3) The wire is now replaced by a conducting rectangular loop as shown. The loop
ID: 2140954 • Letter: 3
Question
3)
The wire is now replaced by a conducting rectangular loop as shown. The loop has length L = 50 cm and width W = 12 cm. At time t = 0, the loop moves with velocity v = 14 cm/s with its left end located a distance d = 73 cm from the y-axis. The resistance of the loop is R = 2.8 ?. What is i(0), the induced current in the loop at time t = 0? Define the current to be positive if it flows in the counter-clockwise direction.
4)
5)
The wire is now replaced by a conducting rectangular loop as shown. The loop has length L = 50 cm and width W = 12 cm. At time t = 0, the loop moves with velocity v = 14 cm/s with its left end located a distance d = 73 cm from the y-axis. The resistance of the loop is R = 2.8 ?. What is i(0), the induced current in the loop at time t = 0? Define the current to be positive if it flows in the counter-clockwise direction.Explanation / Answer
if we know numerical of i1 we may get numerical values to our problems
the near end is at a distnce 0.73 m then magnetic field B at near end is
B1=uo*i1/(2*pi*r1)=
here r1 is 0.73m
B at the far end is B2 =uo*i1/(2*pi*r2)=
r2 is 0.73+0.5=1.23m
A is area of the coil=600*10^-4 m^2...
dB/dt=(d/dt)(B2-B1)=
induced emf is e=-A*dB/dt=0.06*
iR=
currnet i =
4)according to fleming's left hand rule
The current flows counterclockwise
5)dB is B2-B1=[(uo*i2)/(2*pi)][(1/r1)-(1/r2)]
area of the coil is A=600*10^-4 m^2...
induced emf e=i*R=-A*dB/dt....
substitute i value from 3 rd problem...
and we may easily calculate i2...
if you given i1 numerical value i will give you numerical values for the problems
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