Rate of change of current with time, galvanometer deflection

[kuruman]

Φ = B A or Φ = B A N?

Which do you think is correct and why?

 

[moenste]

Which do you think is correct and why?

We have the number of turns, so Φ = B A N.

 

[kuruman]

Correct. Now can you find a general expression for the induced current in the secondary when the flux through it is changing? Just an expression, don't worry about the numbers.

 

[moenste]

Correct. Now can you find a general expression for the induced current in the secondary when the flux through it is changing? Just an expression, don't worry about the numbers.

Something like Φ = B I A N?

 

[kuruman]

Something like Φ = B I A N?

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?

 

[moenste]

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?

 

[kuruman]

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

 

[moenste]

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.

 

[kuruman]

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.)

 

[moenste]

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)?

 

[kuruman]

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?

 

[moenste]

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?

 

[cnh1995]

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.

 

[kuruman]

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?"

 

[moenste]

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:

(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.

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.

 

[cnh1995]

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.

 

[cnh1995]

Q charge = M I / R, where M is standard mutual inductance.

No. What is the general relationship between charge and current?

 

[kuruman]

No. What is the general relationship between charge and current?

If you don't remember look it up.

 

[moenste]

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.

No. What is the general relationship between charge and current?

Q = I t.

 

[kuruman]

No, look it up.

 

[kuruman]

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]

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

 

[cnh1995]

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]

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]

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

 

[cnh1995]

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?