You have a coil with 20 loops inside a magnetic field generator that is temporarily generating a field that has a rate of change of -0.25. The other end of this coil is wrapped around a core with 15 turns. How many turns must a second coil wrapped around the same coil have to produce an output voltage of 25 V?
To solve this question, we need to use Faraday's law of electromagnetic induction, which states that the induced voltage in a coil is directly proportional to the rate of change of the magnetic flux through the coil.
First, let's calculate the magnetic flux through the first coil. The magnetic flux (Φ) is given by the product of the magnetic field (B) and the area (A) through which the field lines pass:
Φ = B * A
Since the number of loops (N) in the first coil is 20, the effective area (A) is multiplied by 20. Hence:
Φ1 = 20 * B
Next, we need to calculate the rate of change of magnetic flux (dΦ/dt) through the first coil. The given rate of change of the magnetic field (dB/dt) is -0.25:
dΦ1/dt = 20 * dB/dt
Now, let's consider the second coil. Since the second coil is wrapped around the first coil, the magnetic flux through the second coil (Φ2) is proportional to the magnetic flux through the first coil (Φ1):
Φ2 ∝ Φ1
Given that the number of turns in the second coil (N2) is unknown, we can express the proportionality as:
Φ2 ∝ N2 * Φ1
Substituting the previously calculated value for Φ1, we get:
Φ2 ∝ N2 * (20 * B)
Finally, we can use Faraday's law to find the voltage (V2) induced in the second coil:
V2 = N2 * (dΦ2/dt)
Substituting the proportionality expression for Φ2, we have:
V2 = N2 * (dΦ2/dt) = N2 * d(N2 * (20 * B))/dt = N2 * N2 * 20 * dB/dt
Now, we can solve for N2 by setting V2 to 25 V and the given value of dB/dt to -0.25:
25 V = N2 * N2 * 20 * (-0.25)
Rearranging the equation:
N2 * N2 = -25 V / (20 * (-0.25))
N2 * N2 = -25 V / (-5)
N2 * N2 = 5 V
Taking the square root of both sides:
N2 = √(5 V) ≈ 2.236
Therefore, the second coil must have approximately 2.236 turns to produce an output voltage of 25 V.