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Variation in EMF of a magnet moving through a Coil |
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| Apr4-12, 03:32 AM | #1 |
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Variation in EMF of a magnet moving through a Coil
NOTE : My Current question and source of debate is in post number 10.
Lets say I have a square coil. I accelerate it in a direction such that it is perpendicular to a magnetic field directed into the page. As it enters the field and before it is completely in the field an INCREASING EMF will be induced. The graph of EMF against time would be a straight line graph of CONSTANT GRADIENT through the origin which decreases to zero instantaneously once it is completely in the magnetic field, correct ? http://www.a-levelphysicstutor.com/f...ro-mag-ind.php Now, here we have a magnet being dropped into a coil. The graph of EMF against time is more of a smooth pulse.... I have two questions : 1.) Why is it not a straight line that has constant gradient that then decreases instantaneously to zero ? 2.) Why does the EMF decrease to zero ? I know its because no lines of force are cut by the coil, but I cannot see how this is the case. I'm having trouble visualising it. |
| Apr4-12, 05:17 AM | #2 |
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hi elemis!
![]() it's not that no lines of force are cut, it's that the number of lines of force that are cut is not changing (or not noticeably changing, when the magnet is a long way away) zero change => zero emf … why should it be?
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| Apr4-12, 05:41 AM | #3 |
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But the magnet is accelerating is it not ? Hence, the number of lines of force cut per unit time is INCREASING..... Hence, there is a change in magnetic flux from one moment to the next.
Well, if you accelerate a coil through a magnetic field you get a straight line of constant gradient that then decreases to zero once the whole coil is in the field, correct ? Why doesnt the same apply here ? |
| Apr4-12, 05:59 AM | #4 |
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Variation in EMF of a magnet moving through a CoilIt clearly shows a long stretch where EMF = 0. The explanation given is as follows: there is no change in the flux linking the coil as the motion is parallel to the magnetic field. |
| Apr4-12, 06:06 AM | #5 |
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hi elemis!
![]() a magnet's magnetic field is highly non-uniform |
| Apr4-12, 06:09 AM | #6 |
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| Apr4-12, 06:19 AM | #7 |
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| Apr4-12, 06:24 AM | #8 |
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Now that I think about it, it should be a smooth change over from a positive emf to a negative emf as shown in the first picture.... |
| Apr4-12, 06:38 AM | #9 |
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ok, in the second picture there's a vertical solenoid (with no battery), and a very small magnet dropping through it so long as the magnet is well inside the solenoid, the number of field lines being cut at any instant (the flux) is constant, irrespective of the speed remember, the speed at which they're cut doesn't matter, what matters is the difference in the number being cut at any instant, from one instant to another |
| Apr4-12, 07:29 AM | #10 |
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http://www.physicsforums.com/attachm...6&d=1333537181
No, you're failing to see my point. EMF = BLV If the magnet is travelling at 0.1 m/s it will cut 3 (just a random number) coils in the solenoid. Since its accelerating it is now at a speed of, lets say, 0.3 m/s. Therefore, in one second it will travel further than previously and hence cut MORE coils in the same time. So why does the EMF fall to zero for so long ? |
| Apr4-12, 02:47 PM | #11 |
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ah, no, BLv only applies when v is in the plane of the wire, ie when the plane of the wire is crossing the B field lines (as opposed to moving along them)
BLv comes from emf = d/dt (flux) = d/dt (BA) where A is area cut by the flux = d/dt (BLW) where L is length inside the B field and W is width now if the loop is entering the B field sideways on (in its own plane), with the L sides at back and front, so that L stays the same while W increases, then dW/dt = v, and so … emf = BLvsee the diagram at http://hyperphysics.phy-astr.gsu.edu...elevol.html#c3 (hyperphysics has the best diagrams! )comparing your example with that diagram, your magnet, which is pointing vertically, is coming out of the page, but your wire isn't moving across the field lines, it's moving along them … the number of dots (field lines) that your wire encloses is staying the same
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| Apr4-12, 11:50 PM | #12 |
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Yes ! I finally understand ! Thank you Tiny Tim ! :)
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