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Found in: Page 861

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# What is the peak emf generated by rotating a 1000-turn, $${\rm{5}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{ T}}$$ diameter coil in the Earth’s magnetic field, given the plane of the coil is originally perpendicular to the Earth’s field and is rotated to be parallel to the field in $${\rm{10}}{\rm{.0 ms}}$$?

Electromotive force is $${\rm{157 mV}}$$.

See the step by step solution

## Step 1: Definition of electromotive force

Electromotive force is defined as the electric potential produced by either an electrochemical cell or by changing the magnetic field. EMF is the commonly used acronym for electromotive force.

A generator or a battery is used for the conversion of energy from one form to another. In these devices, one terminal becomes positively charged while the other becomes negatively charged. Therefore, an electromotive force is work done on a unit electric charge.

## Step 2: Calculate the electromotive force.

The electromotive force is given by

$${\rm{\varepsilon }} = - {\rm{N}}\frac{{{\rm{\Delta \Phi }}}}{{{\rm{\Delta t}}}}$$

$${\rm{\Phi }} = {\rm{BAcos(\theta )}}$$ is the flux, thus the change from when the coil is perpendicular to the magnetic field until it is parallel will just be $${\rm{\Delta \Phi }} = {\rm{AB}}$$.

Therefore, we have

$${\rm{\varepsilon }} = - \left( {\frac{{{\rm{NAB}}}}{{\rm{t}}}} \right)$$

where $$t$$ is the time for the needed change in the flux, given in the exercise. Numerically, neglecting the minus in the formula, we will have

\begin{aligned}{}{\rm{\varepsilon }} &= \frac{{{\rm{1000}} \times \left( {{\rm{\pi }} \times \frac{{{{\left( {{\rm{0}}{\rm{.2}}\;{\rm{m}}} \right)}^{\rm{2}}}}}{4}} \right) \times \left( {{\rm{5}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 5}}}}\;T} \right)}}{{{\rm{10}}{\rm{.0}}\;{\rm{ms}}\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}\;s}}{{1\;{\rm{ms}}}}} \right)}}\\ &= {\rm{157}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}\;{\rm{V}}\left( {\frac{{{\rm{1}}\;mV}}{{{\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}\;V}}} \right)\\ &= 157\;mV\end{aligned}

Hence, the electromotive force is $${\rm{157}}\;{\rm{mV}}$$.