Question

a current in a solenoid is changing at a rate of 0.030 a/s which induces as emf of magnitude 14.2 mv. at the moment that the current is equal to 1.8 a, the average magnetic flux through each turn of the solenoid is 0.0042 wb. how man turns are in the solenoid (round to the nearest turn)?

Answer 1

Using Faraday's law of induction, we find that the solenoid has approximately 3 turns when the induced emf is 14.2 mV and the rate of magnetic flux change is 0.0042 Wb.

Step-by-step explanation:

To calculate the number of turns in the solenoid, we can use Faraday's law of induction, which states that the induced emf (ε) is equal to the rate of change of magnetic flux (Φ) times the number of turns (N). The formula is:

ε = -N(dΦ/dt)

Given that the induced emf is 14.2 mV (or 0.0142 V) and the rate of change of current is 0.030 A/s, we have:

ε = 0.0142 V
dΦ/dt = 0.0042 Wb (since the current at the moment is 1.8 A which is not changing)

Now we solve for N:

N = ε / (dΦ/dt) = 0.0142 V / 0.0042 Wb = 3.38095

So, rounding to the nearest turn, the solenoid has 3 turns.