Question

Are Motional emfs a Practical Source of Electricity? How fast (in \({\bf{m}}/{\bf{s}}\) and \({\bf{mph}}\)) would a \({\bf{5}}.{\bf{00}} - {\bf{cm}}\) copper bar have to move at right angles to a \({\bf{0}}.{\bf{650}} - {\bf{T}}\) magnetic field to generate \({\bf{1}}.{\bf{50}}\,{\bf{V}}\) (the same as a AA battery) across its ends? Does this seem like a practical way to generate electricity?

Short answer

\(46.153\,{\rm{m/s}}\,\,{\rm{or}}\,\,{\rm{103}}{\rm{.24}}\,{\rm{mph}}\)

Step-by-step solution

Identification of given data

Length of copper wire \(L = 5.00\,{\rm{cm}}\,\,{\rm{or}}\,\,0.05\,{\rm{m}}\)

Magnetic field \(B = 0.650\,{\rm{T}}\)

Induced motional emf \(\varepsilon = 1.50\,{\rm{V}}\)

Significance of motional emf

When a conductor moves through a magnetic field, it creates a motional emf. It is expressed as,

\(\varepsilon = vBL\) …(i)

Where, \(v\) is the velocity, \(B\) is the magnetic field and \(L\) is the length of the conductor

 Determining the velocity of copper bar

Rearranging the equation (i)

\(v = \frac{\varepsilon }{{BL}}\)

Substitute all the values in above equation

\(\begin{aligned}{}v = \frac{{1.50\,{\rm{V}}}}{{\left( {0.650\,{\rm{T}}} \right) \times \left( {0.05\,{\rm{m}}} \right)}}\\ = 46.153\,{\rm{m/s}}\,\,{\rm{or}}\,\,{\rm{103}}{\rm{.24}}\,{\rm{mph}}\end{aligned}\)

Hence, the velocity of copper bar is \(46.153\,{\rm{m/s}}\,\,{\rm{or}}\,\,{\rm{103}}{\rm{.24}}\,{\rm{mph}}\)

For generating electricity by using motional emf, there should be large magnetic field requirement of magnetic field in the region. So it is not possible to generate electricity by using the concept of motional emf