A magnetic field increases from 0 to 0.09 T in 3.0 s. How many turns of wire are needed in a circular coil 20 cm in diameter to produce an induced emf of 7.5 V? Please help soon.

dB/dt=.09/3

E=n*area*dB/dt

To solve this problem, you can use Faraday's law of electromagnetic induction, which states that the induced electromotive force (emf) in a coil is equal to the rate of change of magnetic flux through the coil.

First, let's determine the change in magnetic flux. The magnetic flux through a coil is given by the formula:

Φ = B * A * N,

where Φ is the magnetic flux, B is the magnetic field, A is the area of the coil, and N is the number of turns of wire.

Given that the magnetic field increases from 0 to 0.09 T, and the coil has a diameter of 20 cm (or a radius of 10 cm), we can calculate the change in magnetic flux as follows:

ΔΦ = B_final * A - B_initial * A,

where B_final is the final magnetic field, B_initial is the initial magnetic field (which is 0 in this case), and A is the area of the coil.

Substituting the given values:

ΔΦ = 0.09 T * π * (0.1 m)^2 - 0 T * π * (0.1 m)^2.

Simplifying the equation:

ΔΦ = 0.09 T * π * 0.01 m^2.

Next, we need to calculate the induced emf using Faraday's law, which states that the induced emf is equal to the rate of change of magnetic flux:

emf = -dΦ/dt,

where emf is the induced emf, and dt is the change in time.

Given that the induced emf is 7.5 V, and the change in time is 3.0 s, we can equate the induced emf to the rate of change of magnetic flux:

7.5 V = -ΔΦ / dt.

Rearranging the equation to solve for ΔΦ:

ΔΦ = -7.5 V * dt.

Substituting the given value for dt:

ΔΦ = -7.5 V * 3.0 s.

Finally, we can equate the two expressions for ΔΦ and solve for the number of turns N:

0.09 T * π * 0.01 m^2 = -7.5 V * 3.0 s * N.

Simplifying the equation:

N = (0.09 T * π * 0.01 m^2) / (-7.5 V * 3.0 s).

Calculating the equation:

N ≈ 0.012095 turns.

Therefore, approximately 0.012095 turns of wire are needed in a circular coil 20 cm in diameter to produce an induced emf of 7.5 V.