Question 3 of 7: Magnetic Field of a Solid Cylindrical Conductor
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams 04-BS-9 Basic Electromagnetics. Three-hour, closed-book exam (approved Casio/Sharp calculator only). Constants given: $\varepsilon_0=8.854\times10^{-12}$ F/m, $\mu_0=4\pi\times10^{-7}$ H/m, $e=1.602\times10^{-19}$ C. Format: seven questions offered; any five constitute a complete paper and only the first five appearing in the answer book are marked. All seven are solved below for completeness.
Reference texts: Sadiku, Elements of Electromagnetics / Hayt & Buck, Engineering Electromagnetics — Coulomb's law and superposition of point charges, Gauss's law and conductors in electrostatic equilibrium, Ampère's law for cylindrical and bent-wire current distributions, the force between currents, Faraday's law (rotational and transformer EMF), parallel-plate capacitance and the force on a levitated charge, and magnetic-circuit (reluctance) analysis of a gapped core.
Question 3: Magnetic Field of a Solid Cylindrical Conductor (15 marks)
Find.(e) location/value of the greatest $B$. (f) the minimum $B$ inside the conductor. (g) $\vec B$ at $P_1$. (h) the magnetic force on the moving charge at $P_1$.
Cross-section (x-z plane) of the conductor: B rises linearly from zero at the axis to a maximum at the surface r=R, then falls as 1/r outside. P1 is far outside, at radial distance ρ=50 m from the y-axis.
Approach. Ampère's law gives $B(r)=\mu_0Ir/(2\pi R^2)$ inside a uniform solid conductor ($r\le R$) and $B(r)=\mu_0I/(2\pi r)$ outside ($r>R$); both formulas agree at $r=R$, which is therefore the field's overall maximum. Evaluate $\vec B$ at $P_1$'s actual radial distance, then use $\vec F=q\vec v\times\vec B$.
Part (e) — greatest $B$. Inside, $B\propto r$ (rises from the axis); outside, $B\propto1/r$ (falls with distance) — both are maximised at the boundary $r=R$, i.e. anywhere on the conductor's own surface (a full circle of radius 10 mm around the y-axis, not a single point, by symmetry):
$$B_{\max}=\frac{\mu_0I}{2\pi R}=\frac{(4\pi\times10^{-7})(1000)}{2\pi(0.010)}$$
$$\boxed{B_{\max}=0.0200\ \text{T}=20.0\ \text{mT, at }r=R=10\ \text{mm}}$$
Part (f) — minimum $B$ inside. $B(r)=\mu_0Ir/(2\pi R^2)$ inside the conductor vanishes at $r=0$:
$$\boxed{B_{\min}=0,\ \text{at the conductor's own axis (the }y\text{-axis)}}$$
Part (g) — $\vec B$ at $P_1$. Only the $x,z$ coordinates set the radial distance from the current-carrying $y$-axis:
$$\rho=\sqrt{40^2+30^2}=50\ \text{m}\ (\gg R,\text{ so }P_1\text{ is far outside the conductor})$$
$$B=\frac{\mu_0I}{2\pi\rho}=\frac{(4\pi\times10^{-7})(1000)}{2\pi(50)}$$
$$\boxed{B=4.000\times10^{-6}\ \text{T}=4.000\ \mu\text{T}}$$
By the right-hand rule (curling around $+y$), the direction is $\hat\varphi=(z/\rho,\,0,\,-x/\rho)=(0.6,0,-0.8)$, so
$$\boxed{\vec B=(2.400,\,0,\,-3.200)\times10^{-6}\ \text{T}}$$
Part (h) — force on the moving charge. With $\vec v=(0,1,0)$ m/s and $q=10\times10^{-9}$ C:
$$\vec F=q\,\vec v\times\vec B=q\,(0,1,0)\times(2.400,0,-3.200)\times10^{-6}$$
$$\boxed{\vec F=(-3.200,\,0,\,-2.400)\times10^{-14}\ \text{N},\ |\vec F|=4.000\times10^{-14}\ \text{N}}$$
directed along $(-0.8,0,-0.6)$ — i.e. radially toward the wire, the same attraction a parallel current would feel (the moving positive charge behaves, for this instant, like a short current element parallel to the wire's own current).
