A50-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?

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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Explanation:

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