course phy 232
Hey, I have a question with a magnetism problem set. Its been a while since ive submitted anything but i have been working through book, or a was a lot more a few weeks ago. Trying to manage this with other classes but, after a few tests this week ill be able to apply myself to the E&M and hopefully be able to take the test pretty soon. the part that im doubtful of is that that the ""average magnitude of the voltage,
You might need to read the set first ,
but i would think that the average woudl be the change in flux of .816 tesla m^2 to zero would be (.816-0)/2 and then this would be divided by the time for a rotation to find the average voltage,
let me know the reasoning behind this. i might not be thinking it through
The average rate of change is (change in voltage) / (change in clock time). This has nothing to do with dividing by 2.
If you divide the difference of two values of a quantity by two, you get half the difference, which isn't involved with the rate-of-change definition.
The average rate is (change in flux) / (change in clock time) = ( .816 T m ^ 2)/( .06755 sec) = 24.15 T m ^ 2 / sec = 24.15 Volts.
all the way
this is the whole problem set:
A magnetic field of magnitude 5.1 Tesla passes through a square loop with side .4 meters, with the field perpendicular to the plane of the loop. The loop rotates at 3.7 Hz. What is the average magnitude of the voltage produced?
Solution
3.7 Hz means 3.7 cycles per second. This implies a time of 1/ 3.7 seconds = .2702 sec for each cycle.
One cycle corresponds to 360 degrees; a 90 degree turn corresponds to a time of ( .2702 sec)/4 = .06755 sec.
The area of the loop is ( .4 m) ^ 2 = .16 m ^ 2, so the flux when the plane of the loop is prependicular to the field is ( 5.1 T)( .16 m ^ 2) = .816 T m ^ 2.
After being turned 90 degrees, the loop will be parallel to the field, and will intercept none of the field. The flux will then be zero.
The magnitude of the flux change from .816 T m ^ 2 to zero is .816 T m ^ 2.
This occurs in .06755 seconds, so the average rate is ( .816 T m ^ 2)/( .06755 sec) = 24.15 T m ^ 2 / sec = 24.15 Volts.
Since this is the average during each quarter of a cycle, it is equal to the overall average.
Generalized Solution
If a loop rotates at frequency f in a magnetic field, then the time required for a complete rotation is 1/f (think of 10 cycles per second; the time per cycle is obviously 1/10 sec). In a quarter of a rotation the loop will rotate through 90 degrees. This will require 1/4 of the time of a cycle. Since a complete rotation takes time 1/f, the time to complete 1/4 of a cycle is `dt = 1/4 (1/f) = 1 / (4f).
If the loop rotates 90 degrees, starting from a position where the field is perpendicular to the loop, the magnitude of the flux change will be | .7 `phi | = B * A. Since this takes place in time `dt = 1 / (4f), the rate of flux change will be
average rate of flux change = | `d`phi | / `dt = B * A / ( 1 / (4f) ) = 4f * B * A.
This is in fact the average rate, since the magnitude of the flux change will be B * A through every subsequent 90 degree rotation.
"
course phy 232
Hey, I have a question with a magnetism problem set. Its been a while since ive submitted anything but i have been working through book, or a was a lot more a few weeks ago. Trying to manage this with other classes but, after a few tests this week ill be able to apply myself to the E&M and hopefully be able to take the test pretty soon. the part that im doubtful of is that the
You might need to read the set first ,
but i would think that the average woudl be the change in flux of .816 tesla m^2 to zero would be (.816-0)/2 and then this would be divided by the time for a rotation to find the average voltage,
let me know the reasoning behind this. i might not be thinking it through all the way
this is the whole problem set:
A magnetic field of magnitude 5.1 Tesla passes through a square loop with side .4 meters, with the field perpendicular to the plane of the loop. The loop rotates at 3.7 Hz. What is the average magnitude of the voltage produced?
Solution
3.7 Hz means 3.7 cycles per second. This implies a time of 1/ 3.7 seconds = .2702 sec for each cycle.
One cycle corresponds to 360 degrees; a 90 degree turn corresponds to a time of ( .2702 sec)/4 = .06755 sec.
The area of the loop is ( .4 m) ^ 2 = .16 m ^ 2, so the flux when the plane of the loop is prependicular to the field is ( 5.1 T)( .16 m ^ 2) = .816 T m ^ 2.
After being turned 90 degrees, the loop will be parallel to the field, and will intercept none of the field. The flux will then be zero.
The magnitude of the flux change from .816 T m ^ 2 to zero is .816 T m ^ 2.
This occurs in .06755 seconds, so the average rate is ( .816 T m ^ 2)/( .06755 sec) = 24.15 T m ^ 2 / sec = 24.15 Volts.
Since this is the average during each quarter of a cycle, it is equal to the overall average.
Generalized Solution
If a loop rotates at frequency f in a magnetic field, then the time required for a complete rotation is 1/f (think of 10 cycles per second; the time per cycle is obviously 1/10 sec). In a quarter of a rotation the loop will rotate through 90 degrees. This will require 1/4 of the time of a cycle. Since a complete rotation takes time 1/f, the time to complete 1/4 of a cycle is `dt = 1/4 (1/f) = 1 / (4f).
If the loop rotates 90 degrees, starting from a position where the field is perpendicular to the loop, the magnitude of the flux change will be | .7 `phi | = B * A. Since this takes place in time `dt = 1 / (4f), the rate of flux change will be
average rate of flux change = | `d`phi | / `dt = B * A / ( 1 / (4f) ) = 4f * B * A.
This is in fact the average rate, since the magnitude of the flux change will be B * A through every subsequent 90 degree rotation.
Check my note and let me know if that doesn't clear it up.