Can someone help me explain these two problems better?

[motornoob101]

A friend of my has trouble understanding these two questions. I tried to explaining it to her but my explaining skill kind of sucks so I am wondering if someone can explain these questions better.

First question is this.
4- A very long solenoid with a circular cross section and radius r1= 2.80 centimeters with ns= 280 turns/cm lies inside a short coil of radius r2= 3.40 centimeters and Nc= 26 turns.

http://capa.mcgill.ca/res/mcgill/dc...uphysislib/Graphics/Gtype66/prob05a_coils.gif
If the current in the solenoid is ramped at a constant rate from zero to Is= 1.70 Amperes over a time interval of 64.0 milliseconds, what is the magnitude of the emf in the outer coil while the current in the solenoid is changing?

She was wondering why we use the radius/area of the solenoid in calculating the change in magnetic flux rather than say the area of the big coil. She thought that the magnetic field lines from the solenoid would spread out and affect the big coil.

Here is her original question "why don't we use the second radius...the one of the outer coil. I know inside the solenoid the magnetic field is constant, but won't the outer coil feel a smaller magnetic field, and therefore flux, because it is at a certain distance from the inner solenoid? Or does flux only have to do with the change, and the change will be proportionally the same, no matter the distance?"

The way I said is that.. Magnetic field lines inside a solenoid is like this..

http://p3t3rl1.googlepages.com/pquestion1.jpg

So when you put the solenoid inside the big coil (side cross sectional view), it only affects the area that belongs to the solenoid and no where else..

http://p3t3rl1.googlepages.com/pquestion2.jpg

She thought the magnetic field lines would go off like these and affect the big coil...
http://p3t3rl1.googlepages.com/pquestion3.jpg

Well my explanations didn't really work as she is still confused about it.

Can anyone reword what I said so that she can understand it?

Also, she is wondering why we can't use the mutual inductance equation to solve it..

The equation is.. $$E= -L\frac{\Delta I}{\Delta t}$$

If you can find L, then it should work right?

Again, here is her original question.."Apart from the fact that you need the length of the solenoid to calculate the inductance using that method, why, conceptually, can you not do it that way?"

I have no clue. Can anyone explain why conceptually it is a wrong method?

Ok, the second question is something like this..

The magnetic field B at the center of a circular coil of wire carrying current I is

$$B=\frac{\mu NI}{2r}$$

The question asks.. If you use a greater number of turns and this same power supply (120 V), will a greater magnetic field strength result?

In class the prof explained B ~ NI. R ~ N ( more length of wire, higher resistance) therefore I ~ 1/N. Thus, the increase in B field strength caused by greater N turns would be canceled by the reduced current exactly so that the magnetic field stays constant. The thing is though, $$R = \rho\frac{L}{A}$$. Different wires have different resitivity, so wouldn't some wire experience less of a decrease a current when more turns are wrapped around, so that the net affect is increase in B field?

Thanks and have a good day!

 

[motornoob101]

Ok, guess no one can explain it? :(

 

[ctjen]

The lines of flux from the solenoid, the primary coil, would envelope the secondary coil.

Look at this Wiki article on Coupled Inductors. http://endotwikipediadotorg/wiki/Mutual_inductance#Coupled_inductors

Chris

 

[ctjen]

.


Ok, the second question is something like this..

The magnetic field B at the center of a circular coil of wire carrying current I is

$$B=\frac{\mu NI}{2r}$$

The question asks.. If you use a greater number of turns and this same power supply (120 V), will a greater magnetic field strength result?

In class the prof explained B ~ NI. R ~ N ( more length of wire, higher resistance) therefore I ~ 1/N. Thus, the increase in B field strength caused by greater N turns would be canceled by the reduced current exactly so that the magnetic field stays constant. The thing is though, $$R = \rho\frac{L}{A}$$. Different wires have different resitivity, so wouldn't some wire experience less of a decrease a current when more turns are wrapped around, so that the net affect is increase in B field?

Thanks and have a good day!

The link below shows the derivation of:

$$B=\frac{\mu NI}{2r}$$

http://info.ee.surrey.ac.uk/Workshop/advice/coils/air_coils.html#fsl

http://en.wikipedia.org/wiki/Inductor#Q_factor discusses a lossless scenario in the last paragraph.

Chris

 

[motornoob101]

I don't really quite understand these as these pages you linked me to are quite advanced physics.. I am only taking introductory E&M (no calc) so these concepts are quite beyond me but thanks for trying!

