- #1
aokidopi
- 29
- 0
1. http://www.webassign.net/userimages/ch20_prob20_40.jpg?db=v4net&id=110544
A square loop consists of a single turn with a resistance of 5.00 ohms. The loop has an area of 500 cm2, and has a uniform magnetic field passing through it that is directed out of the page. The loop contains a 12-volt battery, connected as shown in the figure above.
(a) At the instant shown in the figure, there is no net current in the loop. At what rate is the magnetic field changing? Use a positive sign if the field is increasing and magnitude, and a negative sign if the field is decreasing in magnitude.
T/s
(b) If the polarity of the battery was reversed, and the magnetic field was still changing at the rate you calculated above, what would the magnitude of the net current through the loop be?
A
2. Faraday's Law and Magnetic Flux
Flux=ABcos(theta)
emf=N*change in flux/change in time
V=IR
3. I was able to solve for part A using a combination of the first two equations to solve for Magnetic field/ time.
However, I am having trouble understanding the second part. If the rate of change in magnetic field is same, isn't emf still 12V and the resistance 5 ohms? I know that 12/5 for the ohm's law does not result in the correct solution.
Thanks for your help.
A square loop consists of a single turn with a resistance of 5.00 ohms. The loop has an area of 500 cm2, and has a uniform magnetic field passing through it that is directed out of the page. The loop contains a 12-volt battery, connected as shown in the figure above.
(a) At the instant shown in the figure, there is no net current in the loop. At what rate is the magnetic field changing? Use a positive sign if the field is increasing and magnitude, and a negative sign if the field is decreasing in magnitude.
T/s
(b) If the polarity of the battery was reversed, and the magnetic field was still changing at the rate you calculated above, what would the magnitude of the net current through the loop be?
A
2. Faraday's Law and Magnetic Flux
Flux=ABcos(theta)
emf=N*change in flux/change in time
V=IR
3. I was able to solve for part A using a combination of the first two equations to solve for Magnetic field/ time.
However, I am having trouble understanding the second part. If the rate of change in magnetic field is same, isn't emf still 12V and the resistance 5 ohms? I know that 12/5 for the ohm's law does not result in the correct solution.
Thanks for your help.