# Magnetic Field Circuit Problem

1. Mar 6, 2012

### aokidopi

1. http://www.webassign.net/userimages/ch20_prob20_40.jpg?db=v4net&id=110544

A square loop consists of a single turn with a resistance of 5.00 ohms. The loop has an area of 500 cm2, and has a uniform magnetic field passing through it that is directed out of the page. The loop contains a 12-volt battery, connected as shown in the figure above.
(a) At the instant shown in the figure, there is no net current in the loop. At what rate is the magnetic field changing? Use a positive sign if the field is increasing and magnitude, and a negative sign if the field is decreasing in magnitude.

T/s

(b) If the polarity of the battery was reversed, and the magnetic field was still changing at the rate you calculated above, what would the magnitude of the net current through the loop be?

A

2. Faraday's Law and Magnetic Flux
Flux=ABcos(theta)
emf=N*change in flux/change in time
V=IR

3. I was able to solve for part A using a combination of the first two equations to solve for Magnetic field/ time.

However, I am having trouble understanding the second part. If the rate of change in magnetic field is same, isn't emf still 12V and the resistance 5 ohms? I know that 12/5 for the ohm's law does not result in the correct solution.

2. Mar 6, 2012

Bump~

3. Mar 8, 2012

### aokidopi

Bump~~

Last edited: Mar 8, 2012
4. Mar 9, 2012

bump

5. Mar 9, 2012

### Staff: Mentor

Voltages in series add or subtract. The loop generates the same emf as before, but now the battery contributes its 12v so that the two voltages add.

6. Mar 11, 2012

### ment2byours

How do you solve for a?

I did the following:

induced emf e = - N dφ/dt

flux φ = B * A ( A is the area of loop)

e = - 1 * d( B * 0.5m^2 ) /dt

-12 = - 0.05 * dB/dt

dB/dt = 12/.05 = 240 T/s

Why is this incorrect? It asks for magnitude so sign shouldnt matter anyways.

7. Mar 13, 2012

### Staff: Mentor

Hi ment2byours, are you in the same class as aokidopi? How do you know what the correct answer is?

Using this relation,
you can find dφ/dt. Beyond that, I don't think there is much you can say. Though you can definitely determine the instantaneous current.

8. Mar 13, 2012

### Staff: Mentor

There are two voltage sources present. The coil is generating a voltage, and the battery is also contributing a voltage. These can either add, or subtract.

9. Mar 13, 2012

### ment2byours

Oh, I am not sure if I'm in the same class or not and my problem is that I DON'T know what the correct answer is BUT I do know mines is the wrong one.

Thanks for the help let me think/try it now.

10. Mar 13, 2012

### ment2byours

I still don't get it. Is my math wrong?

11. Mar 13, 2012

### Staff: Mentor

240 T/s looks right.

12. Mar 15, 2012

### joeshmo

you put down 0.5m^2 for the area when it should be 5m^2 since you are converting from cm to m

13. Mar 15, 2012

### ment2byours

oh yeah i was given 500cm^2 but that still converts to 0.05m^2 so it was just a slip of the hand, I still dont see what else is wrong.

14. Mar 18, 2012

### joeshmo

you need to include the minus sign!

15. Mar 18, 2012