(a) A car generator turns at \({\rm{400 rpm}}\) when the engine is idling. Its\({\rm{300 - turn}}\), \({\rm{5}}{\rm{.00}}\)by \({\rm{8}}{\rm{.00 cm}}\) rectangular coil rotates in an adjustable magnetic field so that it can produce sufficient voltage even at low rpms. What is the field strength needed to produce a \({\rm{24}}{\rm{.0 V}}\) peak emf? (b) Discuss how this required field strength compares to those available in permanent and electromagnets.

a)

The emf produced is given by:

\({\rm{\varepsilon }} = {\rm{NBA\omega }}\)

For a given emf, it is clear that the needed field will have to be

\({\rm{B}} = \frac{{\rm{\varepsilon }}}{{{\rm{NA\omega }}}}\)

Let us now remember that since the area is a rectangle, we have\({\rm{A}} = {\rm{a}} \times {\rm{b}}\).

Also, to convert from rotations per minute to radians per second, we need to multiply by\({\rm{2\pi }}\)and divide by\({\rm{60}}\)that is, multiplying by\(\frac{{\rm{\pi }}}{{{\rm{30}}}}\).

This means that the final expression will be

\({\rm{B}} = \frac{{{\rm{30\varepsilon }}}}{{\left( {{\rm{a}} \times {\rm{b}}} \right) \times {\rm{N}} \times {\rm{\pi }} \times {\rm{rpm}}}}\)

Substituting numerically, we have

\(\begin{aligned} {\rm{B}} &= \frac{{{\rm{30}} \times \left( {{\rm{24}}{\rm{.0}}\;{\rm{V}}} \right)}}{{\left( {{\rm{0}}{\rm{.05}}\;{\rm{m}} \times {\rm{0}}{\rm{.08}}\;{\rm{m}}} \right) \times {\rm{300}}\;{\rm{turn}} \times {\rm{\pi }} \times {\rm{400}}}}\\ &= {\rm{0}}{\rm{.477\;T}}\end{aligned}\)

Therefore, the strength of the field produced is \({\rm{0}}{\rm{.477\;T}}\).

b)

This required field is quite high to be achieved by permanent magnets if the distance between them is considerable, and not as easy to be achieved by electromagnets either. Let us mention that the strongest electromagnets used in magnetic resonance imaging usually produce fields no greater than\({\rm{5 T}}\).