Rate of change of current with time, galvanometer deflection

In summary, a solenoid with an inductance of 2.0 H is connected in series with a resistor and a 2.0 V DC supply. The final current is 4.0 A and the initial rate of change of current with time is 1.0 A s^-1. When the current is 2.0 A, the rate of change of current with time is 0.5 A s^-1. An EMF greater than 2.0 V is produced when the current is switched off. In a separate scenario, a long air-cored solenoid with 1000 turns of wire per metre and a cross-sectional area of 8.0 cm^2 has a secondary coil
  • #36
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?
Because they are close together and have mutual inductance?
 
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  • #37
moenste said:
Because they are close together and have mutual inductance?
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
 
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  • #38
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
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.

Now we need to somehow find the magnetic field from flux, though we don't know time.

And then we use θ = a N B A / R to find the deflection θ.
 
  • #39
moenste said:
We find V first: V = I R = 4 * 60 = 240 V.
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.)
 
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  • #40
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.)
Maybe something like I = E / R = - (1 / R) * (d / d t) * (N Φ), where E = - (d / d t) * (N Φ)?
 
  • #41
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?
 
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  • #42
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?
But we don't know I and B in it Why do we need it?
 
  • #43
moenste said:
But we don't know I and B
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.
 
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  • #44
moenste said:
But we don't know I and B in it Why do we need it?
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?"
 
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  • #45
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.
But no B is given in the problem. And all my calculations here are wrong:
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.

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?"
Q charge = M I / R, where M is standard mutual inductance.
 
  • #46
moenste said:
But no B is given in the problem. And all my calculations here are wrong:
You can calculate B. You have the sufficient data to calculate B and Φ.
 
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  • #47
moenste said:
Q charge = M I / R, where M is standard mutual inductance.
No. What is the general relationship between charge and current?
 
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  • #48
cnh1995 said:
No. What is the general relationship between charge and current?
If you don't remember look it up.
 
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  • #49
cnh1995 said:
You can calculate B. You have the sufficient data to calculate B and Φ.
B = μ0 n I = 4 π * 10-7 * 1000 * 4 = 5 * 10-3 T.

cnh1995 said:
No. What is the general relationship between charge and current?
Q = I t.
 
  • #50
No, look it up.
 
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  • #51
To clarify my previous post, you need a general expression that is valid when the charge and current vary with time.
 
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  • #52
kuruman said:
To clarify my previous post, you need a general expression that is valid when the charge and current vary with time.
I = d Q / d t?
 
  • #53
Correct. So put together to get a differential equation using the alternate expression for I, Post #41.
 
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  • #54
moenste said:
I = d Q / d t?
Right.
 
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  • #55
kuruman said:
Correct. So put together to get a differential equation using the alternate expression for I, Post #41.
d Q / d t = (1 / R) (d Φ / d t)

d (B A N / R) / d t = (1 / R) (d B A N / d t)
d (μ0 N I A N / R) / d t = (1 / R) (d μ0 N I A N / d t).
 
  • #56
Multiply both sides of the first equation by dt and integrate. What do you get?
 
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  • #57
kuruman said:
Multiply both sides of the first equation by dt and integrate. What do you get?
d Q = (d t / R) d Φ
Q = t Φ / R?
 
  • #58
No. You have an extra dt on the right that doesn't belong.
 
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  • #59
moenste said:
d Q = (d t / R) d Φ
Q = t Φ / R?
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.
 
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  • #60
kuruman said:
No. You have an extra dt on the right that doesn't belong.
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.
Q = Φ / R = B A N / R?
 
  • #61
Bingo. Calculate the right side with numbers and you are done.
 
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  • #62
kuruman said:
Bingo. Calculate the right side with numbers and you are done.
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
 
  • #63
moenste said:
Q = Φ / R = B A N / R?
Right.
 
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  • #64
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
That should be 1.34*10-4..
Use sensitivity of the galvanometer to find the number of divisions.
 
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  • #65
cnh1995 said:
That should be 1.34*10-4..
Use sensitivity of the galvanometer to find the number of divisions.
θ = a Q = 2 * 1.34 * 10-4 = 2.68 * 10-4.
 
  • #66
moenste said:
θ = a Q = 2 * 1.34 * 10-4 = 2.68 * 10-4.
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?
 
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  • #67
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?
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.
 
  • #68
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.
Right.
 
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