Classical Electrodynamics: Homework 8 - Forces, Torques on Loop & Charged Sphere Potential, Assignments of Physics

Download Classical Electrodynamics: Homework 8 - Forces, Torques on Loop & Charged Sphere Potential and more Assignments Physics in PDF only on Docsity! Physics 505, Classical Electrodynamics Homework 8 Due Thursday, 4th November 2004 Jacob Lewis Bourjaily Problem 5.11 Let us consider a circular loop of wire carrying a current I located at the center of a Cartesian coordinate system with its normal having spherical angles θ0, φ0. Let us suppose that there is an applied magnetic field Bx = B0 (1 + βy) and By = B0(1 + βx). a) We are to calculate the force acting on the loop explicitly. We are then to show that our tedious calculation reproduces Jackson’s (trivial) result in equation (5.69). Notice that, in general, we know that the force on this system is F = ∫ J×B d3x. However, the ‘most convenient’ coordinates systems for B and J are not the same. Therefore, this problem is essentially an exercise in coordinate transformations. Because all of our transformations are orthogonal ones, they are trivial—but tedious. Let us label our two coordinate systems M and M ′ for the obvious Cartesian system and the one where the current I flows completely in the x′-y′ plane, respectively. It is obvious that we can actively transform M into M ′, obtaining expressions for x̂′, ŷ′, ẑ′ in terms of x̂, ŷ, ẑ, by first applying the rotation in φ0 and then θ0. Specifically, we see   x̂′ ŷ′ ẑ′   =   cos θ0 0 − sin θ0 0 1 0 sin θ0 0 cos θ0     cosφ0 sin φ0 0 − sin φ0 cos φ0 0 0 0 1     x̂ ŷ ẑ   , =   cos θ0 cosφ0 cos θ0 sinφ0 − sin θ0 − sinφ0 cos φ0 0 sin θ0 cosφ0 sin θ0 sin φ0 cos θ0     x̂ ŷ ẑ   , =   cos θ0 cosφ0x̂ + cos θ0 sin φ0ŷ − sin θ0ẑ − sin φ0x̂ + cos φ0ŷ sin θ0 cosφ0x̂ + sin θ0 sin φ0ŷ + cos θ0ẑ   . Similarly, we can transform M ′ into M , obtaining expressions for x̂, ŷ, ẑ in terms of x̂′, ŷ′, ẑ′, by first applying the rotation in −θ0 and then −ϕ0. Specifically, we see that we can write  x̂ ŷ ẑ   =   cosφ0 − sin φ0 0 sin φ0 cos φ0 0 0 0 1     cos θ0 0 sin θ0 0 1 0 − sin θ0 0 cos θ0     x̂′ ŷ′ ẑ′   , =   cos θ0 cos φ0 − sin φ0 sin θ0 cos φ0 cos θ0 sin φ0 cos φ0 sin θ0 sinφ0 − sin θ0 0 cos θ0     x̂′ ŷ′ ẑ′   , =   cos θ0 cos φ0x̂′ − sin φ0ŷ′ + sin θ0 cos φ0ẑ′ cos θ0 sin φ0x̂′ + cosφ0ŷ′ + sin θ0 sin φ0ẑ′ − sin θ0x̂′ + cos θ0ẑ′   . Let us now reexpress J in the coordinate system M and B in terms of the polar angle ϕ around the circular current in its plane. In M ′, the ‘natural coordinate system’ of J, we have that J(ϕ) = aIϕ̂ = aI (− sinϕx̂′ + cos ϕŷ′) , where we have used ϕ to represent the polar angle about the circular loop of current. Inserting our work above for x̂′ and ŷ′, we see that J(ϕ) = aI   − sin ϕ cos θ0 cosφ0 − cosϕ sin φ0 − sin ϕ cos θ0 sinφ0 + cos ϕ cos φ0 sin ϕ sin θ0   . 1 2 JACOB LEWIS BOURJAILY Because to calculate the force, we must integrate over the entire loop of current, it would be very nice to express B as a function of ϕ in M ′. This can be done quite straightforwardly: B(ϕ) = B0   1 + βy(ϕ) 1 + βx(ϕ) 0   = B0   1 + βa (cos ϕ cos θ0 sin φ0 + sin ϕ cosφ0) 1 + βa (cos ϕ cos θ0 cos φ0 − sin ϕ sin φ0) 0   . Now, when we take the cross product, we can use the fact that any single trigonometric function of ϕ will integrate to zero around the loop, unless the term is quadratic in sin ϕ or cos ϕ (or any other even power). Therefore, we see that (J×B)x (ϕ) =a2IB0β sin2 ϕ sin θ0 sin ϕ0 + terms that vanish in the integral over ϕ; (J×B)y (ϕ) =a2IB0β sin2 ϕ sin θ0 cos ϕ0 + terms that vanish in the integral over ϕ; (J×B)z (ϕ) =a2IB0β (− sin2 ϕ cos θ0 sin φ0 cos φ0 + cos2 ϕ cos θ0 sin φ0 sin φ0+ +sin2 ϕ cos θ0 sin φ0 cos φ− cos2 ϕ cos θ0 sin φ0 sinφ0 ) + terms that vanish in the integral over ϕ; =0 + terms that vanish in the integral over ϕ; Therefore, we can perform each of the integrations required for F almost trivially: we will obtain a factor of π from each. We have shown, F = πa2IB0β   sin θ0 sin φ0 sin θ0 cosφ0 0   . ‘óπ²ρ ’²́δ²ι δ²ιξαι Let us compare this result with that we would have obtained using Jackson’s equation (5.69). This simply states that F = ∇ (m ·B) where m is the magnetic dipole moment. In M ′, m′ = πa2I(0, 0, 1). Transforming this according to our trivial results above, we see that in M , m = πa2I   sin θ0 cosφ0 sin θ0 sin φ0 cos θ0   . Now, we have that F−∇ (m ·B) , = B0∇   (1 + βy))mx (1 + βx) my 0   , = B0β ( my mx ) , = πa2IB0β   sin θ0 sin φ0 sin θ0 cosφ0 0   . ‘óπ²ρ ’²́δ²ι δ²ιξαι It is clear that this rather easy calculation exactly reproduces our tedious efforts above. b) We are to calculate the torque to lowest order and describe any observations about higher contributions for the circular loop and for more general shapes. While it is slightly unclear what ‘lowest order’ implies, we can compute the torque exactly. To do this, we will need to integrate over x′ × (J×B). Using our transfor- mation matrices, it is a simple matter to see that the vector x in the integration can be written x = (cos ϕ cos θ0 cosφ0 − sin ϕ sin φ0, cos ϕ cos θ0 sin φ0 + sin ϕ cosφ0,− cos ϕ sin θ0).