How to Solve Physics Electromagnetism Problems

[seiferseph]

I have 3 more physics electromagnetism problems, thanks again for the help (helped tremendously in the last unit for my test).

http://i2.photobucket.com/albums/y15/seiferseph/1.jpg

not sure what equation to use, i can solve for change in flux using the magnetic field and area, but not sure how to get induced emf (which would help me get current)


http://i2.photobucket.com/albums/y15/seiferseph/2.jpg

how do i get the direction of the acceleration?


http://i2.photobucket.com/albums/y15/seiferseph/3.jpg

not sure what to do here, i thought you use lenz's law where it creates a counter b field against the motion of the bar magnet, but that doesn't work for the bottom.

thanks again!

 

[OlderDan]

I have 3 more physics electromagnetism problems, thanks again for the help (helped tremendously in the last unit for my test).

http://i2.photobucket.com/albums/y15/seiferseph/1.jpg

not sure what equation to use, i can solve for change in flux using the magnetic field and area, but not sure how to get induced emf (which would help me get current)


http://i2.photobucket.com/albums/y15/seiferseph/2.jpg

how do i get the direction of the acceleration?


http://i2.photobucket.com/albums/y15/seiferseph/3.jpg

not sure what to do here, i thought you use lenz's law where it creates a counter b field against the motion of the bar magnet, but that doesn't work for the bottom.

thanks again!

For the first one, the induced emf depends on the rate of change of the magnetic flux. Review Faraday's law. Since current depends on the emf, and current is the rate of charge flow, you will find a connection between the flux change and the amount of charge.

The second is just the force on a current carrying wire in a magnetic field.

For the last, review Lenz's Law and the right hand rule for fields produced by moving charges.

 

[seiferseph]

The second is just the force on a current carrying wire in a magnetic field.

For the last, review Lenz's Law and the right hand rule for fields produced by moving charges.

ok, for the first i solve for change in flux, then multiply by number of turns to get emf. then i use I = V/R and get 0.06 A for the current. why is that the same as the charge? and doesn't the formula for emf use time as well?


for the second one, the b-field is up and current left, so i get the force going into the page, how does that make it go right?

for the 3rd, it creates a counter b field at each point opposite, so doesn't it make a north at the top (to keep it there) and a south at bottom (to oppose)? that's the opposite of the answer.

 

[OlderDan]

ok, for the first i solve for change in flux, then multiply by number of turns to get emf. then i use I = V/R and get 0.06 A for the current. why is that the same as the charge? and doesn't the formula for emf use time as well?


for the second one, the b-field is up and current left, so i get the force going into the page, how does that make it go right?

for the 3rd, it creates a counter b field at each point opposite, so doesn't it make a north at the top (to keep it there) and a south at bottom (to oppose)? that's the opposite of the answer.

Assume you change the flux at a constant rate, and it takes time T to make the change. That will produce a constant emf for that time, and a constant current will flow that is proportional to the magnitude of the emf. The current is the charge that flows divided by the time it takes to flow, or T

|emf| = change in flux/T
|I| = |emf|/R = charge/T

Same time on both sides of the equation, so...

For the second, how do you get current to the left? Current is into the page for the moveable bar. The rails are stationary.

For the last one, think in terms of flux. The bottom is trying to increase the flux downward (field lines getting closer together- field strength increasing), so the induced emf has to produce a flux upward (magnetic field points upward. The top is decreasing the flux downward (field lines getting farther apart- field strength decreasing), so the induced emf has to increase the flux downward.

In terms of North and South, the bottom has to repel the falling magnet and the top has to attract it. The induced field must have "north" below pointing upward and above pointing downward.

 

[seiferseph]

Assume you change the flux at a constant rate, and it takes time T to make the change. That will produce a constant emf for that time, and a constant current will flow that is proportional to the magnitude of the emf. The current is the charge that flows divided by the time it takes to flow, or T

|emf| = change in flux/T
|I| = |emf|/R = charge/T

Same time on both sides of the equation, so...

ok, i see now how to solve without the time.

For the second, how do you get current to the left? Current is into the page for the moveable bar. The rails are stationary.

ok, so the current (the one created by the battery or whatever, which is shown to go to the left) goes through the bar, therefore going into the page where the bar is? then i get it.

For the last one, think in terms of flux. The bottom is trying to increase the flux downward (field lines getting closer together- field strength increasing), so the induced emf has to produce a flux upward (magnetic field points upward. The top is decreasing the flux downward (field lines getting farther apart- field strength decreasing), so the induced emf has to increase the flux downward.

In terms of North and South, the bottom has to repel the falling magnet and the top has to attract it. The induced field must have "north" below pointing upward and above pointing downward.

ok i get it now, thanks a lot for the help!