 

[ctjen]

A friend of my has trouble understanding these two questions. I tried to explaining it to her but my explaining skill kind of sucks so I am wondering if someone can explain these questions better.

First question is this.
4- A very long solenoid with a circular cross section and radius r1= 2.80 centimeters with ns= 280 turns/cm lies inside a short coil of radius r2= 3.40 centimeters and Nc= 26 turns.

http://capa.mcgill.ca/res/mcgill/dcmcgill/oldproblems/msuphysislib/Graphics/Gtype66/prob05a_coils.gif
If the current in the solenoid is ramped at a constant rate from zero to Is= 1.70 Amperes over a time interval of 64.0 milliseconds, what is the magnitude of the emf in the outer coil while the current in the solenoid is changing?

She was wondering why we use the radius/area of the solenoid in calculating the change in magnetic flux rather than say the area of the big coil. She thought that the magnetic field lines from the solenoid would spread out and affect the big coil.

Here is her original question "why don't we use the second radius...the one of the outer coil. I know inside the solenoid the magnetic field is constant, but won't the outer coil feel a smaller magnetic field, and therefore flux, because it is at a certain distance from the inner solenoid? Or does flux only have to do with the change, and the change will be proportionally the same, no matter the distance?"

The way I said is that.. Magnetic field lines inside a solenoid is like this..

http://p3t3rl1.googlepages.com/pquestion1.jpg

So when you put the solenoid inside the big coil (side cross sectional view), it only affects the area that belongs to the solenoid and no where else..

http://p3t3rl1.googlepages.com/pquestion2.jpg

She thought the magnetic field lines would go off like these and affect the big coil...
http://p3t3rl1.googlepages.com/pquestion3.jpg

Well my explanations didn't really work as she is still confused about it.

Can anyone reword what I said so that she can understand it?

Also, she is wondering why we can't use the mutual inductance equation to solve it..

The equation is.. $$E= -L\frac{\Delta I}{\Delta t}$$

If you can find L, then it should work right?

Again, here is her original question.."Apart from the fact that you need the length of the solenoid to calculate the inductance using that method, why, conceptually, can you not do it that way?"

I have no clue. Can anyone explain why conceptually it is a wrong method?

Ok, the second question is something like this..

The magnetic field B at the center of a circular coil of wire carrying current I is

$$B=\frac{\mu NI}{2r}$$

The question asks.. If you use a greater number of turns and this same power supply (120 V), will a greater magnetic field strength result?

In class the prof explained B ~ NI. R ~ N ( more length of wire, higher resistance) therefore I ~ 1/N. Thus, the increase in B field strength caused by greater N turns would be canceled by the reduced current exactly so that the magnetic field stays constant. The thing is though, $$R = \rho\frac{L}{A}$$. Different wires have different resitivity, so wouldn't some wire experience less of a decrease a current when more turns are wrapped around, so that the net affect is increase in B field?

Thanks and have a good day!

Almost only counts in horseshoes...I hope this is more helpful.

I wasn't able to log into the website given above regarding the original problem so I might be missing some information but based on the problem as written above there isn't enough information to derive a numerical answer. Without the length of the coils you cannot determine their inductance.


Assuming that the solenoid is an air-core coil the inductive coupling to the secondary coil is weak and neither the mutual inductance formula cited or the well known turns ratio formula will work as they both assume very close coupling between the coils (as is found in coils cored with high permeability magnetic materials). The mutual inductance between the two coils not being unity would have to be determined from the actual inductances of the coils and other empirical data or by direct measurement. The mutual inductance is proportional to the square root of the the product of the two inductances being coupled and a coefficient of coupling which is between one and zero. For air-core coils the coefficient of coupling will not be anywhere close to one as would be the case in a high permeability cored coil. The outside of an air-core coil does see a weak magnetic field so the closeness of the secondary coil to the primary coil is important in regard to the degree of coupling that occurs. Adding extra turns to a single layer air-core solenoid (solenoid meaning pipe shape) is counter productive at some point because of the increase in resistance of the added wire which requires more current (to compensate for resistive losses) which requires thicker wires. Thicker wire increases the length of the coil without increasing the number of turns (lengthening a coil reduces the inductance). As you can see the whole process ends up heading away from the original goal of increasing the magnetic flux density.

Chris