Calculate the induced EMF of a rotating loop

[dainceptionman_02]

Homework Statement

A rectangular loop (area = 0.15m^2) turns in a uniform magnetic field, B=0.20T. When the angle between the field and the normal to the plane of the loop is π/2 rad and increasing at 0.60rad/s, what emf is induced in the loop?

Relevant Equations

ξ=-N(dΦ/dt), Φ=BAcosθ

there are a bunch of problems in this section that ask similar questions, but they ask the amplitude and this doesn't. this is an even problem so i do not have the answer, but my hunch is that it is not an amplitude question. i solved for the amplitude so i am guessing i got this one wrong.

ξ=-N(dΦ/dt) = -NBA(d/dt)(cosθ) = -NBA(-sinθ)dθ/dt, where dθ/dt=ω, which is 0.60rad/s.

the problem states that initially the angle between the field and the normal to the plane is π/2 rad and increasing at 0.60rad/s, which should mean that initially the flux is zero, because Φ=BAcosθ=BAcos(π/2)=0

now, the amplitude, or the maximum emf induced in the loop is ωNBA = (0.60rad/s)(1)(0.20T)(0.15m^2) = 0.018V, but like i said, i think they are asking something else, but i don't know what?

 

[BvU]

Hi,

!​

Homework Statement:: A rectangular loop (area = 0.15m^2) turns in a uniform magnetic field, B=0.20T. When the angle between the field and the normal to the plane of the loop is π/2 rad and increasing at 0.60rad/s, what emf is induced in the loop?
Relevant Equations:: ξ=-N(dΦ/dt), Φ=BAcosθ

i think they are asking something else

They are asking the instantaneous emf, for which you have a relevant equation with all factors known ...

[edit] Oh, and: when in doubt, make a sketch !

$\$

 

[dainceptionman_02]

They are asking the instantaneous emf, for which you have a relevant equation with all factors known ...

[edit] Oh, and: when in doubt, make a sketch !

i have never seen this before! what do i do, or how do i set it up? thanks!

 

[BvU]

You already did !

ξ=-N(dΦ/dt) = -NBA(d/dt)(cosθ) = -NBA(-sinθ)dθ/dt

with all factors known

 

[dainceptionman_02]

You already did !

with all factors known

so the instantaneous emf is the amplitude, or ωNBA?

 

[BvU]

Make a sketch to understand why !

 

[nasu]

so the instantaneous emf is the amplitude, or ωNBA?

What do you think is this ξ in the formula shown by BvU? You seem to know it too.

 

[Steve4Physics]

I'd like to add a few words to what @BvU has already said.

so the instantaneous emf is the amplitude, or ωNBA?

EDIT: amplitude = ωNBA. But instantaneous emf means something else.

That's not true in general. (It can be true in special cases.)

When a coil rotates in a magnetic field, the emf produced varies with time. At different times (different angles) during a rotation, the emf can be positive or negative or zero. For a steady angular speed, a graph of emf against time is a sine curve.

The amplitude is the maximum value of |emf|; it is given by ωNBA for a simple setup.

That means if you want the value of emf at some arbitrary moment of time (some arbitrary angle) you can’t use ωNBA.

Of course if the particular angle of interest is, by coincidence, an angle where |emf| is maximum, then ωNBA gives you |emf| but not the sign (+ or -) of the emf.