Finding Vector Potential of Infinite Solenoid

AI Thread Summary
To find the vector potential of an infinite solenoid with n turns per unit length, radius R, and current I, the magnetic field inside the solenoid is uniform and given by B = (n)(I)/((ε₀)(c²)). The vector potential A can be derived using the relationship B = ∇ × A, leading to the equation A = (n)(I)(R²)/((2)(ε₀)(c²)(r)) for points outside the solenoid. The line integral of A around a closed path equals the flux of B through that path, which simplifies the calculations. The discussion emphasizes the application of Stoke's theorem and the importance of choosing an appropriate closed curve for integration.
heman
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How to find the Vector Potential of an infinite solenoid with n turns per unit length,radius R and current I.
since here current extends to infinty..
How will it be done
Pls Help
 
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Heman, please try to write the equation for vector potential. Then we can better point you in the right direction.

You can use LateX - see the thread on Math applications of LateX - https://www.physicsforums.com/showthread.php?t=8997 (General physics forum).
 
The infinite solenoid has a uniform magnetic field parallel to the solenoid axis. Let that axis be the z axis, for concreteness. So \mathbf {B}=B \mathbf {k}. Now the vector potential is \mathbf {B}=\nabla \times \mathbf {A}. The vector potential is not unique, so any vector-valued function \mathbf {A} whose curl gives you B \mathbf{k} will fit the bill.
 
You may consider using the following:
gradXA = B
then use Stoke's theorem to write surface int(gradXA.dS) = line int (A.dl), with the closed curve chosen suitably.

This gives line int(A.dl) = int(B.ds).

The surface integral is straight forward while a reasonable choic eof the closed curve makes the line integral staright forward.

Of help?
 
heman said:
How to find the Vector Potential of an infinite solenoid with n turns per unit length,radius R and current I.
since here current extends to infinty..
How will it be done
Pls Help

The line integral of A around a closed path is equal to the flux of B through the path. For an infinitely long solenoid, B = (n)(I)/((eps0)(c^2)) at internal points. (B = 0 at all points outside of the solenoid.) Thus the flux of B inside the solenoid is (pi)(R^2)(n)(I)/((eps0)(c^2)). At a distance r>R from the solenoid's axis, (2)(pi)(r)(A)=flux of B. That is,

A=(n)(I)(R^2)/((2)(eps0)(c^2)(r))
 
jimmy neutron said:
Of help?

I don't think so. In fact the thread is 5 years old. It might be of help for further people though.
 
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