# Homework Help: Magnetic field in a solenoid due to induced current in a coil

1. Mar 23, 2013

### CAF123

1. The problem statement, all variables and given/known data
A coil of wire with N turns is placed around the a long solenoid with n turns per unit length carrying a current I as shown below,

A)If a current I flows through the solenoid, derive an expression for the magnetic flux through the coil and explain why the size and shape of the coil is not important.

B)If the current in the solenoid is varied with time as given by the expression $I(t) = I_o + I_1t + I_2t^2$ and the resistance of the coil is R, derive an expression for the induced current in the external coil of wire.

2. Relevant equations
Ampere's Law
Magnetic Flux

3. The attempt at a solution
The magnetic field inside the solenoid is a constant $\mu n I$. So the magnetic flux through the solenoid is $\mu n I (\pi r^2).$ Since we want the magnetic flux through the coil only this becomes $\mu N I (\pi r^2)$(1).

However, the magnetic field caused by the current in the solenoid loop will induce a current inthe coil in the opposite direction to that of the loop in the solenoid.
By Ampere's Law, take a circle of radius smaller than that of the coil. The enclosed current is $N I$ and so $\int B \cdot ds = \mu N I \Rightarrow B = \mu N I/(2 \pi R), R$ the radius of the Amperian loop.

The coil (I assume) is perfectly symmetrical around the solenoid. So all B field contributions from the wire cancel. So B flux is given by (1). Is it correct?

If we take an Amperian loop to coincide with the radius of the solenoid, then no matter what shape the coil is, the enclosed current will always be the same. (But I don't see how this implies the B field would be same though). When they talk about size, do they mean a coil with a radius greater than that of solenoid?

2) Induced emf ,$E = -\dot{\phi} = -\mu N r^2 \pi [I_1 + 2 I_2 t],$ so $I_{ind} = E/R = -\mu N r^2 \pi [I_1 + 2I_2 t]/R$. (minus because current in opp direction)

Many thanks

#### Attached Files:

• ###### Solenoid.png
File size:
578 bytes
Views:
365
Last edited: Mar 23, 2013
2. Mar 23, 2013

### Staff: Mentor

Only if I is changing (or your coil is superconducting).
For (A), the coil does not change the magnetic flux.

A radius, or any similar parameter (side length for a rectangle, ...).

3. Mar 23, 2013

### CAF123

Does this mean that if the current in the solenoid loop was constant, then there would be no induced current in the coil?
Because it does not have an induced current and thus no changing magnetic field and so does not contribute anything to the flux?

I'd just like to check a couple of things:
-There is no changing current in (A) and so no induced current in the coil => no B field from the wire. So the shape of the coil does not matter because by Ampere's Law, I can always take a circle coinciding with the radius of the solenoid. The enclosed current is NI => the flux is μNIπr2. The size of the coil can range from r to ∞ since the only contribution of the flux is from inside the solenoid.
-Now say there was an induced current in the coil. The above does not hold because the magnetic field (generated by coil)will depend on the shape of the coil. Even though I can still take my Amperian loop to be of radius r, different shapes of the coil will result in differing magnetic fields on the surface of the Amperian loop and so affect B.ds

Thanks for the check.

4. Mar 23, 2013

### Staff: Mentor

You can use your answer in (B) to see it yourself ;).

This (I removed one word).

Exactly at the solenoid? Why?
You need a different shape to get an enclosed current.
Right

Right.

5. Mar 23, 2013

### CAF123

I suppose it doesn't have to be. Maybe a radius just a bit bigger than the radius of the solenoid.

That would be a cylinder of radius r (r as noted above, just a bit bigger than the radius of solenoid) and height h, h being such that all N turns are contained.

6. Mar 24, 2013

### CAF123

I have another quick question. (I would have posted it in another thread, but I am sticking with the solenoid).

Problem Statement, known variables and given data
A solenoid with n terns per unit length and cross-sectional area A carries a current I(t) that varies at I(t) = Io exp (−αt). Use the Faraday’s Law (i.e $\oint E \cdot ds = -\dot{\phi}$) to derive an expression for the induced E -field outside the solenoid a distance r from the axis.

Attempt at a solution

If we are looking at a distance r from the axis, and outside the solenoid then r > R, R the radius of the solenoid. I think Faraday's Law in the above form only holds for closed paths, but to get an enclosed current I would need to make a cylinder for example. So take this cylinder and make it at a radius greater than the radius of the solenoid and at a height the length of the solenoid. (This may not actually be needed for this question though).

Since there is a current I(t) in the solenoid coil, there will be changing magnetic field in the wire. I believe this changing magnetic field causes a changing magnetic flux and by Faraday's law, an induced electric field. Now take a circle of radius r>R. Inside the solenoid, I can see that E would be parallel to ds but I am not sure outside the solenoid.

If I approximate E parallel to ds all the way round the circle then by Faradays Law, then $$E (2 \pi r) = \mu n \pi R^2 I_o \alpha exp(-\alpha t),$$ but how can I be sure that E will be parallel to ds outside the solenoid?

7. Mar 24, 2013

### Staff: Mentor

Faraday's Law needs a closed path, not a surface. A circle is fine.
The magnetic field in the solenoid is relevant here, not the field in a wire.
E and ds are aligned, this is given by symmetry.

8. Mar 24, 2013

### CAF123

Okay, so the same holds for Ampere's Law (necessarily a closed path not a closed surface)? But in the previous question, I took a cylinder to get the enclosed current.

I think I can see why E and ds would be parallel for the portion of the circle inside the solenoid, but for the portion where r>R, there is no coil and so I don't see how we still know E is parallel to ds.

9. Mar 24, 2013

### Staff: Mentor

Then you did it wrong ;).

There is no part of the circle inside the solenoid. It is a circle around the solenoid.

10. Mar 24, 2013

### CAF123

In one of your posts, you wrote:
In that question, the circle (which I chose initially) will not enclose the N turns of the coil. So if I can't choose a 3D shape, then how am I supposed to find the enclosed current?

So if there is only a current in the solenoid, and E is tangential to the wire always, how do I necessarily know that E is aligned with ds (ds on circle of r>R) outside the solenoid as well? Since the loops in the solenoid are circular with axis through middle of solenoid, I take a circle also with axis through middle of solenoid, but just with a bigger radius. Can I just translate the E field vectors tangential to the wire to outside the solenoid to the circle and then see from there it is always parallel with ds? (Hope that made sense - if not I will try to explain better)
Thanks!

11. Mar 24, 2013

### Staff: Mentor

To get an enclosed current in an area, you can choose a long rectangle where all turns cross the rectangle once.
Symmetry, and the absence of net charges. A circular E-field is the only option.

12. Mar 24, 2013

### CAF123

So like a rectangle parallel to the solenoid axis with the turns intersecting only once?

It makes sense. At point r, will the E field be pointing into the page then?(assuming a clockwise current direction in the coil in the solenoid)

At the point r, there will be a magnetic flux contribution from the n turns per unit length in the solenoid. I've just noticed that the question only says a cross sectional area A, and so this may not even be circular. (But I believe the above discussion still holds since the coils will still be circular). So $$E(2 \pi r) = A\mu n I_o \alpha exp(-\alpha t),$$ is the expression I want, right?

13. Mar 25, 2013

### CAF123

Does the above all make sense?

Also, in the OP I wrote:

Should this be $\mu n N I (\pi r^2)?$ It seems more reasonable.

14. Mar 25, 2013

### Staff: Mentor

Right, that factor of n was missing there.

Everything else: Looks fine, I think.