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?

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