91Ó°ÊÓ

(a)

The speed of the bicycle will be equal to the perimeter of the wheel that rotates the generator times the frequency (not angular frequency). That is,

\({\rm{v}} = {\rm{Pf}}\)

Since we are given the diameter, the perimeter will be\({\rm{P}} = {\rm{\pi d}}\)and we also should remember that the frequency can be given by the angular frequency as\({\rm{f}} = \frac{{\rm{\omega }}}{{{\rm{2\pi }}}}\).

This means that our expression of the result is

\({\rm{v}} = \frac{{{\rm{\omega d}}}}{{\rm{2}}}\)

Numerically we will have

\(\begin{aligned}{}{\rm{v}} &= \frac{{{\rm{1875}}\;{\rm{rad/s}} \times {\rm{0}}{\rm{.016}}\;{\rm{m}}}}{{\rm{2}}}\\ &= {\rm{15\;m/s}}\end{aligned}\)

Therefore, the value of the speed of the bicycle is \({\rm{v}} = {\rm{15\;m/s}}\).

(b)

Let us remember that the emf is directly proportional to the speed of the bicycle.

This allows us to simply write the following relation

\(\begin{aligned}{l}\frac{{{{\rm{v}}_{\rm{1}}}}}{{{{\rm{v}}_{\rm{2}}}}} &= \frac{{{{\rm{\varepsilon }}_{\rm{1}}}}}{{{{\rm{\varepsilon }}_{\rm{2}}}}}\\{{\rm{\varepsilon }}_{\rm{2}}} &= \left( {\frac{{{{\rm{v}}_{\rm{2}}}}}{{{{\rm{v}}_{\rm{1}}}}}} \right){{\rm{\varepsilon }}_{\rm{1}}}\end{aligned}\)

Applying numerically, we can find this speed to be

\(\begin{aligned}{}{{\rm{\varepsilon }}_{\rm{2}}} &= \left( {\frac{{{\rm{10}}\;{\rm{m/s}}}}{{{\rm{15 m/s}}}}} \right) \times {\rm{18}}{\rm{.0}}\;{\rm{V}}\\ &= {\rm{12}}{\rm{.0\;V}}\end{aligned}\)

Therefore, the emf of the generator is \({\rm{12}}{\rm{.0\;V}}\).

c)

With the given conditions, the speed is\(\frac{{\rm{1}}}{{\rm{3}}}\)of the initial speed, while the electromotive force needs to be\(\frac{{\rm{1}}}{{\rm{2}}}\)of the initial.

This means that the field will have to be\(\frac{{\rm{3}}}{{\rm{2}}}\)of the initial field, yielding

\(\begin{aligned}{}{\rm{B}} &= \frac{{\rm{3}}}{{\rm{2}}} \times {\rm{0}}{\rm{.64}}\;{\rm{T}}\\ &= {\rm{0}}{\rm{.96\;T}}\end{aligned}\)

Therefore, the field strength produced by emf is \({\rm{B}} = {\rm{0}}{\rm{.96\;T}}\)