1. May 9, 2006

### spoonthrower

A piece of copper wire is formed into a single circular loop of radius 12 cm. A magnetic field is oriented parallel to the normal to the loop, and it increases from 0 to 0.55 T in a time of 0.45 s. The wire has a resistance per unit length of 3.3*10^-2 ohms/m. What is the average electrical energy dissipated in the resistance of the wire

I know that P=V^2/R
So i need to find the voltage in the wire.

To find the voltage, can i just find the average induced emf or is this approach wrong??

If i use the equation: average induced emf=BA/T i get the wrong answer simply calculating the area from the radius of the circle .12^2*pi.

2. May 9, 2006

### Hootenanny

Staff Emeritus
I can't see that your doing anything wrong here. Have you mulitplied the resistatance by the length of the wire (the circumference of the circle)? Also have you converted the power into energy (E = P.t)? Perhaps, if you show your calcs...

~H

3. May 9, 2006

### spoonthrower

Since the wire is in a loop i guess i have to use this equation to find the current:

B=N(u)I/(2R) where u=4pi*10^-7 and N=# of turns=1 and R=Radius in meters

solving for I to get I= 105042.2624 A

This is a huge # so i dunno if im doing this right.

Power=I^2R where I is the current through the wire and R is the resistance in ohms. so this leads to my next problem. the problem gives me the resistance in ohms/m so i assume i have to multply this resistance by the length (circumference) of the wire to change it into ohms. so the circumference of my wire is .12(2)pi= .753982237 m. so my resistance would be .753982237 m(3.3*10^-2 ohms/m) to get .024881414 ohms.

now that i know both I and R i can plug them into power=I^2R

=(105042.2624)^2(.024881414)= 274538456.9 W

So power= energy/time so energy=power(time)=274538456.9(.45)=1.24*10^8 J.

This is not the correct answer. Please let me know what i am doing wrong. thanks.

4. May 9, 2006

### Hootenanny

Staff Emeritus
Why are you using current? It would be much easier to find the emf generated as you are given all the required information in the question;

$$emf = N\frac{d\Phi}{dt}$$

You can then sub this value into;

$$P = \frac{V^2}{R}$$

as you originally suggested. You were correct in you method for find the resitance, area of loop and it circumference btw.

~H

5. May 9, 2006