Am i Correct?
2.0 cm*.044T*/.10 s= 0.88 V
Each side of a square loop of wire measures 2.0 cm. A magnetic field of 0.044 T perpendicular to the loop changes to zero in 0.10 s. What average emf is induced in the coil during this change?
A. 1.8 V
B. 0.088 V
C. 0.88 V <--
D. 0.00018 V
area of loop = .02^2 = .0004 m^2
dB/dt = .044/.1 = .44
emf = A dB/dt = .0004 *.44 = 1.76 * 10^-4 = .00018 V
To find the average emf induced in the coil during the change in the magnetic field, you can use Faraday's law of electromagnetic induction. According to Faraday's law, the induced emf is given by:
emf = (N * ΔΦ) / Δt
where N is the number of turns in the coil, ΔΦ is the change in the magnetic flux, and Δt is the change in time.
In this case, the loop of wire has only one turn, so N = 1. The change in magnetic flux, ΔΦ, can be calculated by multiplying the area of the loop by the change in magnetic field:
ΔΦ = A * ΔB
ΔΦ = (2.0 cm)^2 * (0.044 T - 0 T)
ΔΦ = (0.02 m)^2 * (0.044 T)
ΔΦ = 0.0000176 T·m²
The change in time, Δt, is given as 0.10 s.
Now you can substitute the values into the formula for emf:
emf = (1 * 0.0000176 T·m²) / (0.10 s)
emf = 0.000176 T·m²/s
Since 1 V is equivalent to 1 T·m²/s, you can convert the emf to volts:
emf = 0.000176 V
Therefore, the average emf induced in the coil during this change is 0.000176 V.
None of the answer choices provided match this value, so the correct answer is not given.
To find the average emf induced in the coil, you can use Faraday's law of electromagnetic induction, which states that the emf (ε) induced in a coil is equal to the rate of change of magnetic flux through the coil.
First, let's calculate the magnetic flux through the coil. The magnetic flux (Φ) through a coil with N turns of wire is given by the formula Φ = B * A * cos(θ), where B is the magnetic field, A is the area of the loop, and θ is the angle between the magnetic field and the normal vector of the loop.
In this case, the magnetic field B is 0.044 T, the area of the loop A is (2.0 cm)^2 = 4.0 cm^2 = 4.0 * 10^(-4) m^2, and θ is 90 degrees since the magnetic field is perpendicular to the loop.
So, Φ = (0.044 T) * (4.0 * 10^(-4) m^2) * cos(90 degrees) = 0.044 * 4.0 * 10^(-4) = 0.0176 * 10^(-4) Wb.
Next, we need to find the rate of change of magnetic flux, which is the change in magnetic flux divided by the change in time. In this case, the magnetic field changes from 0.044 T to 0 (which means the change in magnetic field is 0.044 T) in 0.10 s.
So, the rate of change of magnetic flux (dΦ/dt) = (0.044 T - 0 T) / 0.10 s = 0.044 T / 0.10 s = 0.44 T/s.
Finally, using Faraday's law, the average emf induced in the coil (ε) is equal to the rate of change of magnetic flux (dΦ/dt):
ε = dΦ/dt = 0.44 T/s.
Therefore, the average emf induced in the coil is 0.44 V. However, none of the given options match this value.