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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2017-resit/4/solution

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Past exam of the physics 2 paper of the University of Oxford / 2017 resit / 4 / Solution by

Codex  0 2026-10-25

The current density is the moving charge density times its velocity:

J = ρ v = ρ v_{0} \hat{z} (r < R), J = 0 (r > R) .

For a cylindrical Gaussian surface of radius

r < R

ℓ

, cylindrical symmetry makes the electric field radial. Gauss's law gives

E (2 π r ℓ) = ρ π r^{2} ℓ / ϵ_{0}

. A circular Ampère's circuital law contour encloses electric current

I (r) = ρ v_{0} π r^{2}

B (2 π r) = μ_{0} I (r)

. Hence

E (r) = \frac{ρ r}{2 ϵ _{0}} \hat{r}, B (r) = \frac{μ _{0} ρ v _{0} r}{2} \hat{ϕ} = μ_{0} ϵ_{0} v \times E (r), r < R .

The right-hand rule fixes the positive azimuthal direction of the magnetic field for positive electric charge moving along

+ z

. Both the electric field and the magnetic field vanish on the axis. These are the steady long-cylinder electric field and magnetic field; end effects and changes of the prescribed beam distribution are excluded.

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