Question 3 of 8: Linear dc Machine on Rails — Force, Speed, Power and Efficiency

Question 3: Linear dc Machine on Rails — Force, Speed, Power and Efficiency (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. A single conducting bar free to slide on two frictionless rails that close a loop through a battery and a series resistor; a uniform field crosses the bar perpendicular to its motion, so the machine is the one-dimensional analogue of a dc motor/generator.

Find. [a] the starting force, starting current and the no-load steady speed; [b] the steady speed, the battery/mechanical/resistive powers and the efficiency for a 20 N retarding load; [c] the same quantities when a 10 N force instead drives the bar, and the efficiency in that mode.

Approach. Two equations govern every part: the electrical loop $V_0 = e + IR$ with the motional emf $e = B\ell v$, and the electromechanical force $F = B I \ell$. In steady state the bar does not accelerate, so the electromagnetic force balances the applied mechanical force; solving the pair gives the current, then the speed, and the three powers follow from $P = VI$, $F v$ and $I^2R$.

1. Starting current (bar still, so no motional emf). At $t=0$ the speed is zero, hence $e = B\ell v = 0$ and the full battery voltage appears across $R$. $$I_0 = \frac{V_0}{R} = \frac{2}{0.05}$$ $$\boxed{I_0 = 40\ \text{A}}$$
2. Starting force. The current-carrying bar sits in the field, so it feels a Lorentz force $F = B I \ell$ directed along the rails. $$F_0 = B I_0 \ell = (1)(40)(1) = \boxed{40\ \text{N, directed to the right}}$$ The force accelerates the bar away from the source end (to the right in Figure 3); by Lenz’s law the resulting motion builds an emf that opposes the battery and throttles the current back.
3. No-load steady speed. With no mechanical load and no friction, the bar accelerates until the force — and therefore the current — falls to zero. Zero current means the motional emf has risen to exactly the battery voltage. $$e = V_0 \;\Rightarrow\; B\ell v_{\text{nl}} = V_0 \;\Rightarrow\; v_{\text{nl}} = \frac{V_0}{B\ell} = \frac{2}{(1)(1)}$$ $$\boxed{v_{\text{nl}} = 2\ \text{m/s}}$$

Parts [b] and [c] are both steady-state problems: the bar moves at constant speed, so the net force is zero and the electromagnetic force $B I \ell$ exactly balances whatever mechanical force is applied. The sign of that balance is what distinguishes motoring from generating.

4. [b] Current for a 20 N retarding load (motoring). The machine must produce 20 N to hold the load, so $$I = \frac{F}{B\ell} = \frac{20}{(1)(1)} = 20\ \text{A}$$
5. [b] Steady speed from the loop equation. The motional emf is what is left of the battery voltage after the $IR$ drop. $$e = V_0 - IR = 2 - (20)(0.05) = 1\ \text{V}, \qquad v = \frac{e}{B\ell} = \frac{1}{(1)(1)} = \boxed{1\ \text{m/s}}$$
6. [b] Power balance and efficiency. The battery supplies $V_0 I$; the mechanical load receives $Fv = eI$; the resistor dissipates $I^2R$. $$P_{\text{batt}} = V_0 I = (2)(20) = 40\ \text{W}$$ $$P_{\text{mech}} = F v = (20)(1) = 20\ \text{W} \; (= eI), \qquad P_R = I^2 R = (20)^2(0.05) = 20\ \text{W}$$ $$40 = 20 + 20 \; \checkmark \qquad \boxed{\eta = \frac{P_{\text{mech}}}{P_{\text{batt}}} = \frac{20}{40} = 50\%}$$

In part [c] the retarding load is gone and an external agent instead pushes the bar in its direction of travel. The bar is driven past its no-load speed, so its emf climbs above the battery voltage, the current reverses, and power now flows from the bar into the battery: the machine has become a generator.

7. [c] Current for a 10 N driving force (generating). Steady state again balances the electromagnetic force against the applied 10 N. $$I = \frac{F}{B\ell} = \frac{10}{(1)(1)} = 10\ \text{A} \quad (\text{now reversed, charging the battery})$$
8. [c] Steady speed. Because the current reverses, the $IR$ term now adds to the battery voltage. $$e = V_0 + IR = 2 + (10)(0.05) = 2.5\ \text{V}, \qquad v = \frac{e}{B\ell} = \frac{2.5}{(1)(1)} = \boxed{2.5\ \text{m/s}}$$
9. [c] Power balance and efficiency. The mechanical source supplies $Fv = eI$; of this, $V_0 I$ is delivered to the battery and $I^2R$ is lost as heat. $$P_{\text{mech,in}} = F v = (10)(2.5) = 25\ \text{W} \;(= eI)$$ $$P_{\text{batt}} = V_0 I = (2)(10) = 20\ \text{W}, \qquad P_R = I^2 R = (10)^2(0.05) = 5\ \text{W}$$ $$25 = 20 + 5 \; \checkmark \qquad \boxed{\eta = \frac{P_{\text{batt}}}{P_{\text{mech,in}}} = \frac{20}{25} = 80\%}$$