Motional EMF

PHYElectromagnetic Induction

When a conductor of length $l$ moves with speed $v$ through a magnetic field $B$ , the free charges experience a magnetic force and an EMF $\varepsilon = Blv$ is set up across its ends. In a closed circuit this drives a current, and an opposing force $F = BIl$ must be overcome to keep the rod moving uniformly.

Students chain EMF, current, force and power in a single multi-part numerical.

Motional EMF

\varepsilon = B l v

B

= field (T)

\perp

to velocity,

l

= rod length (m),

v

= speed (m/s);

B

l

and

v

mutually perpendicular for maximum EMF.

Induced current

I = \dfrac{\varepsilon}{R} = \dfrac{B l v}{R}

R

= total circuit resistance (

\Omega

); current flows only in a closed circuit.

Force to keep rod moving & power

F = B I l = \dfrac{B^2 l^2 v}{R}, \qquad P = F v = I^2 R = \dfrac{B^2 l^2 v^2}{R}

F

= applied force balancing the opposing magnetic force;

P

= power dissipated = mechanical power input (uniform motion).

$\varepsilon = Blv$ assumes $B$ , $l$ and $v$ are mutually perpendicular; otherwise use the perpendicular components.
An open rod develops an EMF (and a charge separation) but NO current flows until the circuit is closed.
The induced current opposes the rod's motion (Lenz), so an external force of magnitude $BIl$ is needed to maintain uniform velocity.
Under uniform velocity the kinetic energy is constant, so all the mechanical work done by the applied force is dissipated as heat: $P_{mech} = P_{dissipated} = I^2R$ .
Power can be written three equivalent ways: $Fv$ , $I^2R$ , or $\varepsilon^2/R$ — pick whichever data you have.
The motional EMF is the same Faraday EMF seen from a moving-charge picture; $\varepsilon = Blv$ also equals $|d\phi/dt|$ , since the area swept per second is $lv$ .
Keep SI units (T, m, m/s, $\Omega$ ) and the answers for $\varepsilon$ (V), $I$ (A), $F$ (N), $P$ (W) come out directly.

Where the marks go

Using $F = BIl$ with the wrong current, or forgetting that the same $B$ and $l$ appear in both the EMF and the force.
Computing $I$ with only the rod's resistance when the problem gives the TOTAL circuit resistance $R$ .
Confusing the applied force-to-keep-moving with the retarding magnetic force — they are equal in magnitude but opposite in direction.
Mixing power expressions, e.g. writing $P = IV_{source}$ instead of $P = I^2R = Fv$ for this dissipated power.

How the board asks it

Numericalthe chained emf-current-force-power relations
A metal rod of length $l = 0.5\,\text{m}$ slides on frictionless rails at $v = 4\,\text{m/s}$ in a uniform field $B = 0.2\,\text{T}$ perpendicular to the plane of the rails. If the total circuit resistance is $R = 2\,\Omega$ , calculate (i) the induced emf, (ii) the current, (iii) the force needed to keep the rod moving uniformly, and (iv) the power dissipated.
Derive / prove $\varepsilon = Blv$ from the magnetic force on free charges
A conducting rod of length $l$ moves with velocity $v$ perpendicular to a uniform magnetic field $B$ . Derive an expression for the motional emf induced across its ends, starting from the magnetic force experienced by the free electrons.
Give reasonslenz's law and energy conservation
A rod moves with constant velocity, and hence constant kinetic energy, along rails in a magnetic field, yet an external force must be applied continuously to keep it moving. Account for this, and state where the work done by the external force goes.
Give reasonsopen rod develops emf but no current
A rod moving in a magnetic field has an emf across its ends but no current flows through it. Give reasons, and state the condition under which a current will be set up.
Applicationequivalence of $\varepsilon = Blv$ and $\varepsilon = |d\phi/dt|$
Show that the motional emf $\varepsilon = Blv$ is consistent with Faraday's law $\varepsilon = |d\phi/dt|$ by considering the area swept by the rod per unit time.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.