A flat coil of wire consisting of 15 turns,each with an area of 40 cm²,is positioned perpendicularly to a uniform magnetic field that increases its magnitude at a constant rate from 1.5 T to 5.1 T in 2.0 s.If the coil has a total resistance of 0.20 Ω,what is the magnitude of the induced current? A)7.65E-1 A B)5.40E+1 A C)5.40E-1 A D)8.64E-2 A E)2.16E-1 A

Question

Quizplus · Accepted Answer

The magnetic flux through the coil is given by:
Φ = NAB
where N = 15 is the number of turns, A = 40 cm² = 0.004 m² is the area of each turn, and B is the magnetic field strength.
During the 2.0 s time interval, the magnetic field strength changes from 1.5 T to 5.1 T at a constant rate, so the average magnetic field strength is:
B_avg = (1.5 T + 5.1 T)/2 = 3.3 T
Therefore, the average magnetic flux through the coil is:
Φ_avg = NAB_avg = 15 x 0.004 m² x 3.3 T = 0.198 Wb
Using Faraday's law of electromagnetic induction, the magnitude of the induced emf in the coil is given by:
ε = -dΦ/dt
where ε is the induced emf and dt = 2.0 s is the time interval during which the magnetic field strength changes.
The rate of change of magnetic flux is:
dΦ/dt = ΔΦ/Δt = (0.198 Wb - 0 Wb)/2.0 s = 0.099 V
Therefore, the magnitude of the induced emf is:
ε = -dΦ/dt = -0.099 V
The induced emf causes a current to flow in the coil, which is given by Ohm's law:
I = ε/R
where R = 0.20 Ω is the total resistance of the coil.
Substituting the values, we get:
I = (-0.099 V)/(0.20 Ω) = -0.495 A
The negative sign indicates that the induced current flows in the opposite direction to the direction of the rate of change of magnetic flux. Therefore, the magnitude of the induced current is:
|I| = 0.495 A = 5.40E-1 A (to two significant figures)

A Flat Coil of Wire Consisting of 15 Turns,each with an Area