A flat loop of wire consisting of a single turn of cross-sectional area 8.80 cm^2 is perpendicular to a magnetic field that increases unifor

Question

A flat loop of wire consisting of a single turn of cross-sectional area 8.80 cm^2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 2.80 T in 1.01 s.
a. What is the resulting induced current if the loop has a resistance of 1.00Ω?

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Diễm Thu 3 years 2021-08-09T14:41:31+00:00 1 Answers 14 views 0

Answers ( )

    0
    2021-08-09T14:42:45+00:00

    Answer:

    0.00122 A

    Explanation:

    Parameters given:

    Area, A = 8.8 cm² = 0

    00088 m²

    Initial Magnetic field, Bin = 0.5 T

    Final magnetic field, Bfin = 2.8

    Time taken, t = 1.01 s

    Resistance, R = 1 ohms

    EMF induced in a loop of wire of cross sectional area, A due to a changing magnetic field is:

    EMF = [-(Bfin – Bin) * N * A] / t

    Where N = number of loops = 1

    EMF = [-(2.8 – 0.5) * 1 * 0.00088] / 1.01

    EMF = -0.00122 V

    We know that

    V = I * R

    Hence, current, I, will be:

    I = V/R

    I = 0.00122/1 = 0.00122 A

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