Magnetostatics III Magnetic Vector Potential (Griffiths Chapter 5: Section 4)

Transcription

1 Dr. Alain Brizard Electromagnetic Theory I PY ) Magnetostatics III Magnetic Vector Potential Griffiths Chapter 5: Section ) Vector Potential The magnetic field B was written previously as Br) = Ar), 1) where the magnetic vector potential is defined in terms of the current density Jr) as Ar) = µ From this definition, Ampère s Law becomes Ar) = µ Jr), V Jr ) dτ r r. ) whose solution is given by Eq. ). Note that, like the magnetic field B, the divergence of the magnetic vector potential vanishes: A =. Example I: Current Loop As a first application of the vector-potential formalism, we consider the case of the vector potential due to a single-turn current loop of radius R carrying current I see Figure below). The current density in this case is expressed in terms of the source point r = R cos ϕ x + sin ϕ ŷ) 1

2 as Jr ) = I R δcos θ ) δr R) sin ϕ x + cos ϕ ŷ), and, by azimuthal symmetry, the vector potential has to be independent of the azimuthal angle ϕ so that we may place the field point on the x, z)-plane: r = r sin θ x + cos θ ẑ). Hence, we find and Eq. ) becomes r r = r + R rr sin θ cos ϕ, Ar, θ) = µ IR π sin ϕ x + cos ϕ ŷ) dϕ r + R rr sin θ cos ϕ. Since the integration is symmetric about ϕ =, only the y-component of the vector potential A remains, which represents the component in the ϕ-direction, i.e., Ar, θ) = A ϕ r, θ) ϕ, where A ϕ r, θ) = µ IR π cos ϕ dϕ r + R rr sin θ cos ϕ. ) Although we are not interested in evaluating this integral explicitly, we can look at two interesting limits: r R and r R. In both limits with r < /r > = R/r or r/r), we use the expansion r + R rr sin θ cos ϕ ) 1/ 1 = 1 1 r > r< r> + r ) < sin θ cos ϕ +, r > to find A ϕ r, θ) = µ IR π dϕ cos ϕ 1 1 r< + r ) < sin θ cos ϕ + r > r> r > µ ) I Rr< sin θ. ) r > The vector potential in the far region r R) is, therefore, given by the approximate expression A ϕ r, θ) µ IπR sin θ, r for r R), 5) while in the near region r R), we find A ϕ r, θ) µ I r R ) sin θ, for r R). 6)

3 In both cases, we easily check that A =, since A ϕ is independent of the azimuthal angle ϕ. We now obtain approximate expressions for the magnetic field from the vector potential A = A ϕ ϕ: r θ B = r sin θ θ sin θa ϕ) r r ra ϕ). In the far region, for example, the magnetic field is B µ m cos θ r + sin θ θ ) = µ [ m r) r m r r ], 7) where m = I a =IπR ) ẑ denotes the magnetic dipole moment here, the direction of the area vector a obeys the right-hand rule). Example II: Spinning Charged Sphere Next, we consider the case of a spinning charged sphere of radius R and carrying a uniform surface charge density σ spinning with constant angular velocity ω. To set up the problem, we assume that the field point is on the z-axis: r = r ẑ and that the rotation axis is on the x, z)-plane and makes an angle ψ with respect to the z-axis: ω = ω sin ψ x+cos ψ ẑ) see Figure below). An arbitrary point r on the surface of the sphere is moving with tangential velocity v = ω r = ωr[ sin ψ sin θ sin ϕ ẑ cos θ ŷ) + cos ψ sin θ cos ϕ ŷ sin ϕ x) ], and, therefore, the surface current density is K = σ v. With this surface current density and r r = r + R rr cos θ, the vector potential is A = µ π π R sin θ dθ σ vθ,ϕ ) dϕ r + R rr cos θ = ŷ [ µ σω R sin ψ π sin θ cos θ dθ ]. r + R rr cos θ

4 Note that the vector potential is in the direction of ω r = ωr sin ψ ŷ and that the integral can be written as π sin θ cos θ dθ = 1 1 r + R rr cos θ rr 1 udu η + η 1 u = rrη ) 1/, where η = r > /r <. Hence, the vector potential inside and outside the sphere is A = µ σrω r) 1 inside sphere) R /r outside sphere) as viewed in the frame in which the field point is on the z-axis and the angular velocity ω is in the x, z)-plane. In the frame in which ω is directed along the z-axis and the field position is arbitrary, on the other hand, we find ω r = ωrsin θ ϕ and, thus, the vector potential A = A ϕ r, θ) ϕ has a single azimuthal component given by A ϕ r, θ) = µ σω R r/r) sin θ inside sphere) R/r) sin θ outside sphere) The magnetic field in this case is expressed in terms of the vector potential as r θ B = r sin θ θ sin θa ϕ) r r ra ϕ), and, thus, the magnetic field inside the spinning charged sphere is B in = µ σrω, i.e., it is constant. The magnetic field outside the spinning charged sphere, on the other hand, is B out = µ m cos θ r + sin θ θ ), r where the maggnitude of the magnetic dipole moment for the spinning charged sphere is m = σ ωr R. Magnetostatic Boundary Conditions From the divergenceless condition B = for the magnetic field, we find that the perpendicular component of the magnetic field is continuous across a surface carrying current density K, i.e., n B + B ) =,

5 where n is the unit vector perpendicular to the surface with n K = ). Ampère s Law applied to an infinitesimally thin volume enclosing the surface, on the other hand, implies that the parallel components of the magnetic field and discontinuous across the surface: Combining these two results, we find n B + B ) = µ K. B + B = µ K n). The vector potential, on the other hand, satisfies the boundary conditions A + = A and A + n A n = µ K. 5