1) At time t = t 1 = 0.018 s, what is I 1 , the induced current in the loop? I 1

Question

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

3)

5)

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

1.) When t =0.018s, the length of loop inside the magnetic field region is 47*0.018=0.846cm, So only the loop S1 is presently in the magnetic field region.

Now, I1 = Blv/R -> l=0.26m, v=4.7 m/s---> 1.2*0.26*4.7/2.9 = -0.505A

Also, this current should be in the clockwise direction as then only will it be against the change of magnetic field.

2.)At t=t2=0.468, the loops have crossed 0.468*47 = 22 cm, so now S1 is outside the magnetic field region and s2 in partially inside the magnetic field region, such that the total area of S2 in magnetic field is decreasing in time,

Now, I2 = Blv/R -> l=0.69m, v=4.7 m/s---> 1.2*0.69*4.7/2.9 = +1.34A

This current is counter clockwise as it wishes to support the change in magnetic field

3.) Force to be applied is -> (B^2*l^2)v/R at t=t2, S1 is again outside, so only s2 will be considered...

F=1.2^2 * 0.69^2 * 4.7 / 2.9 = 1.1N against the direction of velocity

4.) at t=t3=0.383, the loops have crossed 18 cm so loops s2 is completely inside, but loop s1 is partially inside...answer will be I3 = Blv/R -> l=0.26m, v=4.7 m/s---> 1.2*0.26*4.7/2.9 = +0.505A

To support the change in magnetic field

5.)CASE I : generates a current I3 in counter clockwise...see the 4th part, case is similar

CASE II: s1 generates a curret I3 in counter clockwise direction, S2 generates a current in clockwise direction Io, so effective current will be small than the magnitude of I3...

I(case I)>I(case II)

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