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.

To find the average electrical energy dissipated in the resistance of the wire, we can first calculate the total magnetic flux passing through the loop. We can then use Faraday's law of electromagnetic induction to find the average induced emf. Finally, we can calculate the average electrical energy dissipated using Ohm's law.

1. Calculate the magnetic flux (Φ):
The magnetic flux passing through the loop can be calculated using the formula:
Φ = B * A
where B is the magnetic field and A is the area of the loop.
Given that the radius of the loop is 13 cm, the area can be calculated as:
A = π * r^2
A = π * (0.13 m)^2
A ≈ 0.0534 m^2

Now, we can calculate the magnetic flux:
Φ = (0.70 T) * (0.0534 m^2)
Φ ≈ 0.0374 Tm^2

2. Calculate the average induced emf:
According to Faraday's law of electromagnetic induction, the induced emf (ε) can be calculated as the rate of change of magnetic flux with respect to time:
ε = -dΦ/dt
where dt is the time interval.
Given that the magnetic field increases from 0 to 0.70 T in a time of 0.45 s, we can substitute the values:
ε = -(0.0374 Tm^2) / (0.45 s)
ε ≈ -0.0831 V

Note: The negative sign indicates that the induced emf opposes the change in magnetic field.

3. Calculate the average electrical energy dissipated:
Using Ohm's law (V = IR), we can calculate the current (I) flowing through the wire by rearranging the equation:
I = ε / R
where R is the resistance per unit length of the wire.

Given that the resistance per unit length is 3.3 * 10^-2 Ω/m, we can substitute the values:
I = (-0.0831 V) / (3.3 * 10^-2 Ω/m)
I ≈ -2.520 A

Since we're only interested in the magnitude of the current, we disregard the negative sign.

Finally, we can calculate the average electrical energy dissipated in the resistance using the formula:
Average electrical energy = I^2 * R * t
where t is the time interval.

Substituting the values, we get:
Average electrical energy = (2.520 A)^2 * (3.3 * 10^-2 Ω/m) * (0.45 s)
Average electrical energy ≈ 0.092 J

Therefore, the average electrical energy dissipated in the resistance of the wire is approximately 0.092 Joules.