 Solution

ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2014/paper-1/2/c/solution

Direct variation of the Lagrangian gives

m \overset{ρ}{¨} - m ρ \dot{θ}^{2} + k ρ - qE cos θ - qBρ \dot{θ} = 0,

m ρ^{2} \ddot{θ} + 2 m ρ \overset{ρ}{˙} \dot{θ} + qBρ \overset{ρ}{˙} + qEρ sin θ = 0.

For a static configuration,

k ρ = qE cos θ

and

qEρ sin θ = 0

. Since

qE > 0

and the radius is positive,

ρ_{0} = qE / k, θ_{0} = 0 (mod 2 π) .

The magnetic force vanishes at rest and does not shift this equilibrium. In canonical variables it has

p_{ρ} = 0

but

p_{θ} = qB ρ_{0}^{2} /2

, rather than zero canonical angular momentum.

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