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    Yusf's Avatar
    Yusf Posts: 198, Reputation: 3
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    #1

    Oct 19, 2016, 06:48 PM
    Faraday's law problem
    Assume I have a rectangular loop of wire. Now changing magnetic flux through this wire will induce a current in it. But assume I cut one side of the loop and throw it away. So now I have only three sides of the rectangle remaining. Now, if I change the magnific flux through it, will I get any current induced in it?

    Now technically the area of this new shape is zero. So there should be no current induced. But instead of using formulas like a machine, I want to gain a better understanding of things. Please tell me why I got current induced in first place, and why I would/would not get any after removing one side of it.

    Thanks.
    hkstroud's Avatar
    hkstroud Posts: 11,929, Reputation: 899
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    #2

    Oct 20, 2016, 05:19 AM
    Your are mistaken in the concept that the shape of the wire (conductor) is relative.
    If you pass a single conductor through a magnetic field, cutting the lines of flux, you will generate a voltage on the conductor.

    I assume you are looking at the typical drawing of an AC generator found in introductory literature to electronics.

    Note that as one side of the rectangular shape wire (the armature) is approaching the North end of the magnetic field of flux it generates a voltage in a one direction relative to the magnetic field. That is to say that the magnetic force causes electrons to move along the conductor. The direction of movement is relative to polarity of the field of force. The North end will cause movement in one direction, the South will generate movement in the opposite direction. I repeat that direction is relative to the magnetic field to force. It is the same direction along the conductor.

    So as one side of your rectangular shaped wire approaches North the other side is approaching South. Having two sides of the rectangular shaped wire approach opposite ends of the magnetic field (opposite polarity) simply increases the voltage or EMF along the conductor.

    So there is no volume or area to the shape of the conductor. There is only volume or area to the size of the conductor.

    As previously stated a single conductor cutting across the magnetic lines of force will generate a voltage in the conductor.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #3

    Oct 20, 2016, 06:24 AM
    In order to have current flow there must be a complete loop - if the loop is cut there is no way for the electrons to flow (current doesn't flow in an open circuit). However, there is an EMF (i.e. voltage) induced along the length of the wire, so if you hook up a volt meter to the ends of the wire you will see a voltage reading.
    Yusf's Avatar
    Yusf Posts: 198, Reputation: 3
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    #4

    Oct 20, 2016, 07:57 AM
    hkstroud, I didn't mean MOVING the loop through magnetic field. I meant keeping it stationary and passing a changing magnetic field through it. But Your answer was informative. Thanks.
    ebaines, thanks for your answer.

    My problem has now been solved.
    hkstroud's Avatar
    hkstroud Posts: 11,929, Reputation: 899
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    #5

    Oct 20, 2016, 08:26 AM
    I meant keeping it stationary and passing a changing magnetic field
    It is irreverent whether the magnet field is moving or the conductor is moving. The conductor cuts across the magnet lines of force.

    If both the magnetic field is physically stationary and the conductor is physically stationary but the magnetic field is constantly changing the same phenomenon occurs, the conductor cuts the magnetic lines of force.

    What you propose is more similar to how a transformer works than to generating an electrical current.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #6

    Oct 20, 2016, 12:53 PM
    hkstroud: I think you have misunderstood the OP's post. He's not talking about the shape of the conductor - he's talking about the shape of a closed loop (or open loop, once part of the loop is cut away) which has the changing magnetic flux running through the area enclosed. Mathematically he's asking why this integral:



    involves the closed loop L and a surface S bounded by the contour of the closed loop.

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