A coil 4.00 cm in radius, containing 500 turns, is placed in a uniform magnetic field that varies with time according to B=(0.0120T/s)t+(3.0

Question

A coil 4.00 cm in radius, containing 500 turns, is placed in a uniform magnetic field that varies with time according to B=(0.0120T/s)t+(3.00×10−5T/s4)t4. The coil is connected to a 600−Ω resistor, and its plane is perpendicular to the magnetic field. You can ignore the resistance of the coil. (a) Find the magnitude of the induced emf in the coil as a function of time. (b) What is the current in the resistor at time t = 5.00 s?

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MichaelMet 5 months 2021-08-14T23:51:30+00:00 1 Answers 108 views 0

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    2021-08-14T23:53:04+00:00

    Answer:

    A. e = 0.0302 V+ (3.02 x 10-4 V/s)t3

    B. I = 1.13 x 10-4 A

    Explanation:

    (a) Consider a coil with radius of r = 4.00 cm, and containing N = 500

    turns, it is placed perpendicularly in a magnetic field of B, which is given

    by,

    B = (0.0120 T/s)t + (3.00 x 10-5 T/s) 14

    the magnitude of the induced emf is,

    |E 1 =

    NdФВ

    at

    where

    B = B A cos(0) = BA. Since the area is constant, then,

    |E| = NA (B)

    = NA ((0.0120 T/s)t + (3.00 x 10-6 T/s*) 4]

    NA[(0.0120 T/s)+(1920 x 10-4 T/s*) t1

    (500) (0.040)? [(0.0120 T/s)+ (1.20 x 10-4 T/s^) tº]

    = 0.0302 V + (3.02 x 10-4 V/s°) 43

    e = 0.0302 V+ (3.02 x 10-4 V/s) t3

    (b) The induced emf at t = 5.00 s, is,

    e = 0.0302 V +(3.02 x 10-4 V/s) (5.00 s)3

    = 0.0680 V

    if the resistance of the coil is R = 600 2, then the induced current at t = 5.00

    s, is,

    E 0.0680 V – 1.13 x 10-4 A

    I= = 600 12

    I = 1.13 x 10-4 A

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