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21 January, 00:06

A 50-turn circular coil (radius = 15 cm) with a total resistance of 4.0 Ω is placed in a uniform magnetic field directed perpendicularly to the plane of the coil. The magnitude of this field varies with time according to B = A sin (αt), where A = 80 μT and α = 50π rad/s. What is the magnitude of the current induced in the coil at t = 20 ms?

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  1. 21 January, 00:16
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    Faraday-Lenz's law gives the emf induced in the coil due to the changing magnetic field:

    V = - dΦ/dt

    V = induced emf, dΦ/dt = change of magnetic flux over time

    Apply Ohm's law to the coil:

    V = IR

    V = emf, I = current, R = resistance

    Make a substitution:

    -dΦ/dt = IR

    I = - (dΦ/dt) / R

    The magnetic field is uniform and perpendicular to the coil, so the magnetic flux through the coil is given by:

    Φ = BA

    Φ = flux, B = magnetic field strength, A = coil area

    The magnetic field strength B is given by:

    B = Asin (at)

    A = 80*10⁻⁶T, a = 50π rad/s

    Plug in the values:

    B = 80*10⁻⁶sin (50πt)

    The area of the coil is given by:

    A = πr²

    A = area, r = radius

    Plug in r = 15*10⁻²m

    A = π (15*10⁻²) ²

    A = 0.0225π m²

    Substitute B and A:

    Φ = 80*10⁻⁶sin (50πt) (0.0225π)

    Φ = 80*10⁻⁶sin (50πt)

    Φ = 1.8*10⁻⁶πsin (50πt)

    Differentiate both sides with respect to time t. The radius r doesn't change, so treat it as a constant:

    dΦ/dt = 9.0*10⁻⁵π²cos (50πt)

    Now let's calculate the current I. Givens:

    dΦ/dt = 9.0*10⁻⁵π²cos (50πt), R = 4.0Ω

    Plug in and solve for I:

    I = 9.0*10⁻⁵π²cos (50πt) / 4.0

    I = 2.25*10⁻⁵π²cos (50πt)

    Calculate the current at t = 20*10⁻³s:

    I = 2.25*10⁻⁵π²cos (50π (20*10⁻³))

    I = - 2.2*10⁻⁴A

    The magnitude of the induced current at t = 20*10⁻³s is 2.2*10⁻⁴A
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