Homework Sourse1 2 3 4 5 6 An infinite straight wire carries current I1 25
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An infinite straight wire carries current I1 = 2.5 A in the positive y-direction as shown. At time t = 0, a conducting wire, aligned with the y-direction is located a distance d = 70 cm from the y-axis and moves with velocity v = 14 cm/s in the negaitve x-direction as shown. The wire has length W = 11 cm.
1)
What is (0), the emf induced in the moving wire at t = 0? Define the emf to be positive if the potential at point a is higher than that at point b.
V
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2)
What is (t1), the emf induced in the moving wire at t = t1 = 3.3 s? Define the emf to be positive if the potential at point a is higher than that at point b.
V
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3)
The wire is now replaced by a conducting rectangular loop as shown. The loop has length L = 48 cm and width W = 11 cm. At time t = 0, the loop moves with velocity v = 14 cm/s with its left end located a distance d = 70 cm from the y-axis. The resistance of the loop is R = 2.6 . 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.
A
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4)
Suppose the loop now moves in the positive y-direction as shown. What is the direction of the induced current now?
The current flows counterclockwise
The current flows clockwise
There is no induced current now
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5)
Suppose now that the loop is rotated 90o and moves with velocity v = 14 cm/s in the positive x-direction as shown. What is I2, the current in the infinite wire, if the induced current in the loop at the instant shown (d = 70 cm) is the same as it was in the third part of this problem (i.e., when the left end of loop was at a distance d = 70 cm from the y-axis)?
Solution
1.
Magnetic field
B=uoI1/2pid =(4pi*10-7)*2.5/(2pi*0.7)
B=7.14*10-7T
Induced emf at t=0
E(0)=WvB =0.11*0.14*7.14*10-7
E(0)=1.1*10-8 Volts
2.
Magnetic field
B=uoI/2pi(d-vt)
B=(4pi*10-7)*2.5/2pi*(0.7-0.14*3.3)
B=2.1*10-6T
Induced emf
E=WvB =0.11*0.14*2.1*10-6
E=3.24*10-8Volts
3.
Net Magnetic field
B=uoI1/2pi [1/d -1/(d+L)]
B=(4pi*10-7)*2.5/2pi [1/0.7 -1/(0.7+0.48)]
B =2.91*10-7T
Induced Voltage
E=BvW =2.91*10-7*0.14*0.11
E=4.47*10-9V
Induced current
I=E/R =4.47*10-9/2.6
I=1.72*10-9A
4.
There is no induced current now
5.
Net Magnetic field
B=uoI1/2pi [1/d -1/(d+w)]
B=(4pi*10-7)*2.5/2pi [1/0.7 -1/(0.7+0.11)]
B =9.7*10-8T
Induced Voltage
E=BvL =9.7*10-8*0.14*0.48
E=6.52*10-9V
Induced current
I=E/R =6.52*10-9/2.6
I=2.5*10-9A
