An electric generator contains a coil of 100 turns of wire, each turn forming a rectangular loop 50cm
by 30cm. The coil is placed entirely in a uniform magnetic field with magnitude, B = 3.5T and
direction initially perpendicular to the coil's plane. What is the maximum value of the emf produced
when the coil is spun at 1000 rev/min about an axis perpendicular to the direction of the magnetic field.
To find the maximum value of the emf (electromotive force) produced by the spinning coil, we need to use the formula:
emf = N * B * A * ω
where:
- emf is the electromotive force (in volts)
- N is the number of turns in the coil
- B is the magnitude of the magnetic field (in teslas)
- A is the area of one loop of the coil (in square meters)
- ω is the angular velocity (in radians per second)
Let's calculate each component step by step:
1. Number of turns (N):
The problem states that the coil contains 100 turns of wire.
N = 100
2. Magnitude of the magnetic field (B):
The problem states that the magnitude of the magnetic field is B = 3.5 T (teslas).
B = 3.5 T
3. Area of one loop of the coil (A):
Each turn forms a rectangular loop measuring 50 cm by 30 cm. We need to convert these measurements to meters.
Length = 50 cm = 0.5 m
Width = 30 cm = 0.3 m
A = Length * Width
A = 0.5 m * 0.3 m = 0.15 square meters
4. Angular velocity (ω):
The problem states that the coil is spun at 1000 revolutions per minute. To convert to radians per second, we need to use the following conversion:
1 revolution = 2π radians
So,
ω = (1000 rev/min) * (2π radians/1 revolution) * (1 min/60 s)
ω ≈ 104.72 radians/s
Now, we can put all the components together and calculate the maximum value of the emf:
emf = N * B * A * ω
emf = 100 * 3.5 T * 0.15 m^2 * 104.72 radians/s
emf ≈ 5496 V
Therefore, the maximum value of the emf produced when the coil is spun at 1000 rev/min is approximately 5496 volts.