A flat loop of wire consisting of a single turn of cross-sectional area 8.10 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 1.90 T in 0.95 s. What is the resulting induced current if the loop has a resistance of 2.80 ?
To find the resulting induced current in the loop, you can use Faraday's law of electromagnetic induction, which states that the induced electromotive force (emf) in a circuit is equal to the rate of change of magnetic flux.
The magnetic flux is given by the product of the magnetic field strength, the area of the loop, and the cosine of the angle between the magnetic field and the normal of the loop:
Φ = B * A * cos(θ)
In this case, the angle between the magnetic field and the normal of the loop is 90 degrees since the loop is perpendicular to the field. Therefore, cos(θ) = 1.
The rate of change of magnetic flux is given by:
dΦ/dt = (Φ2 - Φ1) / Δt
where Φ2 and Φ1 are the final and initial magnetic fluxes respectively, and Δt is the change in time.
Since the area of the loop, A, is given as 8.10 cm², we need to convert it to square meters by dividing by 10000:
A = 8.10 cm² / 10000 = 0.00081 m²
The final and initial magnetic fluxes can be calculated using the given magnetic field values:
Φ2 = B2 * A
Φ1 = B1 * A
where B2 = 1.90 T and B1 = 0.500 T.
Now we can calculate the rate of change of magnetic flux:
dΦ/dt = (1.90 * 0.00081 - 0.500 * 0.00081) / 0.95
After calculating this expression, you will get the value of dΦ/dt.
Finally, to find the induced current, you can use Ohm's law: V = I * R, where V is the induced emf, I is the induced current, and R is the resistance of the loop.
The induced emf is equal to the rate of change of magnetic flux:
V = dΦ/dt
Now you can calculate the induced current:
I = V / R
Substituting the value of V with the calculated rate of change of magnetic flux and the given resistance of 2.80 Ω, you can determine the resulting induced current in the loop.