A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum emf generated

Question

A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum emf generated is:

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Tryphena 1 month 2021-08-12T08:08:42+00:00 1 Answers 4 views 0

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    2021-08-12T08:10:41+00:00

    Answer:

    Maximum emf = 5.32 V

    Explanation:

    Given that,

    Number of turns, N = 10

    Radius of loop, r = 3 cm = 0.03 m

    It made 60 revolutions per second

    Magnetic field, B = 0.5 T

    We need to find maximum emf generated in the loop. It is based on the concept of Faraday’s law. The induced emf is given by :

    \epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t

    For maximum emf, \sin\omega t=1

    So,

    \epsilon=NB\pi r^2\omega \\\\\epsilon=NB\pi r^2\times 2\pi f\\\\\epsilon=10\times 0.5\times \pi (0.03)^2\times 2\pi \times 60\\\\\epsilon=5.32\ V

    So, the maximum emf generated in the loop is 5.32 V.

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