Calculations of Magnetic Fields from Known Current Distributions. B d B2 r 0I B 2 r

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1 Calculations of Magnetic Fields from Known Current Distributions In the absence of magnetic materials this is a relatively simple problem analogous to finding the electric field with known charge distributions. With appropriate symmetry we can use Ampere s Law. Ampere s Law Solutions There are two simple examples. The first is a long straight wire. The we consider an imaginary circle or radius r in a plane perpendicular to the wire, with center on the wire. Then by symmetry the field can depend only on r. There are two possibilities lines moving radially outward as in the electrostatic case, or circles. Since magnetic field lines do not begin and end it must be circles. Then where θ is along the circle as shown I ˆ B d B r I B r Case The second example is a solenoid of length L, radius R << L, with n turns/length. If the solenoid carries a current I we find the magnetic field differently in each of three regions. The first is inside the solenoid, far from the ends. The second is outside the solenoid and far

2 from it. The third is everywhere else. Note that we could always find the field exactly by integrating the Biot-Savart Law. This is different than the analogous case in electrostatics because here we know the current distribution, whereas there we did not precisely know the charge distribution. Region 1 Since we are inside and far from the ends, the field will lie along the axis. We consider an imaginary circuit Then Bd BBrB BrBr ni in out Since B is parallel to the axis we have B =. Since the field lines are closed loops, there are the same number of lines outside as inside. But inside they are confined to the area πr, whereas outside they are spread over all space. Hence, the number/area is much larger inside than outside. Since B is the # lines/area we have B out B in Thus B n I B niz in ˆ where is given by the right-hand rule as applied to one of the current loops Region Now we are far from the solenoid. It therefore looks like a magnetic dipole mnlr Izˆ

3 Field Due to Current Loop We consider a loop of radius R centered at the origin, lying in the xy plane. Then Thus we need Jr'. It is J r' 3 d r' 4 r r' r' R Jr' Icos ˆ r' To see that this is correct, we integrate over the half plane x. This should give I. We find rr rr rdrdicos r as desired. Then Now r I rr dri r' r r Irdrd sin R ˆ I r' R ' Ar r' sin 'dr'd 'd ' cos ' 4 r' rr' r r ' xˆ r sin cos R cos ' yˆ r sin sin R sin ' r cos zˆ 1/ rr' r R rrsincos ' r

4 IR d ' ˆ ' 4 r R rrsincos ' ' 1/ But ˆ ' xsin ˆ ' ycos ˆ ' Hence IR sin 'd ' cos 'd ' xˆ yˆ 1/ 1/ 4 r R rrsin cos ' r R rrsincos ' Because of the azimuthal symmetry we can evaluate this at any. If we choose =,. Then the term can be done at once Hence sin 'd ' sin 'd ' r R rrsin cos ' r R rrsincos ' 1/ 1/ r R rrsin cos ' ˆ IR cos 'd ' 4 r R rrsincos ' This is an elliptic integral. We note that the integral is even. Hence ˆ IR cos 'd ' r R rrsincos ' 1/ 1/

5 Let n = /. Then ˆ IR / 1/ cos n dn r R rrsincos n But Now let n = π/ ε cos n cos n sin n cos n 1 / cos n 1dn ˆ IR 1/ r R rrsin cos n1 sin 1d ˆ IR 1/ r R rr sin 4rR sin sin / where ˆ / IR 1 sin 1 d r R rr sin 1k sin sin 1/ 1/ k 4rR sin r R rrsin / IR ˆ cos K R 1/ 1/ r R rrsin 1k sin

6 Now / cos 1k sin d E k R 1 k 1/ k K k ˆ r R rrsin k k IR E k Ar K K 1/ ˆ r R rrsin k IR E K k K 1/ IRˆ k KE 1/ r R rrsin k where K is the complete elliptic integral of the first kind, and E is the complete elliptic integral of the second kind. We can now find from ˆ ˆ BrArrˆ A ˆ r r rsin A ˆ A ˆ ˆ rˆ ˆ Arsin ˆ ˆcos r r rsin ˆ A 1 A ˆ A Actn ˆ A A 1 A Actn rˆ rˆ rˆ r r r r r r r r This can then be evaluated by calculating A/ r and A/ θ. A mess, but doable. Of course we can always simply evaluate numerically from the Biot-Savat Law. Problem is that this requires a new integral for each point.