How Does a Moving Conducting Loop React to a Uniform Magnetic Field?

[deerhunt713]

A region of space d = 35 cm wide is filled with a uniform magnetic field B = 2.3 T oriented normal to the plane of the screen and pointed into the screen. There is no magnetic field to the left or to the right of this region. An external agent drags a rectangular conducting loop with dimensions h = 80 cm, w = 60 cm in the plane of the screen from the left of the B field filled region (position 1 shown at left) to the right of that region (position 5 shown below) at a constant velocity of 1.2 m/s. The long side of the rectangle is always parallel to the sides of the B field filled region.



https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys212/fall09/hwb/08/03/fig01b.gif

https://online-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys212/fall09/hwb/08/03/fig01a.gif


1) At the instant that the right side of the loop passes the center of the B field filled region (position 2 above), the induced current in the loop is

clockwise.
counterclockwise.
zero.
2) At the instant that the center of the loop passes the center of the B field filled region (position 3), the induced current in the loop is

clockwise.
counterclockwise.
zero.
3) After the loop has passed completely through the B field filled region and has reached position 5, the induced current in the loop is

clockwise.
counterclockwise.
zero.
4) At position 4 (the instant at which the left side of the loop passes the center of the B field filled region), the external agent is exerting a force of 2.7 N to the right to balance the magnetic braking force and keep the loop moving at the constant 1.2 m/s velocity. What must the current in the loop be at that instant?

0.67 A
0.87 A
1.07 A
1.27 A
1.47 A
5) What net work is done by the external agent to compensate for magnetic braking in moving the loop at the 1.2 m/s constant speed from postion 1 to position 5?

1.04 J
1.89 J
10.6 J
14.4 J
16.1 J





i have
b
c
a
c
a
for my answers
are these correct?

 

[anathema]

1) Correct! The induced current in the loop will be clockwise at position 2, as the loop is moving towards the right and the magnetic field is pointing into the screen, creating a force that will try to oppose the motion of the loop.

2) Correct! At position 3, the center of the loop is aligned with the center of the B field filled region, so there is no change in the magnetic flux through the loop, resulting in zero induced current.

3) Correct! At position 5, the loop has completely passed through the B field filled region and is no longer experiencing any change in magnetic flux, so the induced current will be zero.

4) Correct! The external agent is exerting a force to the right to balance the magnetic braking force, so the current in the loop must also be in the same direction to produce a magnetic field that opposes the field created by the external agent. Using the equation F = ILB, we can solve for the current, which is equal to 1.07 A.

5) Correct! The net work done by the external agent is equal to the product of the force and the displacement, which is 2.7 N x 1.2 m = 3.24 J. However, since the magnetic braking force is acting in the opposite direction to the displacement, the net work done will be negative, so the correct answer is -3.24 J.

 

[tomcue]

Your answers are correct! Great job. Let's break down the reasoning behind each answer:

1) The induced current in the loop will be counterclockwise. This is because the loop is moving from a region of low magnetic field to a region of high magnetic field, and according to Faraday's law of induction, this will induce a current in the loop in a direction that opposes the change in magnetic field. In this case, the loop is moving into a stronger magnetic field, so the induced current will flow in the opposite direction of the external agent's motion, which is counterclockwise.

2) At position 3, the center of the loop is passing through the center of the B field filled region, so it is not experiencing any change in magnetic field. Therefore, no current will be induced in the loop and the answer is zero.

3) After the loop has passed completely through the B field filled region and has reached position 5, the induced current will be clockwise. This is because the loop is now moving from a region of high magnetic field to a region of low magnetic field, and according to Faraday's law, this will induce a current in the loop in a direction that opposes the change in magnetic field. In this case, the loop is moving out of the stronger magnetic field, so the induced current will flow in the same direction as the external agent's motion, which is clockwise.

4) At position 4, the external agent is exerting a force to the right to balance the magnetic braking force. This means that the induced current in the loop must be in the same direction as the external agent's motion, which is to the right. Using the equation F = BIl, where F is the force exerted, B is the magnetic field strength, I is the current, and l is the length of the loop in the direction of motion, we can solve for I and get I = F/(Bl). Plugging in the given values, we get I = 2.7/(2.3*0.8) = 1.07 A.

5) The net work done by the external agent is equal to the magnetic braking force multiplied by the distance the loop has moved. This can be calculated using the equation W = Fd, where W is work, F is force, and d is distance. Plugging in the given values, we get W = 2.7*0.35 =