- #1
peroAlex
- 35
- 4
HI! I'm a student, this question appeared in one of the old tests. I though I had it solved but apparently I must have missed somewhere. Could someone please at least hint me where I went wrong?
Thin wire is wound into circular frame with radius ## r = 0.05m ## which uniformly rotates around axis with ## \omega = 45 rad/s ## in homogeneous magnetic field ## B = 0.1 T ##. Compute induced voltage ## 0.003 s ## after magnetic dipole of a loop is perpendicular to magnetic field lines.
## U_i = B S \omega \sin{(\omega t)} ##
First I used upper equation (only the ## B S \omega ## part) to compute amplitude of induced voltage which yields ## U_0 \approx 0.0353 V##. Next I tried plugging ## t = 0.003 ## into full equation which returned ## \approx 0.00476 V ##. There are 4 options available (you chose the one which is correct), however, I surprisingly found out that this result is not among 4 possible answers.
Two "plausible" answers are ## 0.035 V ## (which is presumably the amplitude) and ## 0.0217 V ## which seems to be the correct result. The rest are ## 0.0637 V ## and ## 0.0438 V##, but they should be excluded because induced voltage cannot take values higher than amplitude, right?
Now, in order for #0.0217# to stand, ## \sin{\omega t} = 0.63 ## thus ## \omega t = 0.68 rad ##. This would occur when ## t \approx 0.015s## and not at given ##0.003s##!
Remember how magnetic field and magnetic dipole of loop are perpendicular to each other? In this situation, induced voltage should be zero since flux is non-existent. How could it possibly increase to 60% of amplitude in such a short time span? I'm really lost here, could somebody please help me?
Homework Statement
Thin wire is wound into circular frame with radius ## r = 0.05m ## which uniformly rotates around axis with ## \omega = 45 rad/s ## in homogeneous magnetic field ## B = 0.1 T ##. Compute induced voltage ## 0.003 s ## after magnetic dipole of a loop is perpendicular to magnetic field lines.
Homework Equations
## U_i = B S \omega \sin{(\omega t)} ##
The Attempt at a Solution
First I used upper equation (only the ## B S \omega ## part) to compute amplitude of induced voltage which yields ## U_0 \approx 0.0353 V##. Next I tried plugging ## t = 0.003 ## into full equation which returned ## \approx 0.00476 V ##. There are 4 options available (you chose the one which is correct), however, I surprisingly found out that this result is not among 4 possible answers.
Two "plausible" answers are ## 0.035 V ## (which is presumably the amplitude) and ## 0.0217 V ## which seems to be the correct result. The rest are ## 0.0637 V ## and ## 0.0438 V##, but they should be excluded because induced voltage cannot take values higher than amplitude, right?
Now, in order for #0.0217# to stand, ## \sin{\omega t} = 0.63 ## thus ## \omega t = 0.68 rad ##. This would occur when ## t \approx 0.015s## and not at given ##0.003s##!
Remember how magnetic field and magnetic dipole of loop are perpendicular to each other? In this situation, induced voltage should be zero since flux is non-existent. How could it possibly increase to 60% of amplitude in such a short time span? I'm really lost here, could somebody please help me?