Linear DC Machine — Motoring and Generating on a Rail Pair: 25-Nav-B6 Ocean Engineering and Offshore Structures, Undated paper, Q3

Question 3: Linear DC Machine — Motoring and Generating on a Rail Pair (equal value)

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.

Given data
Quantity	Symbol	Value
Flux density (into page)	$B$	1 T
Circuit resistance	$R$	0.05 Ω
Rail separation	$l$	1 m
Battery voltage	$V_b$	2 V
Opposing load, part (b)	$F_{load}$	20 N
Driving force, part (c)	$F_{drive}$	10 N

Find. (a) Initial force, current and unloaded terminal speed; (b) loaded steady-state speed and the three power flows plus efficiency; (c) the generating-mode steady-state speed and power flows plus efficiency once an external force drives the bar past its unloaded speed.

The paper labels the battery $V_0$ and the switch S₁; the working below writes the battery as $V_b$ to avoid confusion with the output-voltage symbols used elsewhere on this paper.

Figure 3 (top view) — conducting bar sliding on rails of separation $l$ in a uniform field $B$, driven by $V_b$ through $R$.

Approach. The circuit equation is $V_b = iR + Bul$ (motional back-emf $e=Bul$ opposing the current when $u$ and the driving current are in the same sense); the mechanical equation is $F_{mag}=BIl$ balancing whatever external force acts, with $P_{batt}=P_{mech}+P_{loss}$ at every steady state.

(a) Instant of closing, $t=0$. The bar is still stationary, so $u=0$ and the back-emf is zero; the whole battery voltage appears across $R$: $$i(0)=\frac{V_b}{R}=\frac{2}{0.05}=\boxed{40\ \text{A}},\qquad F(0)=Bil=(1)(40)(1)=40\ \text{N, to the right}.$$ With no mechanical load the bar accelerates until the net force vanishes, i.e. until the current (and hence $F$) falls to zero, so $e=Bul\to V_b$: $$u_{\text{no-load}}=\frac{V_b}{Bl}=\frac{2}{(1)(1)}=\boxed{2\ \text{m/s}}.$$
(b) Loaded steady state. At steady speed the net force is zero, so the electromagnetic force exactly balances the 20 N opposing load: $$I_b=\frac{F_{load}}{Bl}=\frac{20}{(1)(1)}=20\ \text{A},\qquad e_b=V_b-I_bR=2-(20)(0.05)=1\ \text{V},\qquad u_b=\frac{e_b}{Bl}=\boxed{1\ \text{m/s}}.$$ The three power flows follow directly: $$P_{batt}=V_bI_b=(2)(20)=40\ \text{W},\quad P_{mech}=F_{load}u_b=(20)(1)=20\ \text{W},\quad P_{loss}=I_b^{2}R=(20)^2(0.05)=20\ \text{W},$$ which checks ($40=20+20$), and the efficiency is $$\eta_b=\frac{P_{mech}}{P_{batt}}=\frac{20}{40}=\boxed{50\%}.$$
(c) Driven above the no-load speed (generating). With the 20 N load removed and a 10 N force now driving the bar forward, the electromagnetic force must react against it, which reverses the current relative to parts (a)–(b) and makes the bar a generator charging the battery: $$I_c=\frac{F_{drive}}{Bl}=\frac{10}{(1)(1)}=10\ \text{A},\qquad e_c=V_b+I_cR=2+(10)(0.05)=2.5\ \text{V},\qquad u_c=\frac{e_c}{Bl}=\boxed{2.5\ \text{m/s}}$$ (faster than the 2 m/s no-load speed, as it must be for the machine to push current back into the battery). The power flows are $$P_{mech,in}=F_{drive}u_c=(10)(2.5)=25\ \text{W},\quad P_{batt,out}=V_bI_c=(2)(10)=20\ \text{W},\quad P_{loss}=I_c^{2}R=(10)^2(0.05)=5\ \text{W},$$ which again checks ($25=20+5$), giving $$\eta_c=\frac{P_{batt,out}}{P_{mech,in}}=\frac{20}{25}=\boxed{80\%}.$$

Quantity	Result
(a) Initial current, force	40 A, 40 N
(a) Unloaded steady speed	2 m/s
(b) Loaded speed, $P_{batt}$, $P_{mech}$, $P_{loss}$, $\eta$	1 m/s, 40 W, 20 W, 20 W, 50%
(c) Driven speed, $P_{mech,in}$, $P_{batt,out}$, $P_{loss}$, $\eta$	2.5 m/s, 25 W, 20 W, 5 W, 80%

Topic. The elementary linear dc machine: motoring, loaded steady state, and generating action.

Key relations. Motional emf $e=Bul$; electromagnetic force $F=BIl$; circuit law $V_b=iR\pm e$ (sign set by whether $e$ opposes or aids the supply); power balance $P_{elec}=P_{mech}+P_{loss}$ at every steady state.

Why this works. The bar accelerates whenever a net force acts on it, so steady state is exactly the condition of zero net force — this single idea generates all three sub-answers. In (a) that condition is $F=0$; in (b) it is $F_{mag}=F_{load}$; in (c) an externally supplied force pushes the bar past its natural no-load speed, so the back-emf exceeds the source voltage and the current (and power) reverses direction, converting the machine from motor to generator without changing a single circuit component.

Common pitfall. Sign errors in the circuit equation for part (c) — forgetting that $e>V_b$ once the bar outruns its no-load speed, so the current must be written as flowing into the battery ($e=V_b+I_cR$), not out of it as in parts (a)–(b).

Source. Fitzgerald, Kingsley & Umans, Electric Machinery, 7th ed., §3.1, “A Simple Elementary Linear Machine.”

Question 3 of 8: Linear DC Machine — Motoring and Generating on a Rail Pair

Question 3: Linear DC Machine — Motoring and Generating on a Rail Pair (equal value)