Rate of change of current with time, galvanometer deflection

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Homework Statement


(a) What is meant by the statement that a solenoid has an inductance of 2.0 H?

A 2.0 H solenoid is connected in series with a resistor, so that the total resistance is 0.50 Ω, to a 2.0 V DC supply. Sketh the graph of current against time when the crrent is switched on.

What is:
(i) the final current,
(ii) the initial rate of change of current with time, and
(iii) the rate of change of current with time when the current is 2.0 A?

Explain why an EMF greatly in excess of 2.0 V will be produced when the current is switched off.

(b) A long air-cored solenoid has 1000 turns of wire per metre and a cross-sectional area of 8.0 cm2. A secondary coil, of 2000 turns, is wound around its centre and connected to a ballistic galvanometer, the total resistance of coil and galvanometer being 60 Ω. The sensitivity of the galvanometer is 2.0 divisions per microcoulomb. If a current of 4.0 A in the primary solenoid were switched off, what would be the deflection of the galvanometer?

Answers: (a) (i) 4.0 A, (ii) 1.0 A s-1, (iii) 0.5 A s-1, (b) 268 divisions.

2. The attempt at a solution
(a) (i) I = V / R = 2 / 0.5 = 4 A.

(a) (ii) E = L [d I / d t] → d I / d t = 2 / 2 = 1 A s-1.

(a) (iii) This one can't get right. E = L [d I / d t] → 2 = 2 [2 / d t] → 2 d t = 4 → d t = 2 s. d I / d t = 2 / 2 = 1 A s-1.

I also tried: P = I E = 2 * 2 = 4 W. P = L I [d I / d t] → 4 = 2 * 2 [d I / d t] → d I / d t = 1 A s-1.

Maybe it should be like: 2 A is half the 4 A, so 1 A s-1 / 2 = 0.5 A s-1?

(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.
 
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Answers and Replies

  • #2
kuruman
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One question at a time.
iii) the rate of change of current with time when the current is 2.0 A?
Can you find I(t), the current as a function of time?
Can you find the time t2 such that I(t2) = 2.0 A?
Can you find dI/dt as a function of time?
 
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  • #3
cnh1995
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(iii) This one can't get right
You need to study the step response of a series R-L circuit. Look it up. Then try to answer kuruman's questions in #2.
 
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  • #4
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You need to study the step response of a series R-L circuit. Look it up. Then try to answer kuruman's questions in #2.
Can you find I(t), the current as a function of time?
Isn't it part (a) (ii)?
(a) (ii) E = L [d I / d t] → d I / d t = 2 / 2 = 1 A s-1.
d I / d t = 1 A s-1.

Can you find the time t2 such that I(t2) = 2.0 A?
E = L (d I / d t)
2 = 2 (2 / d t)
d t = 2 s.

Can you find dI/dt as a function of time?
d I / d t = 2 / 2 = 1 A s-1.
 
  • #5
cnh1995
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Isn't it part (a) (ii)?
No. Part a)(ii) is the general formula for voltage across an inductor. You need to find current i(t). How does the current grow in a series R-L circuit when excited with a dc voltage?
 
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  • #6
cnh1995
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E = L (d I / d t)
2 = 2 (2 / d t)
d t = 2 s.
How is E=2V? Voltage across the inductor is not 2V when the current is 2A. At which instant is E=2V? Look up 'step response of a series R-L circuit' and post if anything is unclear.
 
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  • #7
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No. Part a)(ii) is the general formula for voltage across an inductor. You need to find current i(t). How does the current grow in a series R-L circuit when excited with a dc voltage?
How is E=2V? Voltage across the inductor is not 2V when the current is 2A. At which instant is E=2V? Look up 'step response of a series R-L circuit' and post if anything is unclear.
The "step response of a series R-L circuit" looks extremely difficult. Is there an easier / different way to approach this?
 
  • #8
cnh1995
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The "step response of a series R-L circuit" looks extremely difficult. Is there an easier / different way to approach this?
I'm afraid there is no other way. But instead of going for the entire derivation, you can use the formulae directly here.
Current in an RL circuit grows as i(t)=(V/R)*(1-e-tR/L).
You can use this equation to find answers to the questions given in your problem.
 
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  • #9
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I'm afraid there is no other way. But instead of going for the entire derivation, you can use the formulae directly here.
Current in an RL circuit grows as i(t)=(V/R)*(1-e-tR/L).
You can use this equation to find answers to the questions given in your problem.
I have that formula in my book.
E = L (d I / d t) = I R. The solution to this equation involves time and therefore depends on what is taken to be the starting point. Two situation are of particular interest.

