Answer :
Final answer:
The voltage induced in the coil can be calculated using Faraday's law of electromagnetic induction. By applying the formula for induced emf and converting the given parameters, the calculated induced peak voltage is approximately 120 V, making option C) 120 V the correct answer.
Explanation:
To calculate the voltage induced in the coil, we can use Faraday's law of electromagnetic induction which states that the induced electromotive force (emf) in any closed circuit is equal to the negative of the time rate of change of the magnetic flux through the circuit. Since the coil is rotating in a magnetic field, the rate of change of flux is at its maximum when the coil is perpendicular to the field.
The formula for the induced emf (E) is E = NABw sin(wt), where:
- N is the number of turns in the coil,
- A is the area of the coil,
- B is the magnetic field strength,
- w is the angular velocity in radians per second, and
- sin(wt) is the sine of the angular displacement which is 1 at the peak voltage.
The diameter of the coil is 0.15 m, so the radius (r) is 0.15 m / 2 = 0.075 m. Thus, the area (A) is πr2, or π(0.075 m)2.
To find the angular velocity (w), we convert the revolutions per minute (rpm) to radians per second. There are 2π radians in a revolution and 60 seconds in a minute:
w = rpm × (2π radians/revolution) × (1 minute/60 seconds)
w = 3700 × (2π / 60)
w ≈ 386.9 radians/second
Now we can calculate the peak voltage (E) induced in the coil:
E = NABw sin(wt)
E = 200 turns × π(0.075 m)2 × 0.95 T × 386.9 radians/second
E ≈ 120 volts
The induced voltage in the coil is approximately 120 V, so the correct answer is C) 120 V.
