Question 4 of 7: Right-Angle Bent Conductor — Field and Self-Force

Question 4: Right-Angle Bent Conductor — Field and Self-Force (15 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) $\vec B$ at $P_1,P_2$. (b) how $B$ varies from the origin to $P_2$. (c) force per length on arm 2 at $P_2$. (d) total force on the origin–to–$P_2$ segment.

Approach. Each field point gets contributions from both semi-infinite arms. A point that is collinear with a straight segment's own line always receives zero field from that segment (the Biot–Savart cross product vanishes identically along the line), so only the other arm ever contributes at a point on the $y$-axis. Both $P_1$ and $P_2$ sit exactly beside the free end of the $+x$ arm ($\rho=1$ m, foot of the perpendicular at the origin), the textbook "semi-infinite wire, point level with the end" case, $B=\mu_0I/(4\pi\rho)$. Part (d) then integrates the resulting non-uniform field along arm 2 from the origin to $P_2$.

1. Part (a) — field at $P_1$ and $P_2$. Arm 2 (the $y$-axis) contributes nothing at either point (both are collinear with it). Arm 1 alone gives, at either point ($\rho=1$ m, level with its own end): $$B=\frac{\mu_0I}{4\pi\rho}=\frac{(4\pi\times10^{-7})(1000)}{4\pi(1)}$$ $$\boxed{|\vec B|=1.000\times10^{-4}\ \text{T}=100\ \mu\text{T at both points}}$$ By the right-hand rule (current in arm 1 flows in $-\hat x$): at $P_1$ ($-y$ side) the field is $+\hat z$; at $P_2$ ($+y$ side) it is $-\hat z$ — equal magnitude, opposite direction: $$\boxed{\vec B(P_1)=+100\,\mu\text{T}\,\hat z,\quad \vec B(P_2)=-100\,\mu\text{T}\,\hat z}$$
2. Part (b) — variation from the origin to $P_2$. For any point $(0,y,0)$ with $y>0$, the perpendicular foot onto arm 1 is still exactly the origin (arm 1's own endpoint), so the "point level with the end" formula applies at every $y$: $$\boxed{B(y)=\frac{\mu_0I}{4\pi y},\ \text{direction }-\hat z\ \text{throughout}}$$ $B$ falls off as $1/y$ moving away from the corner, and grows without bound as $y\to0^+$ — the field of arm 1 is singular right at its own endpoint.
3. Part (c) — force per length on arm 2 at $P_2$. A current element in an external field feels $d\vec F/d\ell=I\hat y\times\vec B$; at $P_2$, $\vec B=-100\,\mu\text{T}\,\hat z$: $$\frac{dF}{d\ell}=I\,B(1)=1000\times1.000\times10^{-4}$$ $$\boxed{\left.\frac{dF}{d\ell}\right|_{P_2}=0.1000\ \text{N/m, in the}-x\text{ direction}}$$ (arm 2 is pulled toward arm 1's side — the two arms of a continuous current loop attract at the bend, consistent with parallel-current attraction.)
4. Part (d) — total force, origin to $P_2$. Integrating the non-uniform force per length along the segment: $$F_{\text{total}}=\int_0^1 I\,B(y)\,dy=\frac{\mu_0I^2}{4\pi}\int_0^1\frac{dy}{y}=\frac{\mu_0I^2}{4\pi}\Big[\ln y\Big]_0^1$$ $$\boxed{F_{\text{total}}\to\infty\ \text{(the integral diverges logarithmically as }y\to0\text{)}}$$ This is not an arithmetic slip: it is the same singularity found in part (b), integrated. Because the segment is taken to start exactly at the bend — precisely where arm 1's own field blows up — the idealised zero-radius filament model predicts an unbounded force there. A real conductor's finite radius $a$ would cut the integral off at $y\approx a$, giving the finite, regularised estimate $F\approx(\mu_0I^2/4\pi)\ln(1/a)$; no conductor radius is given in this problem, so the honest answer is that the force is unbounded under the stated idealisation.