A piece of wire is formed into a circular loop of radius 23 cm. The loop has a resistance of 141 Ω. A magnetic field of 0.9T is applied perpendicular to the plane of the loop and then increased at a constant rate by a factor of 2.2 in 17 s. Calculate the magnitude of the induced emf in the loop during that time.
We are informed that there is a change in magnetic field; this indicates flux, and we apply Faraday's Law:
|ε| = ΔΦ / ΔT where Φ = BAcosθ and cos0 = 1
We don't know B exactly, but we are given a constant factor of increase; therefore we can treat B as 1.
|ε| = (2.2B - B)[π(23.0 cm * 1 m / 100 m)2] / 17 s
|ε| = (1.2 T)[π(0.23 m)2] / 17 s
|ε| = 1.06×10-2 V
Calculate the current induced in the loop during that time.
Simple application of Ohm's Law:
I = ε / R
I = (1.06×10-2 V ) / 141 Ω
I = 7.49×10-5 A
Calculate the average induced emf when the magnetic field is constant at 1.98 T while the loop is pulled horizontally out of the magnetic field region in 6.5 s.
Back to Faraday's Law. This time, the area changes as opposed to the magnetic field, but the same concept applies:
|ε| = ΔΦ / ΔT where Φ = BAcosθ and cos90 = 1
|ε| = (1.98 T)[0 - π(23.0 cm * 1 m / 100 m)2] / 6.5 s
|ε| = 5.06×10-2 V
Rotating Square Coil
What is the peak emf produced by a 74-turn square coil (of side l = 26.0 cm, as shown in the diagram below,) rotating on an axis with a frequency of 37.0 Hz in a uniform magnetic field of 0.695 T perpendicular to the coil's axis of rotation?
Use the equation for a rotating coil (AKA an electric generator), considering the fact that peak emf is when the magnetic field is perpendicular to the coil:
ε = NABωsin(ωt) where ω = 2πf
ε = (74)(26.0 cm * 1 m / 100 m)2 (0.695 T) * (2π*37.0) sin90
ε = 8.08×102 V
Coiled Wire
You have a piece of thin wire that is 15.5m long, a constant uniform magnetic field of 0.155T, and a device that can rotate a coil at a fixed frequency of 87.5Hz. What is the radius of a circular coil made from this length of wire that will produce an AC e.m.f of maximum voltage 116V? (Neglect the amount of wire used in the connections.)
Once again, we use the rotating coil equation:
ε = NABωsin(ωt) where ω = 2πf
However, we don't know N, the number of turns. However, we do know the length of the wire L;
we can equate and find N:
L = 2πrN (Since the length is equal to the circumference times the number of turns)
N = L / 2πr
We can now plug in the second equation into the first, and solve:
ε = NABωsin(ωt) where ω = 2πf
ε = (L / 2πr)ABωsin(ωt)
116 V = (15.5 m /
r = 116 V / (15.5 m * π * 0.155 T * 87.5 Hz)
r = 1.76×10-1 m
Wavetrains
If light is emitted from an atom in little wavetrains, each lasting up to 2.90 × 10-8 s, how long, at most, is such a disturbance in space?
Wavelength is equal to the speed of light multiplied by the time:
L = ct = (3.0E+08 ms-1)(2.90E-8 s)
L = 8.70 m
If we approximate the wavelength as 510 nm, roughly how many waves long is the train?
The number of waves is simply the length divided by the wavelength:
# Waves = L / λ = (8.70 m) / (510 nm * 1.0E-9 m / 1 nm)
# Waves = 1.71×107
Laser Pulses
A laser that emits pulses of UV lasting 2.25 ns has a beam diameter of 2.25 mm. If each burst contains an energy of 2.70 J, what is the length in space of each pulse?
The length in space is simply the speed multiplied by the time (Kinematics):
l = ct = (3.0E+08) * (2.25 ns * 1.0E-9 s / 1 ns)
l = 6.75×10-1 m
What is the average energy per unit volume (the energy density) in one of these pulses?
A laser is a cylindrical beam, so the volume of that beam is the length in space:
V = πr2 l
V = π[(2.25 mm * 1.0 m / 1000 mm) / 2]2 * 6.75×10-1 m
V = 4.77E-3 m3
We can now find energy density by dividing the energy by the volume:
= 2.70 J / 4.77E-3 m3
= 1.01×106 J/m^3
Irradiance of a Candle
The irradiance 1 m from a candle flame is just about 4.65 × 10-3 W/m2. How much energy will arrive in 1.25 s on a disk having a 3.95- cm2 area held as close to perpendicular as possible 1 m from the flame?
Energy is related to irradiance by the following formula:
E = IAt = (4.65 × 10-3 W/m2)(3.95 cm2 * 1.0 m2 / 1.0E+4 cm2)(1.25 s)
E = 2.30×10-6 J
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