By Roel Snieder

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**Example text**

5). Because of the cylinder symmetry of the problem, we know that the magnetic field is in the direction of the unit vector ϕ ˆ and that the field only depends on the distance r = x2 + y 2 to the wire: B =B(r)ˆ ϕ. 5). When the disc is larger than the thickness of the wire the surface integral of J gives the electric current I through the wire: I = J·dS. Problem a: Use these results and Stokes’ law to show that: B= µ0 I ϕ ˆ . 13) is now determined. Note that the magnetic field depends only on the total current through the wire, but that is does not depend on the distribution of the electric current density J within the wire as long as the electric current density exhibits cylinder symmetry.

THE THEOREM OF GAUSS In this expression, ¯h is Planck’s constant h divided by 2π. Problem a: Check that Planck’s constant has the dimension of angular momentum. 626 × 10−34 kg m2 /s. Suppose we are willing to accept that the motion of an electron is described by the Schr¨odinger equation, then the following question arises: What is the position of the electron as a function of time? According to the Copenhagen interpretation of quantum mechanics this is a meaningless question because the electron behaves like a wave and does not have a definite location.

The theorem of Stokes tells us how to do this. 1). 2) that we write in a slightly different form as: dS v · dr = ( × v) · n ˆ dS = ( × v) ·dS . 2) is that in the expression above we have not aligned the zaxis with the vector × v. 1). 1) holds for an infinitesimal surface area. However, this expression can immediately be integrated to give the surface integral of the curl over a finite surface S that is bounded by the curve C: C v · dr = S (∇ × v) ·dS . 2) 60 CHAPTER 7. THE THEOREM OF STOKES This result is known as the theorem of Stokes (or Stokes’ law).

### A Guided Tour of Mathematical Physics by Roel Snieder

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