Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A conducting loop is made in the form of two squares of sides s 1 = 2.1cm and s

ID: 1522407 • Letter: A

Question

A conducting loop is made in the form of two squares of sides s1 = 2.1cm and s2 = 5.7 cm as shown. At time t = 0, the loop enters a region of length L = 14.9 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 = 52 cm/s. The resistance of the loop is R = 1.8 .

2)

At time t = t2 = 0.363 s, what is I2, the induced current in the loop? I2 is defined to be positive if it is in the counterclockwise direction.A

You currently have 5 submissions for this question. Only 10 submission are allowed.
You can make 5 more submissions for this question.

3)

What is Fx(t2), the x-component of the force that must be applied to the loop to maintain its constant velocity v = 52 cm/s at t = t2 = 0.363 s?N

You currently have 0 submissions for this question. Only 10 submission are allowed.
You can make 10 more submissions for this question.

4)

At time t = t3 = 0.3 s, what is I3, the induced current in the loop? I3 is defined to be positive if it is in the counterclockwise direction.A

You currently have 0 submissions for this question. Only 10 submission are allowed.
You can make 10 more submissions for this question.

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.

II < III

II = III

II > III

You currently have 0 submissions for this question. Only 10 submission are allowed.
You can make 10 more submissions for this question.

(Survey Question)

6)

Below is some space to write notes on this problem.

Explanation / Answer

PART 2

At t = 0.363

d = 18.876

I2 = (0.057*1.5*0.52)/(1.8) = 0.0247 A

PART 3

F = BI2*L

F = 0.18*0.024*0.057 = 2.53*10-3 N

PART 4

At t = t3 = 0.3

d = 15.6

I3 = I2 = 0.0247 A

PART 5

II < III

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote