What charge circulates through the coil?

[moenste]

Homework Statement

Describe and account for two constructional differences between a moving-coil galvanometer used to measure current and the ballistic form of the instrument.

An electromagnet has plane-parallel pole faces. Give details of an experiment, using search coil and ballistic galvanometer of known sensitivity, to determine the variation in the magnitude of the magnetic flux density (magnetic induction) along a line parallel to the pole faces and mid-way between them. INdicate in qualitative terms the variation you would expect to get.

A coil of 100 turns each of area 2.0 * 10^-3 m² has a resistance of 12 Ω. It lies in a horizontal plane in a vertical magnetic flux density of 3.0 * 10^-3 Wb m^-2. What charge circulates through the coil of its ends are short-circuited and the coil is rotated through 180 ° about a diametral axis?

Answer: 100 μC.

2. The attempt at a solution
We need to find charge Q. Q = N A B / R. We don't know B. B = (d Φ / d A) * A = 3 * 10^-3 * 2 * 10^-3 = 6 * 10^-6 T.

So Q = 100 * 2 * 10^-3 * 6 * 10^-6 / 12 = 1 * 10^-7 C.

The correct answer is 100 * 10^-6 C or 10^-4 or 0.00 01. While my answer is 10^-7 or 0.00 00 00 1.

What's wrong?

 

[BvU]

Hi Moenste,

You multiply the flux density with the area twice !
1 Wb/m2 = 1 T

And if you turn the coil over 180 degrees, how much does the flux change ?

 

[moenste]

Hi Moenste,

You multiply the flux density with the area twice !
1 Wb/m2 = 1 T

And if you turn the coil over 180 degrees, how much does the flux change ?

Hi BvU,

Hm, you mean that d Φ / d A = B? Update: T = kg / s² A and Wb / m² = kg m² / m² s² A = kg / s² A = T.

Q = 100 * 2 * 10^-3 * 3 * 10^-3 / 12 = 5 * 10^-5 C.

Not sure, but maybe it should become negative?

 

[BvU]

Let me ask stepwise: what's the change in flux if you rotate 90 degrees ? And then another 90 degrees ?

(and yes,: if it becomes negative, what's the total change ?)

(By the way, I now end up a factor of 10 too high compared to the book; my mistake ?)​

 

[moenste]

Let me ask stepwise: what's the change in flux if you rotate 90 degrees ? And then another 90 degrees ?

(and yes,: if it becomes negative, what's the total change ?)

(By the way, I now end up a factor of 10 too high compared to the book; my mistake ?)​

The answer is 100 μC, not 10 μV, I was changing that while you replied. Sorry for that.

 

[BvU]

Factor 2 clear ? (if so : Well done ! )

(hehe o and colon gives unintended egg on face )

 

[moenste]

Let me ask stepwise: what's the change in flux if you rotate 90 degrees ? And then another 90 degrees ?

(and yes,: if it becomes negative, what's the total change ?)

(By the way, I now end up a factor of 10 too high compared to the book; my mistake ?)​

Factor 2 clear ? (if so : Well done ! )

(hehe o and colon gives unintended egg on face )

No, still calculating. Just fixed the book answer.

Φ = A B cos θ = 2 * 10^-3 * 3 * 10^-3 cos 180 ° = - 6 * 10^{- 6} Wb?

 

[BvU]

$\Delta \Phi = ...\cos \pi \ - ...\cos 0$ !

 

[moenste]

$\Delta \Phi = ...\cos \pi \ - ...\cos 0$ !

Maybe every 90 ° the flux is equal to zero?

But in that case Q = N A B / R = N A * 0 / R = 0.

And if we found the charge Q (C) already, why do we need to find flux Φ (Wb)?

Update
Maybe we need to update the formula to: Q = (N A B cos θ) / R = [100 * (2 * 10^-3) * (3 * 10^-3) * cos 180 °] / 12 = - 5 * 10^-5 C?

 

[BvU]

We needed that to find the charge of 10^-4 C that circulates.

 

[moenste]

We needed that to find the charge of 10^-4 C that circulates.

I'm sorry, still don't quite understand. Could you please tell what is wrong and what is correct in my calculations? At this moment I'm not sure whether something is correct or everything is wrong.

I also don't understand what you mean "find charge that circulates". Q = (N A B cos θ) / R isn't it it?

 

[BvU]

Q = (N A B cos θ) / R isn't it it?

No. If you hold the doil still, no charge flows. $$Q =\Delta (N A B \cos \theta) / R$$

 

[moenste]

No. If you hold the doil still, no charge flows. $$Q =\Delta (N A B \cos \theta) / R$$

The only difference is delta: Q = [(N A B cos θ) / R] - [(N A B cos θ) / R]. But what do we plug into the second pair? Only the angle change from 0 to 180 °.

Q = [(N A B cos θ) / R] - [(N A B cos θ) / R] = - 5 * 10^-5 - 5 * 10^-5 = - 1 * 10^-4 C. A negative one...

 

[BvU]

Correct. In fact I should have placed N and R in front of the $\Delta$

 

[moenste]

Correct. In fact I should have placed N and R in front of the $\Delta$

Hm, so it's Q = N A B [Δ (cos θ)] / R?

Q = N A B (cos θ_Initial - cos θ_Final) / R = [100 * (2 * 10^-3) * (3 * 10^-3) * (cos 0 ° - cos 180 °)] / 12 = 1 * 10^-4 C.

Correct?

 

[BvU]

Must be. But you should convince yourself that it is (and thereby no longer have to ask).

PF isn't very good at stamp-approving things (gets them in trouble with teachers and such).

However, if you want an opportunity for improvement: put absolute sign delimiters around things to avoid a red minus sign afterwards

 

[moenste]

Must be. But you should convince yourself that it is (and thereby no longer have to ask).

PF isn't very good at stamp-approving things (gets them in trouble with teachers and such).

However, if you want an opportunity for improvement: put absolute sign delimiters around things to avoid a red minus sign afterwards

It could be rephrased as " did I understand it correctly?" : ).

 

[BvU]

I think you did..