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# Inducing an EMF watch

1. The equation:

emf = dBA / dt

Shows that an emf may be induced by a change in B or A. I'm confused about the situation in which A is changed to induce an emf, though.

My book describes a situation in which a ring is dropped through a magnetic field such that as it falls, the area of the ring is perpendicular to the magnetic field.
The book considers 'A' to be changing here, but I don't understand why you'd say that the area is changing if the area of the ring stays the same throughout the fall. I get the idea that an emf is induced because the circumference of the ring cuts through the field lines, though.
2. (Original post by pineneedles)
The equation:

emf = dBA / dt

Shows that an emf may be induced by a change in B or A. I'm confused about the situation in which A is changed to induce an emf, though.

My book describes a situation in which a ring is dropped through a magnetic field such that as it falls, the area of the ring is perpendicular to the magnetic field.
The book considers 'A' to be changing here, but I don't understand why you'd say that the area is changing if the area of the ring stays the same throughout the fall. I get the idea that an emf is induced because the circumference of the ring cuts through the field lines, though.
Without seeing the full context of the example given in your book, it would seem the ring is dropped starting from outside the field region. i.e. as the ring enters the constant B-field, it must first experience a changing area at the edge of the field.

Since the coil(ring) area at the field boundary (cutting the field lines) changes, then an emf is induced.

When the whole ring is within the uniform part of the B-field, then both dA and dB are constant and no emf will be generated.

Once the coil has passed through the uniform B-field and begins to exit, then once again, the area of the coil will change as it transitions (falls) out if the other end.

In this way we can deduce two emf 'spikes' will be produced. Thr first as the coil enters the field and tge second as it leaves.

Take a look at this diagram (esp. top rhs) which shows the four ways in which emf is induced in accordance with Faraday's law.

3. (Original post by uberteknik)
Without seeing the full context of the example given in your book, it would seem the ring is dropped starting from outside the field region. i.e. as the ring enters the constant B-field, it must first experience a changing area at the edge of the field.

Since the coil(ring) area at the field boundary (cutting the field lines) changes, then an emf is induced.

When the whole ring is within the uniform part of the B-field, then both dA and dB are constant and no emf will be generated.

Once the coil has passed through the uniform B-field and begins to exit, then once again, the area of the coil will change as it transitions (falls) out if the other end.

In this way we can deduce two emf 'spikes' will be produced. Thr first as the coil enters the field and tge second as it leaves.

Take a look at this diagram (esp. top rhs) which shows the four ways in which emf is induced in accordance with Faraday's law.

Thanks for this!

Can I just ask, if you wanted to find the force on this ring due to the current around it, using F = BIL would you use the diameter for l or the circumference? And could you explain why?
4. (Original post by pineneedles)
Thanks for this!

Can I just ask, if you wanted to find the force on this ring due to the current around it, using F = BIL would you use the diameter for l or the circumference? And could you explain why?
L relates to the length of the conductor within the magnetic field.

All individual charges within the conductor repel each other and all will have a force acting on them as a result of the magnetic field. The total force acting is therefore the sum of the individual charge forces acting in the same direction.

Since current is the movement (flow) of charge, we are interested in the charge flowing in and out of the conductor only - hence the length of the conductor.

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