A flat coil of wire consisting of 20 turns,each with an area of 50 cm²,is positioned perpendicularly to a uniform magnetic field that increases its magnitude at a constant rate from 2.0 T to 6.0 T in 2.0 s.If the coil has a total resistance of 0.40 W,what is the magnitude of the induced current? A)70 mA B)140 mA C)500 mA D)800 mA

Question

Quizplus · Accepted Answer

The formula for the induced emf in a coil is:
emf = -N(dΦ/dt)
where N is the number of turns and (dΦ/dt) is the rate of change of magnetic flux through the coil.

The magnetic flux through each turn of the coil is:
Φ = BA
where B is the magnitude of the magnetic field and A is the area of one turn.

As the magnetic field increases from 2.0 T to 6.0 T in 2.0 s, the average rate of change of the magnetic field is:
(dB/dt) = (6.0 T - 2.0 T) / 2.0 s = 2.0 T/s

The total magnetic flux through the coil is:
Φ = BA = (2.0 T)(20)(50 cm^2) = 20000 cm^2 T = 0.2 Wb

The rate of change of magnetic flux is:
(dΦ/dt) = (Φ2 - Φ1) / (t2 - t1) = (0.2 Wb - 0 Wb) / (2.0 s - 0 s) = 0.1 Wb/s

Substituting the values into the formula for the induced emf, we get:
emf = -N(dΦ/dt) = -(20)(0.1 Wb/s) = -2.0 V

The magnitude of the induced current is:
I = emf / R = (-2.0 V) / (0.40 W) = 5 A

However, the question asks for the magnitude of the induced current, so the answer is:
I = 500 mA

A Flat Coil of Wire Consisting of 20 Turns,each with an Area