# Recent content by lloyd21

1. ### Earth's Magnetic Field Lab question

I actually think I figured it out....3 percent error seems reasonable too, thanks a lot for everything though!
2. ### Earth's Magnetic Field Lab question

I did it three different times haha, all different ways, can you show me a picture of what it might look like?
3. ### Earth's Magnetic Field Lab question

I figured out the first part....I went and re did the lab again and got the max with completely different data. This second part though im getting very off data and I think its because I'm not measuring right with the magnetic sensor....when measuring the magnetic inclination how should I hold...
4. ### Earth's Magnetic Field Lab question

I used the same link, thats how I found the somewhat "true" declination to be 1.54 degrees.
5. ### Average back emf induced in coil

I just dont know what to do after that thread ended. If its wrong completely or if it was solved and thats why no one else responded haha. Ive been stuck on that and another question for about a week and a half!
6. ### Average back emf induced in coil

sorry the initial attempt was back emf = (5.6 x 10^-5)[+2.2 - (-2.2)/Δt] T = 1/f --> 1/2T = 1/f --> 2/f -- > T = 2/20000 = 1.0 x 10^-4 secs If this is the case then, by subbing, Δt = 1.0 x 10^-4 secs, into the above equation, I get: back emf = (5.6 x 10^-5)[+2.2 - (-2.2)/1.0 x 10^-4 secs] =...
7. ### Average back emf induced in coil

So the first approach is much more accurate ? back emf = (5.6 x 10^-5)[+4.4 - (-4.4)/5.0 x 10^-5 secs] = (5.6 x 10^-5)[8.8/5.0 x 10^-5 secs] = (5.6 x 10^-5)(176000) = 9.856 V
8. ### Average back emf induced in coil

This thread is the exact question I have, however I approached it differently, and this is what I got: n = 2(pie)fL = 2(pie)(20x10^3Hz)(5.6x10^-6H) = 7.04 (ohms) Then Irms = I/sqrt(2) = 2.2 A / sqrt(2) = 1.56 A Then Vrms = Irms x n = 1.56A x 7.04 (ohms) = 21.9V

10. ### Earth's Magnetic Field Lab question

any suggestions on what I can do to fix it, because it will affect my total field big time. This is the second part of it.
11. ### Average back emf induced in coil

This was an answer added on later to this question in a different thread- - Okay, so if I am considering a whole period than my currents are actually +4.4 and - 4.4. Additionally, my period is actually just T = 1/f = 1/20000 = 5.0 x 10^-5. back emf = (5.6 x 10^-5)[+4.4 - (-4.4)/5.0 x 10^-5...
12. ### Average back emf induced in coil

Homework Statement The coil in a loudspeaker has an inductance of L = 56uH (or 5.6 x 10^-5 H). To produce a sound frequency of 20 kHz, the current must oscillate between peak values of +2.2 A and -2.2 A in one half of a period. What average back emf is induced in the coil during this variation...