Sliding rod on rails: EMF and force

Problem

A rod of length $\ell = 20 \ \mathrm{cm}$ slides at $v = 4{,}0 \ \mathrm{m/s}$ on two horizontal rails closed on the left by a resistor $R = 2{,}0 \ \mathrm{\Omega}$. The system is immersed in a field $B = 0{,}50 \ \mathrm{T}$ pointing out of the plane. (a) Calculate the electromotive force $\mathcal{E}$ and the current $i$. (b) Determine the braking force on the rod. (c) Verify that the mechanical power of the external agent equals the power dissipated in $R$.

Conducting rod sliding on rails in a field pointing out of the plane.

Solution

(a) EMF and current. The linked flux varies because the circuit’s area changes: $\mathcal{E} = B\ell v$. $\mathcal{E} = B\ell v = 0{,}50 \cdot 0{,}20 \cdot 4{,}0 = \ev{0{,}40 \ \mathrm{V}}$ The current follows from Ohm’s law: $i = \frac{\mathcal{E}}{R} = \frac{0{,}40}{2{,}0} = \ev{0{,}20 \ \mathrm{A}}$

(b) Braking force. On the current-carrying rod immersed in the field, the Laplace force acts, opposite to $v$ (Lenz’s law): $F = i\ell B = 0{,}20 \cdot 0{,}20 \cdot 0{,}50 = \ev{0{,}020 \ \mathrm{N}}$

(c) Energy balance. Mechanical power of the external agent (keeping $v$ constant against $F$): $P_{\text{mech}} = F v = 0{,}020 \cdot 4{,}0 = 0{,}080 \ \mathrm{W}$ Power dissipated by the Joule effect in $R$: $P_{\text{diss}} = R\,i^2 = 2{,}0 \cdot (0{,}20)^2 = 0{,}080 \ \mathrm{W}$ The two powers coincide: $\ev{P_{\text{mech}} = P_{\text{diss}} = 0{,}080 \ \mathrm{W}}$, in accordance with energy conservation.

Links

Topics: Electromagnetic induction Concepts: Faraday-Neumann-Lenz law · Laplace force · Joule effect · Ohm’s law Skills: Flux balance