Problem 25.52
A 100-turn, 2.0-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 60 degrees away from vertical increases from 0.50 T to 1.50 T in 0.60 s .
Part A
What is the induced emf in the coil?
Express your answer using two significant figures.
To find the induced emf in the coil, we can use Faraday's law of electromagnetic induction, which states that the magnitude of the induced emf is equal to the rate of change of magnetic flux through the coil.
The magnetic flux through a coil is given by the product of the magnetic field, the area of the coil, and the cosine of the angle between the magnetic field and the normal to the coil.
Given:
Number of turns in the coil (N) = 100 turns
Diameter of the coil (d) = 2.0 cm = 0.02 m
Initial magnetic field (B1) = 0.50 T
Final magnetic field (B2) = 1.50 T
Time interval (Δt) = 0.60 s
Angle between the magnetic field and the vertical (θ) = 60 degrees
The area of the coil (A) can be calculated using the formula: A = π * (d/2)^2
Let's calculate the area of the coil first:
A = π * (0.02 m/2)^2
A ≈ 0.000314 m^2
Next, we can calculate the change in magnetic flux (ΔΦ) using the formula: ΔΦ = N * B2 * A * cos(θ) - N * B1 * A * cos(θ)
ΔΦ = 100 turns * (1.50 T * 0.000314 m^2 * cos(60 degrees) - 0.50 T * 0.000314 m^2 * cos(60 degrees))
ΔΦ ≈ 0.047 T·m^2
Finally, we can find the induced emf (ε) using the formula: ε = ΔΦ / Δt
ε = 0.047 T·m^2 / 0.60 s
ε ≈ 0.079 V
Therefore, the induced emf in the coil is approximately 0.079 V.