AcademyElectromagnetic Induction
Academy
Motional emf
Level 1 - Physics topic page in Electromagnetic Induction.
Principle
Motional emf appears when magnetic force separates charge in a conductor moving through a magnetic field.
Notation
\(\mathcal E\)
motional emf
\(\mathrm{V}\)
\(B\)
magnetic field magnitude
\(\mathrm{T}\)
\(\ell\)
length of conductor crossing field lines
\(\mathrm{m}\)
\(v\)
speed perpendicular to the conductor and field
\(\mathrm{m\,s^{-1}}\)
\(R\)
circuit resistance
\(\mathrm{\Omega}\)
\(I\)
induced current
\(\mathrm{A}\)
Method
Derivation 1: Magnetic force separates charge
Charges in the moving rod feel magnetic force \(q\\vec v\\times\\vec B\). Charge separation creates an electric field until forces balance.
Force balance
\[qE=qvB\]
Internal field
\[E=vB\]
emf across rod
\[\mathcal E=E\ell=B\ell v\]
Derivation 2: Flux-change view
A rod sliding on rails changes the loop area at rate \(dA/dt=\\ell v\).
Area rate
\[\frac{dA}{dt}=\ell v\]
Flux rate
\[\left|\frac{d\Phi_B}{dt}\right|=B\ell v\]
Same emf
\[|\mathcal E|=B\ell v\]
Derivation 3: Mechanical power
With a closed circuit, magnetic drag requires external work to maintain constant speed.
Current
\[I=\frac{B\ell v}{R}\]
Magnetic force
\[F=I\ell B\]
Power balance
\[Fv=I^2R\]
Rules
These forms assume \(\\vec v\), \(\\vec B\), and the rod length are mutually perpendicular.
Motional emf
\[|\mathcal E|=B\ell v\]
Induced current
\[I=\frac{B\ell v}{R}\]
Magnetic drag
\[F=I\ell B\]
Power balance
\[Fv=I^2R\]
Examples
Question
A
\[0.40\,\mathrm{m}\]
rod moves at \[3.0\,\mathrm{m\,s^{-1}}\]
perpendicular to a \[0.50\,\mathrm{T}\]
field. Find \[|\mathcal E|\]
Answer
\[|\mathcal E|=B\ell v=(0.50)(0.40)(3.0)=0.60\,\mathrm{V}\]
Checks
- Only the velocity component perpendicular to the rod and field contributes.
- The rod polarity comes from \(q\\vec v\\times\\vec B\) for positive charge.
- A closed circuit is needed for current, but not for charge separation/emf.
- Maintaining constant speed requires external power when current flows.