A Level Physics 9702: Motional e.m.f. of a Rod Sliding on Rails

A straight conducting rod $PQ$ of length $L=0.30\text{ m}$ lies across two long, parallel, horizontal conducting rails separated by this same distance. The rails are connected together at one end by a fixed resistor of resistance $R=5.0\ \Omega$ . The whole arrangement lies in a uniform magnetic field of flux density $B=0.40\text{ T}$ , directed vertically, perpendicular to the plane of the rails. The rod is pulled along the rails at a constant speed of $v=2.5\text{ m s}^{-1}$ , perpendicular to its own length, so that the area enclosed by the rails, the rod and the resistor increases uniformly with time.

(a) Show that the e.m.f. induced in the circuit is given by $\varepsilon=BLv$ , and calculate its value for this rod. [3]

(b) Calculate the current in the circuit while the rod moves at this constant speed. [2]

(c) State Lenz's law, and use it to explain the direction of the magnetic force that this induced current exerts on the rod, and hence explain why an external force must be applied to the rod to keep it moving at constant speed. [3]

(d) Calculate the magnitude of the external force that must be applied to the rod to keep it moving at this constant speed. [2]

Part (a): Deriving and calculating the motional e.m.f.

In a short time interval $\Delta t$ , the rod moves a distance $v\Delta t$ along the rails, so the area enclosed by the circuit increases by: $\Delta A=L\times v\Delta t$

Since the field is uniform and perpendicular to this area, the magnetic flux through the circuit is $\Phi=BA$ , so the flux increases by: $\Delta\Phi=B\Delta A=BLv\Delta t$

By Faraday’s law, the magnitude of the induced e.m.f. is the rate of change of flux linkage (here there is only one turn, so flux linkage equals flux): $\varepsilon=\left|\frac{\Delta\Phi}{\Delta t}\right|=\frac{BLv\Delta t}{\Delta t}=BLv$

Substituting the values given: $\varepsilon=0.40\times0.30\times2.5=0.30\text{ V}$

Recomputing as a check: $0.40\times0.30=0.12$ , and $0.12\times2.5=0.30\text{ V}$ , the same result.

Part (b): Induced current

Using $\varepsilon=IR$ with the resistor of resistance $R=5.0\ \Omega$ : $I=\frac{\varepsilon}{R}=\frac{0.30}{5.0}=6.0\times10^{-2}\text{ A}$

Recomputing as a check: $0.30\div5.0=0.06=6.0\times10^{-2}\text{ A}$ , consistent.

Part (c): Lenz’s law and why an external force is needed

Lenz’s law states that the direction of an induced e.m.f. (and any resulting current) is always such as to oppose the change producing it.

The induced current $I$ flowing along the rod, in the magnetic field $B$ , itself experiences a magnetic force (the motor effect, $F=BIL\sin\theta$ ). By Lenz’s law, this force must oppose the change that caused the current (namely the rod’s motion, which is increasing the flux) so this magnetic force acts to retard the rod, opposite to its velocity $v$ .

Since the rod moves at constant speed, its acceleration is zero, so by Newton’s first law the net force on it must be zero. As the induced magnetic force continuously acts backward on the rod, an external applied force of equal magnitude, directed forward (in the direction of motion), must be supplied to exactly balance it. This is consistent with energy conservation: the work done by the external force against this magnetic retarding force is the source of the electrical energy dissipated as heat in the resistor $R$ .

Part (d): Magnitude of the external force

At constant speed, the applied force equals the magnetic retarding force on the current-carrying rod, $F=BIL\sin\theta$ , with the rod perpendicular to the field ( $\theta=90°$ ): $F=BIL=0.40\times6.0\times10^{-2}\times0.30$

$F=0.40\times0.06=0.024;\quad0.024\times0.30=7.2\times10^{-3}\text{ N}$

Recomputing as an independent check using power: the electrical power dissipated is $P=I^2R=(6.0\times10^{-2})^2\times5.0=3.6\times10^{-3}\times5.0=1.8\times10^{-2}\text{ W}$ . Since this power equals the mechanical power delivered by the external force, $P=Fv$ , so $F=P/v=1.8\times10^{-2}/2.5=7.2\times10^{-3}\text{ N}$ , the same result both ways.

Final answers

(a) $\varepsilon=BLv$ (derived above); $\varepsilon=\boxed{0.30}\text{ V}$
(b) $I=\boxed{6.0\times10^{-2}}\text{ A}$
(c) By Lenz’s law the induced current creates a magnetic force opposing the rod’s motion; an equal, forward external force is needed to keep the speed constant
(d) $F=\boxed{7.2\times10^{-3}}\text{ N}$