Question 3 of 8: Magnetic Flux Density at the Surface of an Accelerated Electron Beam
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2018 — 04-BS-9 Basic Electromagnetics. Three-hour, closed-book exam (approved Casio/Sharp calculator only). Aids given: $\varepsilon_0=8.85\times10^{-12}$ F/m, $\mu_0=4\pi\times10^{-7}$ H/m, $e=1.6\times10^{-19}$ C. Format: eight questions offered; any five constitute a complete paper and only the first five appearing in the answer book are marked. All eight are solved below for completeness.
Reference texts: Sadiku, Elements of Electromagnetics / Hayt & Buck, Engineering Electromagnetics — Coulomb's law and Gauss's law for spherical and piecewise charge distributions, the Biot–Savart/Ampère law for finite, semi-infinite and arc-shaped conductors, Faraday's law for a loop crossing a spatially bounded field, the differential (point) form of Ampère's law, and series-layered parallel-plate capacitance; Young & Freedman, University Physics with Modern Physics — Snell's law and the geometry of apparent vs. real position across a refracting interface.
Question 3: Magnetic Flux Density at the Surface of an Accelerated Electron Beam (20 marks)
$1\times10^4$ V (electron energy – not needed below)
Current direction
horizontal, flowing north
Find. The magnitude and direction of $\vec B$ at the field point on the TOP surface of the beam (directly above the beam axis, at $r=R$).
Beam modeled as an infinite straight current (current flowing north, out of the cross-section shown); by the right-hand rule the field circles the beam, pointing EAST at the topmost surface point.
Approach. Since the beam's length vastly exceeds its 1 mm diameter, treat it as an infinite straight current and apply Ampère's law, $B(r)=\mu_0I/(2\pi r)$, with the direction fixed by the right-hand rule ($\hat\varphi=\hat I\times\hat r$).
Field magnitude at $r=R$. An Amperian loop at the surface encloses the ENTIRE beam current, regardless of how it is distributed radially inside:
$$B=\frac{\mu_0I}{2\pi R}=\frac{(4\pi\times10^{-7})(1\times10^{-6})}{2\pi(5\times10^{-4})}$$
$$\boxed{B=4.000\times10^{-10}\ \text{T}}$$
Direction at the top surface. With the current north ($\hat I=\hat y$) and the field point directly above the axis ($\hat r=\hat z$), the right-hand rule gives $\hat\varphi=\hat y\times\hat z=\hat x$, i.e. EAST:
$$\boxed{\vec B=4.000\times10^{-10}\ \text{T, directed EAST}}$$
Check: the accelerating potential ($1\times10^4$ V) is extra information — it would only be needed to find the electrons' drift speed or number density, not the field, since Ampère's law depends only on the total enclosed current $I$ and the radius $r$, never on how that current arose.
