A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.30 Ω is in a 1.0 mT magnetic field, with the coil oriented for maxi

Question

A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.30 Ω is in a 1.0 mT magnetic field, with the coil oriented for maximum flux. The coil is connected to an uncharged 3.0 μF capacitor rather than to a current meter. The coil is quickly pulled out of the magnetic field.

Afterward, what is the voltage across the capacitor?

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Trung Dũng 3 days 2021-07-19T02:23:31+00:00 1 Answers 1 views 0

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    2021-07-19T02:25:16+00:00

    Answer:

    The voltage across the capacitor = 0.8723 V

    Explanation:

    From the question, it is said that the coil is quickly pulled out of the magnetic field. Therefore , the final magnetic flux linked to the coil is zero.

    The change in magnetic flux linked to this coil is:

    \delta \phi  = \phi _f - \phi _i

        = 0 – BA cos 0°

    \delta \phi  =-BA \\ \\ \delta \phi  =-B( \pi r^2) \\ \\ \delta \phi  =(1.0  \ mT)[ \pi ( \frac{d}{2} )^2]

    \delta \phi  =(1.0  \ mT)[ \pi ( \frac{0.01}{2} )^2]

    \delta \phi  = -7.85*10^8 \ Wb

    Using Faraday’s Law; the induced emf on N turns of coil is;

    \epsilon = N |\frac{ \delta \phi}{\delta t} |

    Also; the induced current I = \frac{ \epsilon}{R}

    \frac{ \delta q}{ \delta t}  = \frac{1}{R} (N|\frac{\delta \phi}{\delta t } |)

    \delta q = \frac{1}{R} N|\delta \phi |

    \delta q = \frac{1}{0.30} (10)| -7.85*10^8 \ Wb |

    \delta q = 2.617*10^{-6} \ C

    The voltage across the capacitor can now be determined as:

    \delta \ V =\frac{ \delta q}{C}

    = \frac{ 2.617*10^{-6} \ C}{3.0 \ *10^{-6}  F}

    = 0.8723 V

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