Show what the magnitude of induced emf

AI Thread Summary
The discussion centers on calculating the magnitude of induced electromotive force (emf) in a magnetic field defined by B = K(x^3z^2, 0, -x^2z^3)sin(ωt). Participants are working through the integration process to derive the emf around a circle in the plane z = a. Key points include the need to correctly identify and integrate the magnetic field component Bz, and the importance of using polar coordinates for the integration limits. There is confusion regarding the variables involved in the double integral, with clarification that only two variables should be considered since z is constant. The final expression for the induced emf is ε = (K/4)πa^3R^4ωcos(ωt).
auk411
Messages
54
Reaction score
0
Someone who knows what they are talking about: Show what the magnitude of induced emf

Homework Statement



Consider a magnetic field B = K(x3z2,0, -x2z3)sinωt in the region of interest, where K and ω are positive constants and t is variable time. Show that the magnitude of the induced emf around a circle R in the plane z = a with its center at x = 0, y = 0, z = a is:
ε = (K/4)∏a3R4ωcosωt

Homework Equations



Fluxb = ∫B . dA

The Attempt at a Solution



Since the normal vector points in the k direction, we only have to worry about Bz.

∫Bzdydx. So -∫∫(sinwt)x2a3dydx.

The make the change to polar:

-aK3∫∫(sinwt)(rcosθ)2r dr dθ = -(K/4)a3R4∫cosθsin(wt) dθ.

This doesn't get me anywhere. I'm not really sure what I'm supposed to be integrating over, which is probably why I'm stuck.
 
Last edited:
Physics news on Phys.org
Hi auk411! :smile:

Let's start with Bz.
Can you write it down separately?
It seems you did not copy it correctly.

Secondly you did not bring the constant out of the integral properly.

Furthermore in polar coordinates you would integrate r from 0 to R, and theta from 0 to 2pi.

And for the tip: rewrite (cosθ)2 using cos2θ.
 
I like Serena said:
Hi auk411! :smile:

Let's start with Bz.
Can you write it down separately?
It seems you did not copy it correctly.

Secondly you did not bring the constant out of the integral properly.

Furthermore in polar coordinates you would integrate r from 0 to R, and theta from 0 to 2pi.

And for the tip: rewrite (cosθ)2 using cos2θ.

(cosθ)2 using cos2θ.[/QUOTE] ... huh, what trig identity are you using.

this still doesn't answer the most pressing question. t varies, theta varies and r varies. we have 3 varying variables in a DOUBLE integral. I see no way to reduce them to two. How do I get around this.
 
First things first.
You appear to have skipped my question, so I'll answer it myself:

Bz = -Kx2a3sinωt

This is not what you used.

auk411 said:
(cosθ)2 using cos2θ. ... huh, what trig identity are you using.

cos(2θ) = 2 cos2θ - 1


auk411 said:
this still doesn't answer the most pressing question. t varies, theta varies and r varies. we have 3 varying variables in a DOUBLE integral. I see no way to reduce them to two. How do I get around this.

No, you have 2 variables.
You appear to be thinking spherical coordinates, but you should be thinking in cylindrical coordinates.

z is constant at z=a.
Only the other 2 vary.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top