(t) decreases (c)1const but the frame moves left d) the frame moves right (e) the frame moves up (f) the frame moves down (g) (*) for the case 2a find the magnitude of the induced current. Assume known size a × b of the frame, distance d from the wire, resistance R of the frame and dl/dt (h) (*) the same for case 2f, if the frame moves with a known speed v

Explanation / Answer

2.

a. when I1 increases, flux into the loop increases, so induced flux is out of the loop, hence counter clockwise currnet

b. when I1 decreases, flux into the loop decreases, so induced flux is into the loop, hence clockwise current

c. I1 is constant, the frame moves to left, flux through the frame is constant, hence no induced current

d. the frame moves right, again no current

e. the frame moves up, magnetic field through the frame strengthens, hence flux into the frame increase, so induced flux is out of the frmae, hence counterclockwise induced current

f. the frmae moves down, from the last part ht ecurrnet is clockwise in this case

g. from the given data

at some point in the loop

flux d(phi) = B*dA

B = 2kI/(d + y)

dA = b*dy

hence

d(phi) = 2kIb*dy/(d + y)

integrating from y = 0 to y = a

phi = 2kIb*ln(a/d + 1)

hence

induced emf, V = 2k(dI/dt)b*ln(a/d + 1)

current = V/R = 2k(dI/dt)b*ln(a/d + 1)/R

h. phi = 2kIb*ln(a/d + 1)

induced emf = 2kIbav/d(a + d)

i = V/R = 2kIab*v/Rd(a + d)

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