# Homework Help: Anyone know anything about induced electric fields?

1. May 2, 2005

### shikagami

Here's the problem: Can someone help me start it off?

A long straight solenoid of cross-sectional area 6 cm^2 is wound with ten turns of wire per centimeter, and the windings carry a current of 0.25 A. A secondary winding of two turns encircles the solenoid. When the primary circuit is opened, the magnetic field of the solenoid becomes zero in 0.05s. What is the average induced emf in the secondary?

2. May 2, 2005

### StatusX

The EMF induced around a loop is equal to the time rate of change of the magnetic flus through the loop. So the average induced EMF would be equal to the total change in flux divided by the time this change took. The flux is just the integral of B.dA over the surface bounded by the loop.

3. May 2, 2005

### quasar987

To do this problem, you will need...

1) ...to understand the concept of the magnetic flux through a surface. If you haven't mastered this, go back to studying your notes or book.

2) The equation for the magnetic field inside a long solenoid in terms of the number of turns of wire per unit lenght. This should be in your notes or book somewhere.

3) The equation relating the induced emf in a circuit in terms of the rate of change of flux:

$$\mathcal{E}_{induced}=-\frac{\Delta \Phi_B}{\Delta t}$$

This equation is saying that for each Tesla per second by which the flux through the loop making the circuit changes , there is an induced emf of 1 Volt.

mmmh.

Post your progress if you need further assistance.

4. May 2, 2005

### shikagami

can i use this equation to solve for the magnetic field?

B=MIn

5. May 2, 2005

### quasar987

If my M you mean $$\mu_0$$, then yes, this is exactly the expression of the magnitude of B inside a long soleoid.

6. May 2, 2005

### shikagami

how do i figure out what n is if I don't have length. Should I use the area to find the length which is equal to 2r?

7. May 2, 2005

### shikagami

well... i got a magnetic field of 1.14 x 10^-6 T

8. May 3, 2005

### quasar987

n is given to you in the problem: 10 turns per centimeters = 1000 turns per meter.

Here, "lenght" applies to the lenght of the solenoid itself, not of the wire composing it.

So 1000 turns/m does not mean that the people who constructed the soleoid made 1000 turns out of 1 meter of wire. It means that for each meter of the height of the solenoid (considered as a big cylinder), there is 1000 turns of wire.

9. May 3, 2005

### shikagami

ohhh ok. here's what i got so far 3.15 x 10 ^ -4 T for the magnetic field.

10. May 3, 2005

### quasar987

Yes, good! Actually the exact value is $\pi \times 10^{-4} T$ .

11. May 3, 2005

### shikagami

yay... so I use the emf equation and get -3.78 x 10 ^-6 V. does that sound good?

12. May 3, 2005

### quasar987

I don't think. The next step after finding the field is finding the magnetic flux through the secondary winding. (We assume that the secondary winding has the same radius as the primary one)

How did you go about that?

(sorry for the delay, I had to kill the nastiest spider)

13. May 3, 2005

### shikagami

so i multiply my emf by 2 which gives me -7.56 x 10 ^ -6 V

14. May 3, 2005

### quasar987

?!

What value did you get for $\Phi_B$ through the secondary winding?

15. May 3, 2005

### shikagami

umm.... 3.78 x 10^-7 V

16. May 3, 2005

### quasar987

Volts!? I think you might be confusing emf and magnetic flux! Go look up the definition of magnetic flux through a surface.

(going to sleep now, good luck)