Generator rotation speed, motor counter emf, and induced current in coil

[gillgill]

1. A simple generator has a 200-loop square coil 25.0cm on a side. How fast must it turn in a 0.550T field to produce a 120V peak output?

area=(0.25)^2
=6.25x10^-2

E=-N(BA)/t
t=-(300)(6.25x10^-2)(0-0.550T)/120V
t=8.6x10^-2 s
do you need to times it by 4 to get the time for one rotation??

2. A motor has an armature resistance of 3.40ohms. If it draws 8.70A when running at full speed and connected to a 120V line, how large is the counter emf?
V-Ir-Emf=0
Emf=V-Ir
=120-(8.7)(3.4)
=90.4V
Is this right?

3. A 30cm diameter coil consists of 30 turns of circular copper wire 2.8mm in diameter. A uniform magnetic field, perpendicular to the plane of the coil, changes at a rate of 8.65x10^-3T/s. Determine the current in the loop.

I have no idea how to start this question.
Plz help...thx...

 

[wais]

1. Yes, you are correct. To find the time for one rotation, you would need to multiply by 4. This is because the coil has 4 sides, so it will go through 4 rotations for each full rotation of the generator.

2. Your calculation for the counter emf is correct. Keep in mind that the units for resistance are ohms, not volts, so your final answer should be 90.4V.

3. To find the current in the loop, you can use Faraday's Law of Electromagnetic Induction: E=-N(dB/dt), where E is the induced emf, N is the number of turns in the coil, and (dB/dt) is the rate of change of the magnetic field. In this case, E=0 since the loop is open, so we can rearrange the equation to solve for the current: I=-E/R. The resistance of the loop can be calculated using the formula R=ρL/A, where ρ is the resistivity of copper, L is the length of the wire, and A is the cross-sectional area. Once you have the resistance, you can plug it into the equation for current to find the answer.