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Mutual Inductance

  1. Mar 22, 2007 #1
    In illustrating mutual inductance, there is a primary coil and a secondary coil, which houses an induced current. In applying Lenz’s Law to the direction of the secondary coil’s induced current, the status of the switch must be taken into account. If the switch is closed, does the secondary coil’s induced current flow in a counterclockwise direction (much like in the case when a bar magnet’s north pole is positioned towards a solenoid)? If the switch is opened, does the induced current flow in a clockwise direction (much like in the case when a bar magnet’s north pole is traveling away from the solenoid)?

    Thanks.
     
  2. jcsd
  3. Mar 22, 2007 #2
    This really depends on the construction of the system.

    But if you apply a potential across the primary a current will flow through the first coil. The direction of the resulting current will dictate the direction the B-field swirls around the first coil. The direction of the B-field can be found by simply rotating your fingers in the direction of the current and the direction your thumb points is the direction of the B-field. The direction the B-field swirls around the first coil will tell you the direction the B-field swirls through the second coil. When the B-field is changing due to the switch being thown at the primary the B-field will also be changing through the second coil. Lenz's law (actually, Faraday's law) just states that there will be an emf induced in the second coil such that a current is driven around the second coil that creates a B-field that fights (goes against) the changing B-field.

    The magnitude of the emf induced in the second coil is equal to the time rate of change of the magnetic flux through the second coil (remember that since the coil will have N loops you need to multiply the flux through one loop by N to get the total flux through the secondary solenoid). The polarity of the emf will be oriented in such a way as to drive current around the coil so that a B-field is produced so that the original changing B-field is opposed.

    Clear? meh, I don't think I explained that clearly.

    Oh, and there is a change B-field in the first coil (and as a result in the second coil) only when the DC voltage source is switched on. After a while, any transients will die out and you will get a DC steady state voltage and the B-field will be constant, and you will get no voltage at the secondary.

    This is why transformers are generally used in AC conditions, where the voltage waveform is constantly changing. Also, convince yourself that the amplitude of the voltage waveform at the input is proportional to the amplitude of the voltage waveform at the output. The proportionality constant is the ratio of the number of turns at the secondary to the number of turns at the primary.
     
    Last edited: Mar 23, 2007
  4. Mar 23, 2007 #3
    For the direction of the current through the primary coil that is connected to the emf, does the current flow from a lower potential to a higher potential?

    Thanks again.
     
  5. Mar 23, 2007 #4
    The current flows from high potential to low potential. It flows from positive to negative.
     
  6. Mar 23, 2007 #5
    In the primary coil, the current flows from the positive terminal to the negative terminal.

    If the switch is closed, the current increases? So does the secondary coil's induced emf create a current that has a direction that is opposite to the primary coil's current (if this current flows from positive to negative)?

    If the switch is opened, the current drops? So does the induced emf create a current that has the same direction as the primary coil's current (if it flows from positive to negative)??
     
  7. Mar 23, 2007 #6
    This depends on the construction of the transformer.
     
  8. Mar 23, 2007 #7
    For instance, say you have two conducting loops, loop one and loop 2, sitting next to one another and the loops lie in the x-y plane. Then you turn on a voltage source that drives current around loops 1, in the counter clockwise direction. When you throw the switch at loop 1 there is an increasing current in the counter clockwise direction and this results in an increasing B-field in the +Z direction. This increasing flux pointing in the +Z direction will "loop around" and go into loop 2 in the -Z direction. So, in loop 2, you have an increasing magnetic flux in the -Z direction. Loop 2 will have an induced current in the direction that opposes the change in flux, so the induced current will have to produce a B-field that points in the +Z direction, which means the current in loop 2 will be in the counterclockwise direction, just like loop 1. So, in this case, when the switch at loop 1 is closed that drives a current around the loop in the counterclockwise direction a current will be induced in loop 2 also in the counterclockwise direction. However, eventually a steady state current will be reached and no current will be induced in loop 2.

    A similar analysis can be applied to the case where the switch is opened.

    Keep in mind that faraday's law states that a changing magnetic flux through a conducting loops induces an emf such that a current is produced in a direction that produces a B-field that opposes the changing B-field. Or, more precisely, Vemf = -d(mag. flux)/dt. The negative sign in Faraday's law is Lenz's law.

    When you calculate flux, you need to integrate B.dS over the surface. B and dS are both vectors, and of course the magnitude of dS is a differential element of surface area and the direction of dS is a vector that defines the surface orientation. If B and dS are in the same direction then the flux through the surface, as defined, is positive. The negative sign in faraday's law simply states that the the direction of circulation the EMF drives current is the opposite of what is defined by the surface vector. To determine the current circulation defined by the surface vector use the right hand rule. Point your thumb in the direction of the surface vector. The direction your fingers curl is the direction of the circulation of current defined by the surface vector.

    If the B-field and surface vector point in opposite directions then the integral of the dot product results in a negative quantity and therefore Vemf is positive, by lenz's law. This means that current will circulate in the current direction defined by the surface vector, by the right hand rule.
     
    Last edited: Mar 23, 2007
  9. Mar 23, 2007 #8
    In the "open switch" case for the given example, there is a magnetic flux decay, so the induced current in loop 2 has a flux in the clockwise direction, so from the right hand rule the induced current will be in a clockwise direction??
     
  10. Mar 23, 2007 #9
    In the open switch case, the magnetic flux through loop 2 will decay. Loop 2 will resist the changing flux, or fight the change. Therefore, yes, current will be induced clockwise in loop 2, since this direction fights, or opposes, the change.
     
  11. Mar 23, 2007 #10
    For the setup of loop 1 in the given example, I am unsure of determining the polarity. Loop 1 is like an electromagnet/bar magnet, correct? The position of its north and south poles does not change when the switch's status alters? If the current flows counter-clockwise, does the current encounter the south pole first and then the north pole? (Exactly what is the relationship of the poles as the flux changes?)
     
  12. Mar 23, 2007 #11
    An electromagnet behaves like a bar magnet, yes. However, the north and south magnetic poles are not useful to talk about when dealing with magnetic induction.

    The poles will flip when the current direction is flipped though. Imagine a loop of wire with a current flowing, and then determine the direction the B-field swirls around the loop using the right hand rule. Now place another loop on top of the first loop, but with current flowing in the opposite direction. Are you aware of the Lorentz force, which describes the force a magnetic field exerts on a moving charge (or current)? Convince yourself using the Lorentz law that the two loops with opposite current directions repel one another.

    This stuff is way too hard for me to explain over the the internet. Go pick up a good electromagnetics book and read through it.
     
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