> Two friends A and B (each weighing 40 kg) are sitting on a frictionless platform some distance d apart. A rolls a ball of mass 4 kg on the platform towards B which B catches. Then B rolls the ball towards A and A catches it. The ball keeps on moving back and forth between A and B. The ball has a fixed speed of 5 m/s on the platform.
Why is speed of A after a round trip an integral multiple of speed gained in one round trip i.e $\dfrac{10}{11}$ ? And for B also?
I don't know if the is correct but I'll try. this is a frictionless surface so when I person A, my velocity can change in three scenarios.
1) I throw the ball
2) I receive the ball
3) I throw the ball again, but this would be considered as part of the second trip, so I'll ignore it
When I throw the ball it's velocity is fixed 5 m/s. By conservation of momentum 5*4=40*v
So my velocity after 1) scenario becomes 1/2 m/s
In second scenario I assume I bring the ball to a halt before throwing it back. Once again the change in momentum is 4*5 and so my increase in velocity would be again 1/2. And so after one trip my total increase in 1 m/s
@JohnRennie good morning! Is there electric field inside the wire?
I was exploring whether there is a displacement current in the wire along with conduction current and I saw that If there was an electric field inside wire, then the change in that electric field would naturally bring about a displacement current right?
The displacement current is proportional to dE/dt so for there to be a displacement current we need: 1. an electric field must exist 2. that field must be changing with time
In an ideal (no resistance) wire there is no electric field so (1) is not satisfied.
In a real wire (with resistance) there is an electric field, but if the current is constant the field is constant so (2) is not satisfied.
In a real wire with a changing current then we would have dE/dt ≠ 0 and only in that case would we have a displacement current.
@Wolgwang Hi :-)
The ball travels at 5 m/s relative to the platform, so every time it is caught then thrown again its momentum changes from +5m to -5m, or from -5m to +5m. So the change is 10m or 40 Nm.
And by conservation of momentum that means every time A or B catch then throw the ball their momentum changes by 40 Nm i.e. their velocity changes by 1 m/s.
Yes, though we don't usually consider displacement currents in AC circuits except for components like capacitors. Now you have me wondering how the displacement current for a resistor behaves in an AC circuit ...
A perfect capacitor has an infinite DC resistance so the conduction current would be zero. But yes real or faulty capacitors would have a finite resistance so you'd get both a current and a displacement current.
The displacement current isn't really a current, but it kind of behaves like a current and has the same units as current.
Then yes. For an ideal wire ρ = 0 so E = 0, but for a real wire ρ≠ 0 and we do get an electric field E = jρ in the wire.
A while back someone posted a YouTube video talking about displacement currents and I thought it was pretty good. I'll post a link in case you're interested.
I had a theory that, if you want flow of electrons there must be a potential difference, and if there's a potential difference there should be an electric field
> if you want flow of electrons there must be a potential difference
In real life that is true, but in an ideal case it isn't. Remember that electrons have a mass, and if you get them moving at some speed v then if there is no resistance they will just keep moving at the same speed.
So you can have the electrons moving with a constant velocity, i.e. a current, even when there is no potential difference. In fact this can happen in superconductors.
The way to understand problems like this is to remember that a plate in a capacitor has an inner and outer surface i.e. it looks like this (diagram incoming):
The usual rules for plates apply, so the outer charges are the same and the inner charges are equal and opposite.
The charge on the capacitor is then the value of the inner charge.
As you suggested above the internal charge is q = (Q₁ + Q₂)/2 (taking Q₂ as a positive number i.e. the charge on the right plate is -Q₂)
> Figure shows a small body of mass m placed over a larger mass M whose surface is horizontal near the smaller mass and gradually curves to become vertical. The smaller mass is pushed on the longer one at a speed v and the system is left to itself. Assume that all the surfaces are frictionless.
> (a) Find the speed of the larger block when the smaller block is sliding on the vertical part. (b) Find the speed of the smaller mass when it breaks off the larger mass at height h. (c) Find the maximum height (from the ground) that the smaller mass ascends. (d) Show that the smaller mass will again land on the bigger one. Find the distance traveled by the bigger block during the time when the smaller block was in its flight under gravity.
I calculated the distance in (d) part by finding the velocity of small block in y-axis from the resultant velocity (from (b)), and then calculating time of flight. Is there some other way?
At the top of the curved block the surface is vertical, so when the small block reaches the top of the curved block the horizontal component of its velocity relative to the curved block must be zero. Yes?
This means the small block will move vertically upwards, slow to a halt and then fall vertically downwards again and land in the same place i.e. at the top of the big block.
This is not analytically solvable but we can start with F = -ax and this is solvable. Then we can add in the x³ term and calcuate how this changed the solution we get when only a is non-zero.
@sonicsid Indeed, I doubt you need this for the JEE. Though it is a widely used technique in physics.
I've got a basic idea of it now. Had it been a physical concept I would have been interested, I'll learn techniques that are not useful for me right now when the time comes instead. Thank you anyway!
Is it "widely used" enough to be of use in the JEE somewhere? For some complicated problem in, say, rigid body dynamics could I take a simpler case for approximation and then do something to make the answer more accurate? Or would it be helpful only in cases where the motion of the particle is being governed by some long polynomial?
@JohnRennie Good evening! I want to expand on my initial doubt earlier, does an electric field exist outside the current-carrying wire? Does it depend on whether it is AC or DC?
Imagine the nature of the electric field if the wires are not next to each other. In fact imagine the wires are going in the opposite direction away from the source for miles and then at the right angle for miles and then again at a right angle back around to the load somewhere and then keeps go...
In the modern electromagnetism textbooks, electric fields in the presence of stationary currents are assumed to be conservative,$$
\nabla \times E~=~0
~.$$ Using this we get$$
E_{||}^{\text{out}}~=~E_{||}^{\text{in}}
~,$$which means we have the same amount of electric field just outside of the wi...
These two questions strongly suggest the presence of a radial field, especially the first one with the diagram
Someone told it's due to surface charge and someone else told "As an electronics engineer, this seems trivially true. Consider HV power lines... What's the big deal?"
But anna v says "I am an experimentalist. There do not exist appreciable electric fields outside current carrying wires. We are all doing the experiment continually just by typing on the computer to communicate on this page. There would be sparks continually. So we can only talk of fields smaller than the ionization energy of air, or our skin"
If you consider a current carrying conductor, every instant an electron enters the conductor, another electron will be leaving the conductor. Thus, the current carrying conductor will not be charged (i.e. it would not have any net positive or negative charge). Remember a dipole has zero net charg...