Calculating magnetic flux of a rod

AI Thread Summary
The discussion revolves around calculating the electromotive force (EMF) and velocity related to magnetic flux in a rod. The given magnetic field strength, length of the rod, and angular velocity are used to derive solutions for two problems. The user explores two methods: one using electric principles and another involving an imaginary area to calculate magnetic flux. The main inquiry is about the validity of using an imaginary area to arrive at the same solution for both problems, despite the absence of a physical area in the scenario. Clarifications are sought on the underlying principles that justify this approach.
yahelra
Messages
2
Reaction score
0

Homework Statement


B = 0.02T
L = 40 cm
w = 10 rad/s
a: Electromotive force between O and C = ?
1 - Copy.jpeg


b: If electromotive force is the same - so what is the velocity?
2.jpeg
I found 2 ways of solution. See "3. The Attempt at a Solution "

Homework Equations


φ = BA
ε = -dφ(t)/dt

The Attempt at a Solution


So I found 2 ways:
1. What every one would do - by Electric:
sol.jpeg


2. By the magnetic flux. But wait - There is no area.
So I found that if I would take in problem a the area of what the rod had passed (part of a disc) I would get the exact solution.
1 - Copy.jpeg

1.jpeg


For problem b if I would take the area of a triangle I will get the solution too.My question - Why does the second way is also true? How can it be explained (For both problem a and b)?

Thanks a lot!
 
Last edited:
Physics news on Phys.org
Is that a line or a rod? The way the picture is presented it doesn't look like there is any flux, let alone a change in flux. Did you give us the entire problem statement?
 
Last edited:
I gave you the entire statement.
And you are right, as I wrote there are no magnetic flux because there is no area.

I meant that I've found that we can take an imaginary area and it will give us the same solution. ( See picture -0.016 V)

My questions are why is it true and how can it be explained?
 
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}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top