(i) Make. In this case I = 0 when t = 0 and the solution to the equation can be shown to be I = E / R (1 - e- R t / L).

(ii) Break. In this case, if we assume the current has already reached its equilibrium value, I = E / R when t = 0, then the solution to equations can be shown to be I = (E / R) e- R t / L.
I used the second formula, since we don't have I = 0 but I = 2 A and go t = 2.7 s and then d I / d t = 2 / 2.7 = 0.72 A s-1, not 0.5 though...

Update
Hm, I used the first formula and got the same result. Maybe I'm wrong somewhere. Here are both steps:

We have: L = 2 H, R = 0.5 Ω, V = 2 V, I = 2 A.

(i) I = E / R (1 - e- R t / L)
2 = 2 / 0.5 (1 - e- 0.5 t / 2)
0.5 = 1 - e- 0.5 t / 2
-0.5 = - e- 0.5 t / 2
0.5 = e- 0.5 t / 2
ln 0.5 = - 0.5 t / 2
- 0.69 = - 0.5 t / 2
1.39 = 0.5 t
t = 2.77 s.

d I / d t = 2 / 2.77 = 0.72 A s-1.

(ii) I = (E / R) e- R t / L
2 = (2 / 0.5) e- 0.5 t / 2
0.5 = e- 0.5 t / 2
ln 0.5 = - 0.5 t / 2
- 0.69 = - 0.5 t / 2
- 1.39 = - 0.5 t
t = 2.77 s

d I / d t = 2 / 2.77 = 0.72 A s-1.
 
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  • #10
cnh1995
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d I / d t = 2 / 2.77 = 0.72 A s-1.
That's not how you compute dI/dt.
You have the expression for i(t). Differentiate it w.r.t time and then plug in the values of R, L and E in it.
 
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  • #11
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That's not how you compute dI/dt.
You have the expression for i(t). Differentiate it w.r.t time and then plug in the values of R, L and E in it.
t = 2.77 s

d I / d t = n * 2.77n - 1 A s-1.

Like this?
 
  • #12
cnh1995
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t = 2.77 s

d I / d t = n * 2.77n - 1 A s-1.

Like this?
Since you have not studied calculus yet in school, I think I will have to do it for you here.
I(t)=(E/R)*(1-e-tR/L).
∴ dI/dt= (E/L)*(e-Rt/L).
Plug in the values of E, R, L and t. You have already calculated t.
 
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  • #13
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Since you have not studied calculus yet in school, I think I will have to do it for you here.
I(t)=(E/R)*(1-e-tR/L).
∴ dI/dt= (E/L)*(e-Rt/L).
Plug in the values of E, R, L and t. You have already calculated t.
Ha, that indeed gets 0.5 A s-1. But how did you change the first formula into the second one?
 
  • #14
cnh1995
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Ha, that indeed gets 0.5 A s-1. But how did you change the first formula into the second one?
Derivative of Aekt w.r.t. time= Ak*ekt and derivative of a constant=0. You have the equation in the form,
i(t)=A(1-ekt)=A-A*ekt.
Derivative of the first term will be zero and derivative of the second term will be as above. This all will be taught to you in the calculus class, from the very basic concepts.
 
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  • #15
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Derivative of Aekt w.r.t. time= Ak*ekt and derivative of a constant=0. You have the equation in the form,
i(t)=A(1-ekt)=A-A*ekt.
Derivative of the first term will be zero and derivative of the second term will be as above. This all will be taught to you in the calculus class, from the very basic concepts.
Thank you.

What's wrong with the last part?
(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.
 
  • #16
cnh1995
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Also, if you see the graph of the current as a function of time, you'll see it is an exponentially increasing curve starting from i=0 at t=0. At t=0, the current is zero, meaning the inductor acts as an open circuit. Then the current starts rising slowly and at t=infinity (or after a long enough time) it settles to a value I=E/R. Inductor opposes a sudden change in current through it.
 
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  • #17
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Also, if you see the graph of the current as a function of time, you'll see it is an exponentially increasing curve starting from i=0 at t=0. At t=0, the current is zero, meaning the inductor acts as an open circuit. Then the current starts rising slowly and at t=infinity (or after a long enough time) it settles to a value I=E/R. Inductor opposes a sudden change in current through it.
Yes, this part I understand. The iron-cored coil slows down the current (the current is opposed by the back EMF in the coil) and when the circuit is closed the current is zero at first.

