Solving Induced EMF Problem with Jordan's Help

[Jordanosaur]

Hi guys -

Here's the problem I'm stuck on:

In a physics laboratory experiment, a coil with 160 turns enclosing an area of
13.7 (cm^2) is rotated during the time interval 4.60×10−2(s) from a position in which its plane is perpendicular to Earth's magnetic field to one in which its plane is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 6.50×10−5 (T).


I am not able to figure out how they came up with that figure of induced EMF as

3.10*10^-4

If you use NwBAsin(wt), you come out with 4.86*10^-4 V. Am I missing something in my calculations? Unless I am misunderstanding, this is a rotational motion question requiring the calculation of angular speed for the change in flux.

(PI/2) / (4.60*10^-2) = w (angular velocity)

sin(wt) = 1, therefore change in flux = NwBA

Any help or advice would be much appreciated

Thanks

Jordan

 

[tyco05]

The answer 3.1E-4 V is correct.

Don't think of it in terms of equations. I hate this "Formula physics" that I'm seeing everywhere!

The flux goes from a maximum (when perpendicular) to a minimum (zero when parallel).

The change in flux is then simply whatever the maximum is (because the final flux is zero).

Flux is the amount of field cutting a unit area. Don't forget to convert from cm^2 to m^2.

Induced emf is given by the rate of change of flux.
Average induced emf is given by the change in flux per change in time.

You know the change in the flux (simply the max flux), and the change in time for this.

Then, since there are 160 coils, the total emf is 160 times the emf for one coil.

 

[ogb p]

,

Hi Jordan,

It seems like you are on the right track with your calculations. The induced EMF (electromotive force) can be calculated using the equation E = NABωsin(ωt), where N is the number of turns, A is the area, B is the magnetic field strength, ω is the angular velocity, and t is the time interval.

In this problem, we know that N = 160, A = 13.7 cm^2 = 0.00137 m^2, B = 6.50×10−5 T, and t = 4.60×10−2 s. We also know that the change in flux (ΔΦ) is equal to NABωsin(ωt).

Substituting the known values, we get:

ΔΦ = 160 * (0.00137 m^2) * (6.50×10−5 T) * ω * 1

Since the coil is rotated from a position perpendicular to Earth's magnetic field to one parallel to the field, the change in flux will be equal to the maximum flux, which occurs when the coil is parallel to the field. This means that sin(ωt) = 1.

Solving for ω, we get:

ω = ΔΦ / (NAB) = (3.10×10−4 V) / (160 * 0.00137 m^2 * 6.50×10−5 T) = 0.357 rad/s

So, the angular velocity of the coil is 0.357 rad/s. This is the value that you were missing in your calculations.

I hope this helps. Let me know if you have any further questions.

Best,