A piece of copper wire is formed into a single circular loop of radius 13 cm. A magnetic field is oriented parallel to the normal to the loop, and it increases from 0 to 0.70 T in a time of 0.45 s. The wire has a resistance per unit length of 3.3 10-2 /m. What is the average electrical energy dissipated in the resistance of the wire.
The change in B divided by the time (0.45 s), multiplied by the loop area in square meters, is the average voltage in the loop during the interval. The wire resistance is 2 pi R * 3.3 10-2 ohms(?)/m.
Coumpute the average current (Vav/R) and multiply it by the average voltage, Vav, computed by the method above.
Average Power =
[(delta B)(pi R^2)/(delta t)]^2/(2 pi R)
= (1/2) pi* R^3 [(delta B)/(delta t)]^2
Check my math and my logic. I could be wrong
A thin tube stretch across a street counts the number of pairs of wheels that pass over it. A vehicle classified as Type A with two axles registers two counts. A vehicle classified as Type B with three axles registers three counts. During a 1 hour period a traffic counter registered 35 counts. How many type A vehicles and type b vehicles passed over the traffic?
To find the average electrical energy dissipated in the resistance of the wire, we need to use the formula for electrical energy dissipated in a resistor:
E = I^2Rt
Where:
E is the electrical energy dissipated (in Joules)
I is the current flowing through the resistor (in Amperes)
R is the resistance of the wire (in Ohms)
t is the time (in seconds)
To calculate the current, we need to use Faraday's law of electromagnetic induction, which states that the induced electromotive force (EMF) in a circuit is equal to the rate of change of magnetic flux through the circuit:
EMF = -dφ/dt
Where:
EMF is the induced electromotive force (in Volts)
dφ/dt is the change in magnetic flux (in Weber/s or Volts)
In this case, the change in magnetic flux is equal to the product of the magnetic field strength (B) and the area enclosed by the loop (A):
dφ/dt = B * A/t
Since the loop is circular, the area can be calculated using the formula for the area of a circle:
A = πr^2
Substituting these values into the equation for the change in magnetic flux, we get:
dφ/dt = B * πr^2 / t
Now, we can substitute this equation for the induced EMF into Ohm's law, which states that the current flowing through a resistor is equal to the voltage across it divided by its resistance:
I = EMF / R
Substituting the equation for the induced EMF, we get:
I = (B * πr^2 / t) / R
Finally, we can substitute the values given in the question to calculate the average electrical energy dissipated:
I = (0.70 T * π * (0.13 m)^2) / (0.45 s * 3.3 * 10^-2 Ω/m)
First, calculate the value inside the parentheses:
I = (0.70 T * π * 0.0169 m^2) / (0.045 s * 3.3 * 10^-2 Ω/m)
I = (0.037688 T*m^2) / (0.1485 Ω)
Now calculate the current:
I = 0.2536 A
Finally, calculate the average electrical energy dissipated:
E = (0.2536 A)^2 * (3.3 * 10^-2 Ω/m) * 0.45 s
E = 0.00969468 J
Therefore, the average electrical energy dissipated in the resistance of the wire is 0.00969468 Joules.