I know I have posted this question before but I have tried everything under the sun to no avail. Please help a clueless student!!!!
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.
__________J
Find the emf with faradays law:
emf=-N*area*dB/dt
then, power= emf^2/resistance
where resistance= resistance/length*2PI*radius
To find the average electrical energy dissipated in the resistance of the wire, we can use the formula:
Average electrical energy dissipated = (1/2) * I^2 * R * t,
where I is the current, R is the resistance, and t is the time.
To find the current, we can use Faraday's law of electromagnetic induction:
Emf = -dϕ/dt,
where Emf is the electromotive force, dϕ is the change in magnetic flux, and dt is the change in time.
We know that the magnetic field is increasing from 0 to 0.70 T, and the change in time is given as 0.45 s. The magnetic flux through a circular loop is given by:
ϕ = B * A,
where B is the magnetic field and A is the area of the loop.
The area of a circle is given by:
A = π * r^2,
where r is the radius of the loop.
Substituting the values, we can calculate the change in flux (dϕ):
dϕ = B * A - 0,
= B * π * r^2 - 0,
= 0.70 T * π * (0.13 m)^2.
Now we can calculate the electromotive force (Emf) using the above-determined change in flux and change in time:
Emf = -dϕ/dt,
= - (0.70 T * π * (0.13 m)^2) / 0.45 s.
Next, we need to find the current (I) using Ohm's law:
Emf = I * R * L,
where L is the length of the wire. Here, we are given the resistance per unit length (3.3 * 10^-2 Ω/m) and the length is not specified. Therefore, we assume the length of the wire to be 1 meter.
Substituting the values, we can solve for the current (I):
Emf = I * R * L,
- (0.70 T * π * (0.13 m)^2) / 0.45 s = I * (3.3 * 10^-2 Ω/m) * 1 m.
Now we can solve for the current (I):
I = - (0.70 T * π * (0.13 m)^2) / (0.45 s * (3.3 * 10^-2 Ω/m)).
Once we have the current (I), we can substitute it back into the formula for average electrical energy dissipated to find the answer:
Average electrical energy dissipated = (1/2) * I^2 * R * t,
= (1/2) * (I^2) * (3.3 * 10^-2 Ω/m) * 0.45 s.
Plugging in the calculated current (I) in the above expression will give you the average electrical energy dissipated in the resistance of the wire in joules (J).