Another Faradays Law of Magnetic Induction

[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.

Please help! what am i doing wrong?

 

[lippyka]

The correct equation for the average induced emf is: E = Blv, where B is the magnetic field, l is the length of the wire, and v is the velocity of the change in the flux. The area of the loop does not matter here. Therefore, the average induced emf can be calculated as: E = (0.55 T)(2*pi*12 cm)(0.45 s^-1) = 33.7 V.Now, we can calculate the average electrical energy dissipated as: P = (33.7 V)^2/(3.3 * 10^-2 ohms/m * 2*pi*12 cm) = 20.1 W.

 

[kyle8921]

I can say that your approach is incorrect. The equation you are using, average induced emf=BA/T, is for a single turn of wire. In this scenario, we have a circular loop of wire, which means we have multiple turns of wire. Therefore, we need to use the equation for the total induced emf, which is given by NBA/T, where N is the number of turns of wire.

In this case, N=1, so the equation becomes BA/T. Now, we can calculate the average induced emf by plugging in the values given in the problem: B=0.55 T, A=πr^2=π(0.12)^2=0.045 m^2, and T=0.45 s. This gives us an average induced emf of 0.82 V.

Now we can use the equation P=V^2/R to find the power dissipated in the resistance of the wire. Plugging in the values for V and R (remember, we are given the resistance per unit length, so we need to multiply it by the length of the wire), we get a power dissipation of 0.62 W.

Finally, to find the average electrical energy dissipated, we need to multiply the power by the time, so 0.62 W * 0.45 s = 0.279 J.

Therefore, the average electrical energy dissipated in the resistance of the wire is 0.279 J. I hope this helps clarify your approach and the correct equations to use in this scenario.