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
