A conducting loop is made in the form of two squares of sides s 1 = 3.7cm and s

Question

A conducting loop is made in the form of two squares of sides s1 = 3.7cm and s2 = 6.2 cm as shown. At time t = 0, the loop enters a region of length L = 18.6 cm that contains a uniform magnetic field B = 1.5 T, directed in the positive z-direction. The loop continues through the region with constant speed v = 33 cm/s. The resistance of the loop is R = 2.9 .

1)At time t = t1 = 0.037 s, what is I1, the induced current in the loop? I1 is defined to be positive if it is in the counterclockwise direction.

_______A

2)At time t = t2 = 0.738 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)What is Fx(t2), the x-component of the force that must be applied to the loop to maintain its constant velocity v = 33 cm/s at t = t2 = 0.738 s?

________N

4)At time t = t3 = 0.601 s, what is I3, the induced current in the loop? I3 is defined to be positive if it is in the counterclockwise direction.

_________A

5)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.

1* II < III

2* II = III

3* II > III

Explanation / Answer

s1 = 3.7cm

s2 = 6.2 cm

At time t = 0, the loop enters a region of length L = 18.6

Uniform magnetic field B = 1.5 T

constant speed v = 33 cm/s

Resistance R = 2.9

a)time t=0.037 s

Emf is induced in the right side s1 moving into the field,

E=vBL=vB(S1)=0.33 m/s*1.5*0.037=0.018 V

Current,I=V/R=0.018/2.9=0.0063 A

The induced current creates a magnetic field that opposes the change in the flux. Flux is increasing out of page, and so current is clockwise (creates field into the page) so negative

At time,t=0.738 s

Same process with new s2=0.062 m

E=vBs2=vB(S2)=0.33 m/s*1.5*0.062=0.030 V

Current,I=V/R=0.030/2.9=0.0106 A

Changing flux through loop is decreasing now. Induced current is now counter clockwise

c)constant velocity v = 33 cm/s ,

t = t2 = 0.738 s

power =F.v=v2/r

F=(1/v)vbl2/r=vb2s22/R

F=0.00098 N

4)

E=vBL=vB(S1)=0.33 m/s*1.5*0.037=0.018 V

Current,I=V/R=0.018/2.9=0.0063 A

Decreasing flux creates counter clockwise current

5)Case 1,E=vBs2

case 2,E=vBs3-vBs2

Since, s3-s2>s2, I3>I2

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