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Magnetic Field: Linear Correction

Consider a particle of mass m and charge q in an arbitrary magnetic field B at position

r◦ (t◦ ) = [x◦ , y◦ , z◦ ] with velocity v◦ (t◦ ) = [vx◦ , vy◦ , vz◦ ]. For r(t) = [x(t), y(t), z(t)] and

v(t) = r˙ (t), the particle’s trajectory satisfies:

q

d

v = (v × B)

dt

m

(1)

We may Taylor expand the magnetic field about r◦ as follows:

B(r) ≈ B(r◦ ) + (r − r◦ ) · ∇B(r◦ )

(2)

Where r · ∇B(r◦ ) = (x∂x + y∂y + z∂z )B(r)|r=r◦ . Define B0 = B(r◦ ) along with B1 =

B10 + B11 = −r◦ · ∇B(r◦ ) + r · ∇B(r◦ ). Further separate r = r0 + r1 and v = v0 + v1 ,

where v0 satisfies (1) with field B0 , the explicit solution of which is known since the

field is simply a constant vector. By subbing v and B into (1) and canceling the known

solution, we have

d

q

q

v1 = (v0 × B1 ) + (v1 × (B0 + B1 ))

(3)

dt

m

m

Then note that we must have r1 (0) = v1 (0) = 0 since r0 (0) = r◦ and v0 (0) = v◦ .

Therefore, the lowest order correction in time must be, for y = [α, β, γ], r1 = yt2 =⇒

v1 = 2yt =⇒ dtd v1 = 2y; by subbing these into (3), we can solve for y. The zeroth-order

solution discards all terms with t dependence, and therefore we should set v1 ≈ 0. Also,

applying this idea to B, we have B11 = r · ∇B(r◦ ) = (r0 + r1 ) · ∇B(r◦ ) ≈ r0 · ∇B(r◦ ), so

that B1 ≈ (r0 − r◦ ) · ∇B(r◦ ). By subbing these approximations into (3), we obtain

y=

q

(v0 × (r0 − r◦ ) · ∇B(r◦ ))

2m

(4)

Therefore, for small times, the position and velocity vectors of the particle will be adjusted

by adding r1 = yt2 and v1 = 2yt to the known vectors r0 and v0 , with y given by (4).

1

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