Question

The flux through the coils of a solenoid changes from 2.57 x 10⁻⁵ wb to 9.44 x 10⁻⁵ wb in 0.0154 s. if 4.08 v of emf is generated, how many loops does the solenoid have?

Answer 1

Using Faraday's law of electromagnetic induction, the number of turns in the solenoid is calculated to be 960 turns.

Step-by-step explanation:

The flux change (∆Φ) in the solenoid is given by the difference in flux, which is 9.44 x 10⁻⁵ Wb - 2.57 x 10⁻⁵ Wb = 6.87 x 10⁻⁵ Wb. With 4.08 V of EMF generated and a time interval (∆t) of 0.0154 s, we can use Faraday's law of electromagnetic induction, which states that the magnitude of the induced EMF (E) in a coil is equal to the rate of change of magnetic flux through the coil times the number of turns (N) in the coil (E = N × (∆Φ / ∆t)).

So, rearranging the formula to solve for N, we get:

N = E × (∆t / ∆Φ) = (4.08 V) × (0.0154 s / 6.87 x 10⁻⁵ Wb) = 960 turns.