Magnetic field of a long curent carrying tube

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

To calculate magnetic field at 1 and 3 current flowing through sections A and C can be assumed as if current is flowing in a long straight wire along the dotted line . Now since the current and distances are same for 1 and 3 , B1 = B3 .There is only one such option i.e option 1) which is indeed the correct answer .

This left me thinking , what if the second option would have been 2) B₁ ≠ B₂ = B₃ .Then I would have to choose between option 1) and 2) . I need some argument to believe that the magnetic field at 2 would have been equal to 1 and 3 . So how would I deal with the current flowing through the conical part B ?

 

[cnh1995]

. So how would I deal with the current flowing through the conical part B ?

Have you studied Ampere's circuital law?
(Before that, what can you say about the currents at A, B and C?)

 

[Jahnavi]

(Before that, what can you say about the currents at A, B and C?)

Currents are same .

Have you studied Ampere's circuital law?

You are right . This calls for Ampere's Law .Assuming a circular loop around the conical part , I was actually thinking about some sound reasoning as to why we could take magnetic field B out of the integral .

Two things I need to reason .

1) Direction of magnetic field

2) Is magnetic field constant around the loop .

I think I need to exploit some symmetry arguments .

 

[Charles Link]

@Jahnavi I looked at this one, and I think @cnh1995 has a good answer. Ampere's law is really sufficient for this one, even though the conic section disrupts the symmetry. I do think some lengthy symmetry arguments involving Biot-Savart's law might show that $B_1=B_2=B_3$. On the other hand, there is a current feeding into the center, and in general any time there is a current, there is generally a magnetic field from it. Does the contribution from the current that is moving radially inward have zero magnetic field everywhere? I'm not completely sure, but the question is really far more advanced than I think was intended where they simply want you to apply Ampere's law. Ampere's law has $\oint \vec{B} \cdot d \vec{l}$ with a dot product, so it gives no information about a component of $\vec{B}$ that could potentially exist perpendicular to $d \vec{l}$, i.e. in the forward direction. $\\$ It's an interesting question, but giving a conclusive answer to whether this cone might create a small additional magnetic field does not seem to have an obvious answer.

 

[Jahnavi]

Consider a circular loop around the conductor .It might be through point 1 , 2 or 3 .

Now consider any two points on this circular path . For both the points , the conductor as well as current will look like same . This means magnitude of magnetic field has to be equal at each point of the circular path and the direction has to be tangential at each point of the path . From this simple argument we can take B out of the integral in the Ampere's Law . Now , when we calculate the magnetic field it will be equal at 1 , 2 and 3 .

@Charles Link what do you think ?

 

[Charles Link]

Consider a circular loop around the conductor .It might be through point 1 , 2 or 3 .

Now consider any two points on this circular path . For both the points , the conductor as well as current will look like same . This means magnitude of magnetic field has to be equal at each point of the circular path and the direction has to be tangential at each point of the path . From this simple argument we can take B out of the integral in the Ampere's Law . Now , when we calculate the magnetic field it will be equal at 1 , 2 and 3 .

@Charles Link what do you think ?

Ampere's law only gives the $\vec{B} \cdot d \vec{l}$ component. I'm not convinced there is no $\hat{z}$ component, especially for case 2, but taking on such a task to determine whether such a component might exist, I think, is a very formidable task.

 

[Merlin3189]

... Now since the current and distances are same for 1 and 3 , B1 = B3 .There is only one such option i.e option 1) which is indeed the correct answer .

... if the second option would have been 2) B₁ ≠ B₂ = B₃ .Then I would have to choose between option 1) and 2) . ...

Not what you asked, but B1=B3 means B1 ≠ B2 = B3 is false and you're back to the first option.

 

[Jahnavi]

Ampere's law only gives the $\vec{B} \cdot d \vec{l}$ component.

From symmetry , haven't we argued that B is tangential at all points of the circular path ?

 

[Charles Link]

From symmetry , haven't we argued that B is tangential at all points of the circular path ?

The cone disrupts the symmetry in the $\hat{z}$ direction. There is still $\phi$ symmetry, but not the $\hat{z}$ symmetry. For a single long wire carrying a current, the Biot-Savart equation (with z anti-symmetry) is actually needed (edit: or can be used) to show $B_z=0$ $\\$ (Alternatively, all the current is in the z-direction for a single wire with no cone, so $\vec{B}$ must be perpendicular to $J_z \hat{z}$ . With a cone, you start introducing $J_x$ and $J_y$. And I don't know what the precise answer is. The component $B_{\phi}$ remains the same, but I can't immediately rule out the possibility of a $B_z$ component).

 

[Jahnavi]

Not what you asked, but B1=B3 means B1 ≠ B2 = B3 is false and you're back to the first option.

Nice !

Thanks .

What if option was B1= B3 ≠ B2 ?

 

[Merlin3189]

No problem then.

But I'll have to leave it to others to show what B2 is.