- #36
moenste
- 711
- 12
Because they are close together and have mutual inductance?kuruman said:
Because they are close together and have mutual inductance?kuruman said:We have already agreed that Φ = B A N. You can't have Φ = B I A N at the same time.
Answer me this: The secondary is not connected to a battery or power supply. Why should there be a current in it?
Yes they have mutual inductance, but the mutual inductance by itself does not cause a current to flow in the secondary. What causes the induced current flow?moenste said:Because they are close together and have mutual inductance?
We find V first: V = I R = 4 * 60 = 240 V. Then we find d Φ / d t: E = - N (d Φ / d t) → d Φ / d t = 240 / 2000 = 0.12 Wb s-1.kuruman said:Yes they have mutual inductance, but the mutual inductance by itself does not cause a current to flow in the secondary. What causes the induced current flow?
Please read and understand this, then try answering the question again.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/farlaw.html
You can't say this. The 4.0 A is the current in the primary and the 60 Ω is the total resistance in the secondary. You don't know how much current flows in the secondary but you can write an algebraic expression for it in terms of the induced emf. What is that expression? (No numbers please, just symbols.)moenste said:We find V first: V = I R = 4 * 60 = 240 V.
Maybe something like I = E / R = - (1 / R) * (d / d t) * (N Φ), where E = - (d / d t) * (N Φ)?kuruman said:You can't say this. The 4.0 A is the current in the primary and the 60 Ω is the total resistance in the secondary. You don't know how much current flows in the secondary but you can write an algebraic expression for it in terms of the induced emf. What is that expression? (No numbers please, just symbols.)
But we don't know I and B in it Why do we need it?kuruman said:That's what you want, ## I = \frac{1}{R} \frac{d\Phi}{dt}##. The negative sign can be dropped because we looking for magnitudes. Remember, we are looking for the total charge that flows in the secondary. We just found an expression for the current in the secondary. How is the current related to the charge in the secondary?
I think you do know B (and Φ) in the secondary. The secondary has the same area of 8cm2 as you said in an earlier post.moenste said:But we don't know I and B
But no B is given in the problem. And all my calculations here are wrong:cnh1995 said:I think you do know B (and Φ) in the secondary. The secondary has the same area of 8cm2 as you said in an earlier post.
moenste said:(b) As I understand we need to find θ: θ = a N A B / R, where a is the division per microcoulomb, N = turns, A = cross-sectional area, B = magnetic field and R = resistance. If this formula is correct, I'm not sure how to find B. Maybe something like: B = μ0 n I = 4 π * 10-7 * 1000 * 4 = 5 * 10-3 T. Then plug into the formula: θ = 2 * 2000 * (8 * 10-4) * (5 * 10-3) / 60 = 2.68 * 10-4. It has the 2.68 in it as in the answer 268 divisions. But still I'm not sure whether I'm in the right direction.
Q charge = M I / R, where M is standard mutual inductance.kuruman said:As @cnh1995 remarked, you know B and therefore Φ. Maybe, just maybe, you don't need I, so just answer my question "How is the current related to the charge in the secondary?"
You can calculate B. You have the sufficient data to calculate B and Φ.moenste said:But no B is given in the problem. And all my calculations here are wrong:
No. What is the general relationship between charge and current?moenste said:Q charge = M I / R, where M is standard mutual inductance.
If you don't remember look it up.cnh1995 said:No. What is the general relationship between charge and current?
B = μ0 n I = 4 π * 10-7 * 1000 * 4 = 5 * 10-3 T.cnh1995 said:You can calculate B. You have the sufficient data to calculate B and Φ.
Q = I t.cnh1995 said:No. What is the general relationship between charge and current?
I = d Q / d t?kuruman said:To clarify my previous post, you need a general expression that is valid when the charge and current vary with time.
Right.moenste said:I = d Q / d t?
d Q / d t = (1 / R) (d Φ / d t)kuruman said:Correct. So put together to get a differential equation using the alternate expression for I, Post #41.
d Q = (d t / R) d Φkuruman said:Multiply both sides of the first equation by dt and integrate. What do you get?
Time term should not show up in the equation. That's why you were asked to multiply both the sides by dt. That will eliminate the time variable.moenste said:d Q = (d t / R) d Φ
Q = t Φ / R?
kuruman said:No. You have an extra dt on the right that doesn't belong.
Q = Φ / R = B A N / R?cnh1995 said:Time term should not show up in the equation. That's why you were asked to multiply both the sides by dt. That will eliminate the time variable.
B = μ0 n I = 4 π * 10-7 * 1000 * 4 = 5 * 10-3 T.kuruman said:Bingo. Calculate the right side with numbers and you are done.
Right.moenste said:Q = Φ / R = B A N / R?
That should be 1.34*10-4..moenste said:B = μ0 n I = 4 π * 10-7 * 1000 * 4 = 5 * 10-3 T.
Q = 2000 * (8 * 10-4) * (5 * 10-3) / 60 = 1.33 * 10-4 C
θ = a Q = 2 * 1.34 * 10-4 = 2.68 * 10-4.cnh1995 said:That should be 1.34*10-4..
Use sensitivity of the galvanometer to find the number of divisions.
Well, 1.34*10-4 is the actual charge passed through the galvanometer. You have sensitivity as 2 divisions/ microcoulomb. How would you calculate the no of divisions from this?moenste said:θ = a Q = 2 * 1.34 * 10-4 = 2.68 * 10-4.
θ = a Q = deflections of the galvanometer.cnh1995 said:Well, 1.34*10-4 is the actual charge passed through the galvanometer. You have sensitivity as 2 divisions/ microcoulomb. How would you calculate the no of divisions from this?
Right.moenste said:θ = a Q = deflections of the galvanometer.
Update
1.34 * 10-4 C → 1.34 * 10-4 / 10-6 = 134 μC.
134 * 2 = 268. I think this should be right.