But how do we approach part (b)?
(b) A long air-cored solenoid has 1000 turns of wire per metre and a cross-sectional area of 8.0 cm2. A secondary coil, of 2000 turns, is wound around its centre and connected to a ballistic galvanometer, the total resistance of coil and galvanometer being 60 Ω. The sensitivity of the galvanometer is 2.0 divisions per microcoulomb. If a current of 4.0 A in the primary solenoid were switched off, what would be the deflection of the galvanometer?
Is some part of my initial calculation correct?
(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.
 
  • #18
cnh1995
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But how do we approach part (b)?


Is some part of my initial calculation correct?
I think you have mistaken B for the field inside the galvanometer. It is actually the magnetic field in the inductor and has nothing to do with the galvanometer. As the inductor opposes sudden change in current through it, it means it opposes a sudden change in flux linkage too.
Hence, when you turn off the primary supply, the secondary should maintain the initial B-field at the instant of switching off the primary current.
I have not verified the math yet but this should be the approach IMO.
 
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  • #19
cnh1995
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A secondary coil, of 2000 turns, is wound around its centre
Does it have the same area of 8cm2? I guess it does because IMO there's no way we can solve the question without knowing the area or length of the second coil.
 
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I think you have mistaken B for the field inside the galvanometer. It is actually the magnetic field in the inductor and has nothing to do with the galvanometer. As the inductor opposes sudden change in current through it, it means it opposes a sudden change in flux linkage too.
Hence, when you turn off the primary supply, the secondary should maintain the initial B-field at the instant of switching off the primary current.
I have not verified the math yet but this should be the approach IMO.
Does it have the same area of 8cm2? I guess it does because IMO there's no way we can solve the question without knowing the area or length of the second coil.
In terms of A that's how it is said in the first post. I'd guess it's the same and is equal to 8 cm2.

I used the formula for the magnetic flux of coil. If it's not applicable, how do we find the magnetic field then?

θ = a N A B / R -- is this formula correct? I mean we do need to find theta, right? The deflection of the galvanometer?

I used N I = N I to find the current in the coil: 4 * 1000 = I * 2000 so I = 2 A. Then B = μ0 n I = 4 π * 10-7 * 2000 * 2 = 5 * 10-3 T. With this in mind I'll get θ = 2.68 * 10-4 (same as in the first post).
 
  • #21
cnh1995
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Maybe I don't know much about the ballistic galvanometer. I need to study it before I can say anything further.
 
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  • #22
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Any help on (b) please?
(b) A long air-cored solenoid has 1000 turns of wire per metre and a cross-sectional area of 8.0 cm2. A secondary coil, of 2000 turns, is wound around its centre and connected to a ballistic galvanometer, the total resistance of coil and galvanometer being 60 Ω. The sensitivity of the galvanometer is 2.0 divisions per microcoulomb. If a current of 4.0 A in the primary solenoid were switched off, what would be the deflection of the galvanometer?
(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.
 
  • #23
kuruman
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The ballistic galvanometer measures the total charge that goes through it.
1. Find the magnetic flux through the secondary coil before the current in the primary is switched off.
2. Find an expression relating the total change in magnetic flux ΔΦM to the total charge that flows through the secondary (I am not telling you what it is, but I or someone else could help you find it).
3. Answer the question in terms of galvanometer divisions.
 
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  • #24
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1. Find the magnetic flux through the secondary coil before the current in the primary is switched off.
ICoil 1 NCoil 1 = ICoil 2 NCoil 2 so 4 * 1000 = 2000 * ICoil 2 and we have ICoil 2 = 2 A.

B = μ0 nCoil 2 ICoil 2 = 4 π * 10-7 * 2000 * 2 = 5 * 10-3 T (where n = number of turns or N) -- this should be the magnetic flux through the secondary coil when the current in the primary is switched off. I also used the current in the secondary coil which is 2 A and not 4 A.

This should be correct.
 
  • #25
kuruman
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ICoil 1 NCoil 1 = ICoil 2 NCoil 2
This doesn't make sense. Where did you get it? There is no current in the secondary until the magnetic flux through it changes when the primary is switched off. Look at suggestion 1, post #23. What is the magnetic flux through the secondary when current flows only in the primary and before it is switched off?
 
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