# Rate of change of current with time, galvanometer deflection

kuruman
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To clarify my previous post, you need a general expression that is valid when the charge and current vary with time.

moenste
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?

kuruman
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Correct. So put together to get a differential equation using the alternate expression for I, Post #41.

moenste
cnh1995
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I = d Q / d t?
Right.

moenste
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).

kuruman
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Multiply both sides of the first equation by dt and integrate. What do you get?

moenste
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?

kuruman
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No. You have an extra dt on the right that doesn't belong.

moenste
cnh1995
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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.

moenste
No. You have an extra dt on the right that doesn't belong.
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?

kuruman
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Bingo. Calculate the right side with numbers and you are done.

moenste
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

cnh1995
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Q = Φ / R = B A N / R?
Right.

moenste
cnh1995
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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.

moenste
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.

cnh1995
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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?

moenste
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.

cnh1995
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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.

moenste