Sph. Solenoid Homework: Find Magnetic Field at Center

[kevinnn]

Homework Statement

A large number, N, of closely spaced turns of wire are wound in a single layer upon the surface of a wooden sphere of radius a. The planes of the sphere are perpendicular to the axis of the sphere (take axis to be horizontal), and turns are uniformly spaced per unit circumference and completely cover the surface. The current in the winding is I. Calculate the magnetic field at the center of the sphere.

Homework Equations

Ampere's Law.

The Attempt at a Solution

I have done a few things. Right now I have made my boundary curve go through the center of the sphere and then up and off to infinity. The part I have concerns about is the shape of the solenoid. Is there any integration I have to do when considering the current enclosed? Is the surface integral just (B)(2a)? Thanks.

 

[Charles Link]

The solution to this is a Biot-Savart integral. Ampere's Law will not work because you don't have the symmetry to do a line integral. They ask for the magnetic field at the center of the sphere in this problem, and the field is not uniform throughout the sphere. Setting up and solving the Biot-Savart integral for the magnetic field $B$ is a good exercise using vectors in spherical coordinates.

 

[Charles Link]

And I will give one helpful hint for this problem-it's a little bit of work but not as difficult as it might first appear- because symmetry makes it so that you only need to compute the z-component of the vector cross product in the Biot-Savart integral.

 

[Charles Link]

This one is an interesting calculation for anyone else who wants to give it a try. The answer I got for it turns out to be a little less than the $B$ field in a long cylindrical solenoid. The Biot-Savart integral for it can be readily evaluated with a little calculus.

 

[kevinnn]

Thanks for the reply. I think I almost have it. But, when I am evaluating the integral, I think I'm having a problem with dl. The length, l, is 2(pi)r right? so dl would be 2(pi)dr? When I plug this into the integral and evaluate it from (0 to a) I'm off by a factor of 0.5. What am I missing here?

 

[Charles Link]

Thanks for the reply. I think I almost have it. But, when I am evaluating the integral, I think I'm having a problem with dl. The length, l, is 2(pi)r right? so dl would be 2(pi)dr? When I plug this into the integral and evaluate it from (0 to a) I'm off by a factor of 0.5. What am I missing here?

It's a surface integral in this case, instead of a volume integral. In the volume type integral of Biot-Savart, $J(x') \times (x-x') d^3x'$ appears in the numerator where $J(x')$is the current density term in a volume type integral for Biot-Savart. In this problem though, it is a surface integral of the form $K(x') \times (x-x') dA'$ (in the numerator) where $K(x')$ is the surface current per unit length. ( $x$ and $x'$ are the coordinates of the observation point and the coordinates of the moving electrical charge respectively). The surface current per unit length gets integrated in one direction to get the total current and then integrated along the wire (as in the single wire form of Biot-Savart which is a term like $I (x') \times (x-x') ds'$ (where $I(x')$ is the current)). Thereby a surface integral of the $K(x') \times (x-x') dA'$ form is what you are needing in the numerator. There will not be any "dr" integration. You should be familiar with the form that $dA'$ has in spherical coordinates.

 

[Charles Link]

This one is a good exercise in spherical coordinates along with x,y,z coordinates. The determinant method is probably the easiest way to compute the vector cross product in the numerator of the Biot-Savart integral. Hopefully the OP was able to successfully compute the surface integral.