A piece of copper wire is formed into a single circular loop of radius 11 cm. A magnetic field is oriented parallel to the normal to the loop, and it increases from 0 to 0.65 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 calculate the average electrical energy dissipated in the resistance of the wire, we can use the formula:

\[ E = I^2 \times R \times t \]

Where:
- E is the electrical energy dissipated
- I is the current flowing through the wire
- R is the resistance of the wire
- t is the time for which the current flows

To find the current flowing through the wire, we can use Faraday's law of electromagnetic induction, which states that the induced electromotive force (EMF) is equal to the rate of change of magnetic flux through the loop. The induced EMF can be calculated by multiplying the rate of change of magnetic field with time, with the area enclosed by the loop.

Let's calculate the induced EMF first:

\[ EMF = - \frac{d\Phi}{dt} \]

Where:
- EMF is the induced electromotive force
- \(\Phi\) is the magnetic flux

The magnetic flux through the loop is given by:

\[ \Phi = BA \]

Where:
- B is the magnetic field
- A is the area of the loop

The area of the loop can be calculated using the formula:

\[ A = \pi r^2 \]

By substituting the given values:

\[ A = \pi \times (0.11 \, m)^2 \]

Now we can calculate the magnetic flux:

\[ \Phi = (0.65 \, T) \times \pi \times (0.11 \, m)^2 \]

Next, we can calculate the rate of change of magnetic flux:

\[ \frac{d\Phi}{dt} = \frac{\Delta \Phi}{\Delta t} \]

\[ \frac{d\Phi}{dt} = \frac{\Phi_{final} - \Phi_{initial}}{\Delta t} \]

\[ \frac{d\Phi}{dt} = \frac{(0.65 \, T) \times \pi \times (0.11 \, m)^2 - 0}{0.45 \, s} \]

With the induced EMF calculated, we can now calculate the current flowing through the wire using Ohm's Law:

\[ EMF = I \times R \]

\[ I = \frac{EMF}{R} \]

\[ I = \frac{\frac{d\Phi}{dt}}{R} \]

Now we can substitute the values and calculate the current:

\[ I = \frac{\frac{(0.65 \, T) \times \pi \times (0.11 \, m)^2 - 0}{0.45 \, s}}{3.3 \times 10^{-2} \, \Omega/m} \]

Finally, we can calculate the average electrical energy dissipated:

\[ E = I^2 \times R \times t \]

\[ E = \left(\frac{\frac{(0.65 \, T) \times \pi \times (0.11 \, m)^2 - 0}{0.45 \, s}}{3.3 \times 10^{-2} \, \Omega/m}\right)^2 \times (3.3 \times 10^{-2} \, \Omega/m) \times 0.45 \, s \]