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Doc Al said:Think about which way the induced current would flow.
Doc Al said:How's this. The induced field points from M to N. Thus the induced current would flow from M to N. You'd get the same effect by replacing the induced EMF with a voltage source oriented as shown in the diagram: a battery with that orientation would move current from M to N.
Make sense?
Can you post the full problem (the previous page), so I can understand the full context.jegues said:Can you help me make sense of this question of a previous year exam?
See figure attached.
That minus sign indicates that the induced EMF opposes the change in flux.Our text defines,
[tex]e_{ind} = -\frac{d\psi}{dt}[/tex]
Doc Al said:Can you post the full problem (the previous page), so I can understand the full context.
That minus sign indicates that the induced EMF opposes the change in flux.
I think you are misinterpreting the meaning of that minus sign in Faraday's law. Don't apply it mechanically. Instead, figure out which way the flux is increasing, then use that to determine which way the induced EMF must be directed to oppose that changing flux.jegues said:What I had put in the red box tells me that,
[tex]e_{ind} = -\frac{d\psi}{dt} = -v(t)[/tex]
[tex]e_{ind} = -v(t)[/tex]
Doc Al said:I'm afraid I'm not seeing the issue that disturbs you. There are two ways of getting the voltage drop across the resistor. Each gives the same answer.
The first way is using Faraday's law. Since the flux through the loop is increasing into the page as the rod moves counterclockwise, the induced EMF must be counterclockwise to oppose it. Thus the voltage drop across the resistor has the sign indicated in the diagram.
The second way is to treat the moving rod using motional EMF. The magnetic force on the (positive) charge carriers in the rod would be towards the center, thus the induced EMF in the rod would again drive a counterclockwise current through the loop. The voltage drop across the resistor is the same as before.
Doc Al said:I think you are misinterpreting the meaning of that minus sign in Faraday's law. Don't apply it mechanically. Instead, figure out which way the flux is increasing, then use that to determine which way the induced EMF must be directed to oppose that changing flux.
Doc Al said:Answer this: Which way does the induced current flow through the resistor?
No matter how you look at it, the induced current flows across the resistor in the +x direction.
Yes. The voltage drop is always opposite to the current flow: Current always flows from high to low voltage across the resistor. (Imagine that the rotating rod has a battery--a voltage generator equal to the EMF--which points inward. Then treat the resulting circuit like any other. The voltage drop across the resistor must point opposite to that of the battery if you view them as being in series.)jegues said:The induced current will flow in a counterclockwise direction, into the + terminal of the resistor.
Is this what tells me that the voltage must be positive?
Doc Al said:Yes. The voltage drop is always opposite to the current flow: Current always flows from high to low voltage across the resistor. (Imagine that the rotating rod has a battery--a voltage generator equal to the EMF--which points inward. Then treat the resulting circuit like any other. The voltage drop across the resistor must point opposite to that of the battery if you view them as being in series.)
Make sense?
I agree with your reasoning. The induced EMF would be clockwise and thus the voltage drop across the resistor (with respect to the indicated signs) would be negative.jegues said:Here is an example where my reasoning with Lenz Law fails me.
See figure attached.
The flux will flow in the positive z direction.
Thus if i(t) is increasing, the flux is increasing in the z direction.
The induced current in the loop will be such to oppose this increasing flux.
For this, the induced current would flow clockwise if you were looking down onto the loop in the xy plane.
This would leave me to believe that v(t) is negative! (induced current entering the negative terminal of the resistor)
The answer says it should be positive!
Doc Al said:I agree with your reasoning. The induced EMF would be clockwise and thus the voltage drop across the resistor (with respect to the indicated signs) would be negative.
Induced emf, or electromotive force, is the voltage that is created when there is a change in the magnetic field passing through a conductor.
Induced emf is calculated by multiplying the rate of change of magnetic flux by the number of turns in the conductor. It can also be calculated by multiplying the magnetic flux density by the area of the loop and the sine of the angle between the magnetic field and the normal to the loop.
Quick induced emf occurs when there is a sudden change in the magnetic field, while slow induced emf occurs when there is a gradual change in the magnetic field. Quick induced emf results in a larger voltage, while slow induced emf results in a smaller voltage.
Lenz's law states that the direction of the induced emf in a circuit will always oppose the change in the magnetic field that created it. This means that induced emf will always try to maintain the status quo and resist any changes in the magnetic field.
Induced emf is used in many everyday devices, such as generators, transformers, and electric motors. It is also used in renewable energy sources like wind turbines and hydroelectric dams. It is also the basis for technologies like wireless charging and electromagnetic induction stovetops.