1) At time t = t 1 = 0.018 s, what is I 1 , the induced current in the loop? I 1
ID: 2261865 • Letter: 1
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1) At time t = t1 = 0.018 s, what is I1, the induced current in the loop? I1 is defined to be positive if it is in the counterclockwise direction.
2) At time t = t2 = 0.468 s, what is I2, the induced current in the loop? I2 is defined to be positive if it is in the counterclockwise direction. _____________A
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A conducting loop is made in the form of two squares of sides s1 = 2.6cm and s2 = 6.9 cm as shown. At time t = 0, the loop enters a region of length L = 17.1 cm that contains a uniform magnetic field B = 1.2 T, directed in the positive z-direction. The loop continues through the region with constant speed v = 47 cm/s. The resistance of the loop is R = 2.9 Ohm At time t = t1 = 0.018 s, what is I1, the induced current in the loop? I1 is defined to be positive if it is in the counterclockwise direction. At time t = t2 = 0.468 s, what is I2, the induced current in the loop? I2 is defined to be positive if it is in the counterclockwise direction. What is Fx(t2), the x-component of the force that must be applied to the loop to maintain its constant velocity v = 47 cm/s at t = t2 = 0.468 s? At time t = t3 = 0.382 s, what is I3, the induced current in the loop? I3 is defined to be positive if it is in the counterclockwise direction. Consider the two cases shown above. How does II, the magnitude of the induced current in Case I, compare to III, the magnitude of the induced current in Case II? Assume s2 = 3s1.Explanation / Answer
When a wire moves in a magnetic field, Potential is developed across it given by,
V = B*v*l ; l=length of wire and v = velocity
a) t=0.018 sec, Distance = 47*0.018 = 0.846 cm < 2.6 cm
So only the small arm moves.
V = 1.2 * 0.47 * 0.026 = 0.014664 volts.
I = V/R = -5.056 * 10^-3 A (Clockwise)
b) t=0.468 sec, Dis = 22 cm
Only the biggest and left most arm will be inside field.
I = V/R = Bvl/R = 1.2 * 0.47 * 0.069 / 2.9
= -13.42 * 10^-3 A (clockwise)
c) Force acting = B*I*L = 1.2 * -13.42 * 10^-3 * 0.069 = 1.11 * 10^-3 N (Negative x direction)
d) t=0.382, Dis = 17.954 cm
Both the biggest arm and the the two intermediate arms will be moving in the field.
their I will be in opposite direction, (+ve potential connected to +ve)
In biggest arm, I = -13.42 * 10^-3 A (clockwise)
In the intermediate arms, I = Bv(0.069-0.026)/R = 8.36 * 10^-3 A (counterclockwise)
Hence, Net I = -5.06*10^-3 A (clockwise)
e) A, I2>I1
In I2, current is due to the 2 intermediate arm, say I and I added to 2I.
In I1, Current is due to the bigger arm as well, but in opposite direction, Since bigger arm is 3 times, current = -3I. Hence, net = -I.
Magnitude wise 2I > -I
Hence Case 2 has greater current.
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