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College Physics (Urone)
Found in: Page 860

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Short Answer

An \({\rm{emf}}\) is induced by rotating a \({\rm{1000 - }}\)turn, \({\rm{20}}{\rm{.0 cm}}\) diameter coil in the Earth’s \({\rm{5}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{ T}}\) magnetic field. What average \({\rm{emf}}\) is induced, given the plane of the coil is originally perpendicular to the Earth’s field and is rotated to be parallel to the field in\({\rm{10}}{\rm{.0 ms}}\)?

The average \({\rm{emf}}\) induced is obtained as \({\rm{1}}{\rm{.57}}\;{\rm{mV}}\).

See the step by step solution

Step by Step Solution

Step 1: Define Electromagnetic Induction

The creation of an electromotive force across an electrical conductor in a changing magnetic field is known as electromagnetic or magnetic induction. Induction was discovered in \({\rm{1831}}\) by Michael Faraday, and it was mathematically characterized as Faraday's law of induction by James Clerk Maxwell.

Step 2: Evaluating the average emf induced

The electromotive force is induced when a coil with the value \({\rm{N}}\) turns experiences a flux change of the value \({\rm{\Delta \Phi }}\) in time \({\rm{\Delta t}}\) is given by:

\({\rm{E = - N}}\frac{{{\rm{\Delta \Phi }}}}{{{\rm{\Delta t}}}}\)……………….(1)

The flux reaches the point zero from the maximal value under the given time, we can simply write the change in flux as:

\(\begin{aligned}{}{\rm{\Delta \Phi = 0 - BA}}\\{\rm{ = - BA}}\end{aligned}\)

The emf then will be:

\({\rm{\varepsilon = }}\frac{{{\rm{NBA}}}}{{\rm{t}}}\)……………………….(2)

Substituting the area such that:

\(\begin{aligned}{}{\rm{A = \pi }}{{\rm{R}}^{\rm{2}}}\\{\rm{ = }}\frac{{{\rm{\pi }}{{\rm{D}}^{\rm{2}}}}}{{\rm{4}}}\end{aligned}\)

Therefore the analytical result is,

\({\rm{\varepsilon = }}\frac{{{\rm{\pi NB}}{{\rm{D}}^{\rm{2}}}}}{{\rm{4}}}\)………………………(3)

The numerical value is then evaluated as:

\(\begin{aligned}{}{\rm{\varepsilon }} &= \frac{{{\rm{\pi \times 1000 \times }}\left( {{\rm{5 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}\;{\rm{T}}} \right){\rm{ \times }}{{\left( {{\rm{0}}{\rm{.2}}\;{\rm{m}}} \right)}^{\rm{2}}}}}{{\rm{4}}}\\ &= {\rm{1}}{\rm{.57\;}} \times {\rm{1}}{{\rm{0}}^{ - {\rm{3}}}}\;{\rm{V}}\left( {\frac{{1\;{\rm{mV}}}}{{{\rm{1}}{{\rm{0}}^{ - {\rm{3}}}}\;{\rm{V}}}}} \right)\\ &= {\rm{1}}{\rm{.57}}\;{\rm{mV}}\end{aligned}\)

Therefore, the average \({\rm{emf}}\) induced is obtained as \({\rm{1}}{\rm{.57}}\;{\rm{mV}}\).

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