Question 3 of 7: Magnetic Field of a Cylindrical Current-Carrying Pipe

Question 3: Magnetic Field of a Cylindrical Current-Carrying Pipe (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. (a) how $B$ varies for a point on the $x$-axis from $x=0$ to $100$m. (b) $B$ at $P_0$. (c) $\vec B$ at $P_1$. (d) the Lorentz force on the moving charge at $P_1$.

Approach. By cylindrical symmetry Ampère's law gives $B(r)=\mu_0I_{\text{enc}}(r)/(2\pi r)$, with $I_{\text{enc}}$ the current enclosed within radius $r$ (zero inside the bore, a fraction of $I$ proportional to enclosed area inside the annulus, all of $I$ outside the pipe); then apply the Lorentz force law for the moving charge.

1. Part (a) — $B$ along the $x$-axis, $x=0$ to $100$m. A point $(x,0,0)$ is a perpendicular distance $r=x$ from the pipe's axis (the $y$-axis). Three regions: $$B(r)=\begin{cases}0 & r \lt R_{\text{in}}=6\text{mm (inside bore, no enclosed current)}\\[2pt]\dfrac{\mu_0I}{2\pi r}\cdot\dfrac{r^2-R_{\text{in}}^2}{R_{\text{out}}^2-R_{\text{in}}^2} & 6\text{mm} \lt r \lt 10\text{mm (rising through the annulus)}\\[6pt]\dfrac{\mu_0I}{2\pi r} & r>10\text{mm (falling as }1/r\text{)}\end{cases}$$ Numerically: $B=0$ up to $x=6$mm; rises to a peak of $0.0200$ T at $x=R_{\text{out}}=10$mm (all $1000$A now enclosed); then falls monotonically as $1/x$ — $B=2.00\times10^{-4}$ T at $x=1$m, down to $$\boxed{B(100\text{m})=2.00\times10^{-6}\ \text{T at }x=100\text{m (still falling as }1/x)}$$
2. Part (b) — $B$ at $P_0=(2,2,2)$mm. Perpendicular distance from the $y$-axis uses only the $x,z$ components: $r_{P_0}=\sqrt{2^2+2^2}=2.828$mm, which is $\lt R_{\text{in}}=6$mm, so $P_0$ sits inside the hollow bore: $$\boxed{B(P_0)=0}$$
3. Part (c) — $\vec B$ at $P_1=(40,10,30)$m. Perpendicular distance $r_{P_1}=\sqrt{40^2+30^2}=50$m (well outside the pipe, so all $1000$A is enclosed): $$B=\frac{\mu_0I}{2\pi r_{P_1}}=\frac{(4\pi\times10^{-7})(1000)}{2\pi(50)}$$ $$\boxed{|\vec B(P_1)|=4.000\times10^{-6}\ \text{T}}$$ Direction, by the right-hand rule ($\hat B\propto\hat y\times\hat r$ with $\hat r=(0.8,0,0.6)$ the unit vector from the axis to $P_1$): $\hat B=(0.6,0,-0.8)$, i.e. $$\vec B(P_1)=(2.400\times10^{-6},\,0,\,-3.200\times10^{-6})\ \text{T}$$
4. Part (d) — force on the moving charge. $q=10$nC, $\vec v=(0,1,0)$ m/s: $$\vec F=q\,\vec v\times\vec B=(10\times10^{-9})(0,1,0)\times(2.400\times10^{-6},0,-3.200\times10^{-6})$$ $$\boxed{\vec F=(-3.200\times10^{-14},\,0,\,-2.400\times10^{-14})\ \text{N},\ |\vec F|=4.000\times10^{-14}\ \text{N}}$$