a simple generator has 150 loop circular coil of radius 8.0cm how fast must it turn in a 0.50T field to produce a 100 peak output
Answers
Answer:
Electric generators induce an emf by rotating a coil in a magnetic field, as briefly discussed in Induced Emf and Magnetic Flux. We will now explore generators in more detail. Consider the following example.
EXAMPLE 1. CALCULATING THE EMF INDUCED IN A GENERATOR COIL
The generator coil shown in Figure 1 is rotated through one-fourth of a revolution (from θ = 0º to θ = 90º ) in 15.0 ms. The 200-turn circular coil has a 5.00 cm radius and is in a uniform 1.25 T magnetic field. What is the average emf induced?
The figure shows a schematic diagram of an electric generator. It consists of a rotating rectangular coil placed between the two poles of a permanent magnet shown as two rectangular blocks curved on side facing the coil. The magnetic field B is shown pointing from the North to the South Pole. The two ends of this coil are connected to the two small rings. The two conducting carbon brushes are kept pressed separately on both the rings. The coil is attached to an axle with a handle at the other end. Outer ends of the two brushes are connected to the galvanometer. The axle is mechanically rotated from outside by an angle of ninety degree that is a one fourth revolution, to rotate the coil inside the magnetic field. A current is shown to flow in the coil in clockwise direction and the galvanometer shows a deflection to left.
Figure 1. When this generator coil is rotated through one-fourth of a revolution, the magnetic flux Φ changes from its maximum to zero, inducing an emf.
Strategy
We use Faraday’s law of induction to find the average emf induced over a time Δt:
emf
=
−
N
Δ
Φ
Δ
t
.
We know that N = 200 and Δt = 15.0 ms, and so we must determine the change in flux ΔΦ to find emf.
Solution
Since the area of the loop and the magnetic field strength are constant, we see that
Δ
Φ
=
Δ
(
B
A
cos
θ
)
=
A
B
Δ
(
cos
θ
)
.
Now, Δ (cos θ) = −1.0, since it was given that θ goes from 0º to 90º . Thus ΔΦ = −AB, and
emf
=
N
A
B
Δ
t
.
The area of the loop is A = πr2 = (3.14…)(0.0500m)2 = 7.85 × 10−3 m2. Entering this value gives
emf
=
200
(
7.85
×
10
−
3
m
2
)
(
1.25
T
)
15.0
×
10
−
3
s
=
131
V