# Solve Physics Problem: Average Electrical Energy Dissipated

• osustudent2010
In summary, the problem discussed is a physics problem involving a piece of copper wire formed into a single circular loop, with a changing magnetic field and a given resistance per unit length. The conversation discusses using Faraday's law to calculate the induced EMF, treating the wire loop as a circuit, and calculating the average power dissipated in the resistance of the wire. The final answer is found using the equation E=Pt, where E is the average electrical energy and P is the average power.
osustudent2010
Hi, I am struggling with this physics problem:

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 0.50 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 (in J) dissipated in the resistance of the wire.

I really don't know where to start in this problem. I can find the circum of the loop and area of the loop. But I'm not sure how to relate resistance, magentic flux, and electrical enery- any help would be appricated!

Thanks

Faraday's law will allow you to calculate the induced EMF due to the changing magnetic flux through the loop of wire. Start there.

I found the EMF to be -.0422
(-NA delta B)/ delta t
where N= the number of loops
A = area
B= magnetic field
t= time

OK. What are the units of EMF?

Now treat the wire loop as a circuit. What's the average power dissipated in the resistance of the wire?

the units are V, so .0422 V

P= (V^2)/R (via P=IV and V=IR)

so I took P= (.0422^2)/.0228

The R came from the circum of the loop times the resistance per unit of length

The answer I got was 1.85 J, but this isn't the right answer-- is there something I'm missing?

osustudent2010 said:
the units are V, so .0422 V

P= (V^2)/R (via P=IV and V=IR)

so I took P= (.0422^2)/.0228
Looks OK. That's the average power. (What units?)

The R came from the circum of the loop times the resistance per unit of length
OK.

The answer I got was 1.85 J, but this isn't the right answer-- is there something I'm missing?
How did you get this answer?

I got it! I made a calculation error earlier

I got the answer by using this equation:
E=Pt

where P= equals the average power
t= time

THANKS so much for helping me through this problem, you are the best!

## What is the formula for calculating average electrical energy dissipated?

The formula for calculating average electrical energy dissipated is E = P x t, where E is the energy dissipated in joules (J), P is the power in watts (W), and t is the time in seconds (s).

## How do I determine the power in a circuit?

To determine the power in a circuit, you can use the formula P = VI, where P is the power in watts (W), V is the voltage in volts (V), and I is the current in amperes (A).

## What is the difference between average and instantaneous electrical energy dissipated?

Average electrical energy dissipated is the total energy dissipated over a period of time, while instantaneous electrical energy dissipated is the energy dissipated at a specific moment in time. Average electrical energy dissipated takes into account fluctuations in power, while instantaneous electrical energy dissipated only considers the power at a single point in time.

## How do I convert units when calculating average electrical energy dissipated?

To convert units when calculating average electrical energy dissipated, you can use the following conversions: 1 watt = 1 joule/second and 1 watt-hour = 3600 joules. For example, if you have a power of 100 watts and a time of 2 hours, you would convert the time to seconds (7200 seconds) and then use the formula E = P x t to calculate the energy dissipated (720,000 joules).

## Can average electrical energy dissipated be negative?

Yes, average electrical energy dissipated can be negative if the power is negative, meaning that energy is being supplied to the circuit rather than dissipated. This can occur in certain circuits, such as those with capacitors or when using alternating current (AC).

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