# Homework Help: Quick induced emf question (exam tomorrow)

1. Dec 11, 2011

### jegues

1. The problem statement, all variables and given/known data

Are these two things not contradicting?

2. Relevant equations

3. The attempt at a solution

See figure attached.

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2. Dec 11, 2011

### Staff: Mentor

Think about which way the induced current would flow.

3. Dec 11, 2011

### jegues

From positive to negative?

(N to M)

Things still aren't clear, can you explain it as bluntly as possible?

4. Dec 11, 2011

### Staff: Mentor

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?

5. Dec 11, 2011

### jegues

Can you help me make sense of this question of a previous year exam?

See figure attached.

Our text defines,

$$e_{ind} = -\frac{d\psi}{dt}$$

The red box tells me that eind = -v(t),

while the blue box tells me that,

eind = v(t)

Where is the missing negative sign or what am I missing?

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6. Dec 11, 2011

### Staff: Mentor

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.

7. Dec 11, 2011

### jegues

Here you are, the full solution and only the problem statement are in the 2 figures attached to this post.

Your next reply will be the last reply I am likely to see before my exam, so if you could make it as clear as you possibly can it would be greatly appreciated!

Thanks again!

EDIT:

What I had put in the red box tells me that,

$$e_{ind} = -\frac{d\psi}{dt} = -v(t)$$

$$e_{ind} = -v(t)$$

and what's in the blue box tells me that,

$$e_{ind} = \oint_{l} (\vec{v} \times \vec{B}) \cdot \vec{dl} = v(t)$$

$$e_{ind} = v(t)$$

What am I mixing up?

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• ###### Entirething.JPG
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8. Dec 11, 2011

### Staff: Mentor

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.

9. Dec 11, 2011

### Staff: Mentor

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.

10. Dec 11, 2011

### jegues

The flux is increasing into the page as the the rod moves.

By Lenz' law a current will be induced to oppose this flux.

The induced current will flow in a counterclockwise fashion.

We can think of this induced current as flowing out of a voltage source know as eind, where the induced current flows out of the positive terminal of the voltage source.

KVL gives the following,

eind = -v(t).

From which I deduce,

$$v(t) = \frac{d\psi}{dt}$$

If we are to use motional emf,

from the definition,

$$e_{ind} = \oint_{l} (\vec{v} \times \vec{B}) \cdot \vec{dl}$$

In the first case I get a positive, and the second case a negative.

11. Dec 11, 2011

### jegues

I can get that every time, I know I am using and understanding Lenz Law correctly, it's just getting the sign of a specific voltage across a specific resistor!

Is there a way I can always get the sign either using Faradays law or motional emf by only thinking about Lenz Law?

12. Dec 11, 2011

### Staff: Mentor

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.

13. Dec 11, 2011

### jegues

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?

14. Dec 11, 2011

### jegues

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!

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15. Dec 11, 2011

### Staff: Mentor

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?

16. Dec 11, 2011

### jegues

This example makes sense.

Can you clarify where I am wrong in my reasoning for the 2nd example I've posted? (Solenoid with a square loop around it)

I'm fairly certain my logic with Lenz Law is correct.

17. Dec 11, 2011

### Staff: Mentor

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.

18. Dec 11, 2011

### jegues

THANK YOU VERY MUCH FOR YOUR HELP!

One mistake can cause so much confusion!!!