Consider a metal ring and a region with a magnetic field that points out of the screen.
Suppose the ring is completely inside the field region initially and is moving to the right.
(a) Is the induced current in the ring clockwise, counterclockwise, or zero?
(b) Is the magnetic force on the ring to the left, to the right, or zero?
The ring now begins to emerge from the field region, still moving to the right.
(c) Is the induced current in the ring clockwise, counterclockwise, or zero?
d) Is the magnetic force on the ring to the right, to the left, or zero?
----
I think that parts a and b are zero, as there's no change...I may be wrong. Can someone help out?
Suppose the ring is completely inside the field region initially and is moving to the right.
(a) Is the induced current in the ring clockwise, counterclockwise, or zero?
(b) Is the magnetic force on the ring to the left, to the right, or zero?
The ring now begins to emerge from the field region, still moving to the right.
(c) Is the induced current in the ring clockwise, counterclockwise, or zero?
d) Is the magnetic force on the ring to the right, to the left, or zero?
----
I think that parts a and b are zero, as there's no change...I may be wrong. Can someone help out?
-
You are correct about parts a and b. There is no change in flux, so there is no induced emf and no induced current.
When the ring begins to emerge, though, the magnetic flux through the ring will begin to change. You'll be getting less and less flux out of the screen through the ring. Lenz' Law tells you that the induced current will flow to oppose the change in flux. Since we're getting less flux out of the screen through the ring, the induced field in the middle of the ring must be out of the screen. Using the Right-Hand-Rule shortcut for the Biot-Savart Law, a current counterclockwise will make the induced field point out of the screen.
Here's the "brute force" way to look at the force: If the current in counterclockwise around the ring, it is going down the screen on the left side of the ring that is still in the field. The force on a current-carrying wire in a magnetic field (the external field pointing out of the screen) is
F = ILxB
and LxB is to the left (Right-Hand-Rule for a cross-product). So, the force is to the left.
Note, though, that once you realize the force isn't zero (because there is a current-carrying wire in a magnetic field), then the force can't possibly be to the right. If it were to the right, the ring would accelerate out of the field. The kinetic energy of the ring would have increased for no obvious reason, violating the principle of conservation of energy.
When the ring begins to emerge, though, the magnetic flux through the ring will begin to change. You'll be getting less and less flux out of the screen through the ring. Lenz' Law tells you that the induced current will flow to oppose the change in flux. Since we're getting less flux out of the screen through the ring, the induced field in the middle of the ring must be out of the screen. Using the Right-Hand-Rule shortcut for the Biot-Savart Law, a current counterclockwise will make the induced field point out of the screen.
Here's the "brute force" way to look at the force: If the current in counterclockwise around the ring, it is going down the screen on the left side of the ring that is still in the field. The force on a current-carrying wire in a magnetic field (the external field pointing out of the screen) is
F = ILxB
and LxB is to the left (Right-Hand-Rule for a cross-product). So, the force is to the left.
Note, though, that once you realize the force isn't zero (because there is a current-carrying wire in a magnetic field), then the force can't possibly be to the right. If it were to the right, the ring would accelerate out of the field. The kinetic energy of the ring would have increased for no obvious reason, violating the principle of conservation of energy.
-
You are very welcome.
Report Abuse
-
Tes