A square loop of copper wire is initially placed perpendiclar to the lines of a constant magnetic field of 5\times10^3 T.The area enclosed by the loop

ossidianaZ 2021-02-17 Answered
A square loop of copper wire is initially placed perpendiclar to the lines of a constant magnetic field of \(\displaystyle{5}\times{10}^{{3}}\) T.The area enclosed by the loop is 0.2 square meter. the loopis then turned through an angle \(\displaystyle{90}^{\circ}\) so thatthe plane of the loop is parallel to the field lines. the turntakes 0.1 second. what is the average emf induced in the loop during the turn?

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Expert Answer

yagombyeR
Answered 2021-02-19 Author has 23653 answers
\(\displaystyle{E}_{{{a}{v}{g}}}=-{N}{\frac{{\triangle\phi}}{{\triangle{t}}}}\)
N is number of turns of loop = 1
\(\displaystyle\triangle\phi\) is change of flux linked with the loop
\(\displaystyle\triangle{t}\) is time in which the change of flux has taken place
\(\displaystyle\phi={B}{A}={5}\times{10}^{{3}}\times{0.2}={1}\times{10}^{{3}}\) when plane of loop is perpendicular tofield \(\displaystyle\phi={0}\) when the plane of loop is parallel to field.
\(\displaystyle{E}={1}\times{\frac{{{10}^{{3}}}}{{{0.1}}}}={1}\times{10}^{{4}}\) v
The negative sign in formula simply shows the direction of induced emf is such that it opposes the change influx according to Lenz's Law.
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