Recent content by Ryoblck

Homework Statement The distance between the movable mirror and the beam splitter in a Michelson interferometer is increased a small amount. When this happens, you see 200 dark fringes move across the field of view. If the incident light was 600nm, by how much was the mirror moved (in...

Wait but when I converted 0.1m to cm it actually was 10cm... Edit: Agh I'm getting confused in my own math. I'm lost but I'll just try and follow what you're doing. With 100/circumference, the equation would be 100/1 making it 100 turns. That sounds correct to me now.

Oh my gosh. I feel so dumb right now. With the given information, the original radius is .080cm. With this, the cross-sectional area is doubled. The cross-sectional area was .0201cm and now is .0402cm. Now that I have the new area and the same length, I found that 100cm/.0402cm would equal...

The length of wrapped wire is 100cm. With 200 turns, we can determine that is takes .5cm for each turn to wrap the wire. How would I include the length of the wire into the equation? I'm not seeing this at all.

1/(piR^2) is the answer. Meaning I subsititute that into the circumference formula?

The formula I know to find the circumference of the cylinder involves the radius. 2(pi)r is the circumference... So what am I missing?

What formula do I use? I don't know any formula that finds the radius with the length and the number of turns given.

But there is no radius given. So how will I find it? No area given either but the length of the uncoiled wire and coiled wire.

Ok I understand part B and D now. Regarding the cross-sectional area, would the result of the larger circumference equal to 100 turns exact? Because since we doubled the cross-sectional area, the number of turns have to be divided by two. Or am I looking in the wrong direction? I don't think I...

So the coil in part C would result in only 100 loops due to the bigger radius? And for part B, I think I was using the wrong angle. So possibly the magnetic field points upwards?

Homework Statement A 1.0 m piece of wire is coiled into 200 loops and attached to a voltage source as shown. A. Find the strength of the magnetic field inside the coil if V = 100 V and R = 40 Ω. B. Which direction does the magnetic field point? C. The wire is then uncoiled and re-wrapped so...

Oh ok that makes sense. Thanks for all of your help. :)

Thank you so much! That actually made sense in the end. Now about the third question, its about the voltage drop across R2. I used the V=IR equation substituting R2 with 4 ohms and I2 with 2.455 A and found that it was 9.82 V. Does that sound right?

I labeled top right resistor R1, the middle battery E1, middle resistor R2, bottom left battery E2, and bottom resistor R3. I didn't include Va or Vb anywhere because I never understood that stuff

After trying some equations for an hour, I'm still lost. The only problem with my equations is the unknowns of the current. I don't know where to begin to find it. I realized that I actually need three equations for the three unknown currents. I2=I1+I3 i) I1R1-E1-I2R2 ii)E2-I3R3-E1-I2R2 I...