How does the current flow in the sheet? The drawing shows a strip i.e. a sheet that is long and narrow. Does the current flow along the long direction?
In that case I would treat the sheet as made up from infinitesimal strips of width $dx$. Each string carries a current $dI = I/b dx$ and generates the same field as a wire.
@Abcd possibly, though the field lines around the strip will be vaguely oval shaped and the circuital law works best when you can choose a path that is either parallel or normal to the field lines.
@Abcd in setups like these it's not obvious what the total flux means i.e. what you are suppose to calculate to get the total flux. In the case of the two wires you take the region between the two wires and integrate the flux over that area.
Consider a simple situation:
Of an infinite wire and a loop kept in front of it. We are to find the mutual inductance of the system.
$\phi = Mi$, where M denotes mutual inductance.
The method we adopt to solve such questions is:
Make current flow through the object you like and calculate t...
I think what is being calculated here is the partial inductance. That is, you write the inductance of a loop as a sum of the partial self inductance and partial mutual inductances of the elements of the loop.
If we call the four edges a, b, c and d (a and c long) then the inductance of the loop can be written as $L = L_a + L_b + L_c + L_d + M_{ac} + M_{bd}$ where $L_a$ etc are the partial self inducatnces and $M_{ac}$ etc are the partial mutual inductances.
In the limit of very long a and c the contributions from the short edges become negligible and we get $L \approx L_a + L_c + M_{ac}$
If you Google partial inductance you'll find loads of articles on this. I hadn't heard of the term either, so I'm essentially thinking aloud here.
The point is that taking our example of the infinitely long loop we can define the partial inductances as the components of the total inductnace, and the total inductance is well defined.
@harambe it's a slightly artificial situation because we're assuming whatever collision supplied the impulse lasted zero time. Since it lasted zero time the angle rotated and distance moved during the collision are both zero, and the linear and angular momentum changed discontinuously at time zero.
Since the changes at time zero are discontinuous we can't, strictly speaking, talk about the value of any physical property at zero.
When the string is cut two things happen: 1. the rod starts accelerating downwards i.e. the rod centre of mass stars accelerating downwards 2. the rod starts rotating clockwise
And we know that the left end if the rod stays fixed because it's connected to the string, so we know that the downward acceleration of the left end of the rod due to gravity must be the same as the upward acceleration due to the rotation.
With physics problems there are always two steps: 1. set up the equations of motion 2. solve them
You seem to be very good at step 2 - obviously your maths is very good. It's just figuring out what the equations of motion should be that's the problem.
Yes, with complicated motion like that we tend to use a technique called Hamiltonian or Langrangian mechanics to calculate the motion. But that's far more advanced than you'd do for JEE. You won't study that unless you do physics at university.
I wrote the translational equation as well as torque about centre of mass but I am stuck at one step. I don't know the magnitude of which torque is greater -the normal reaction or spring force
@harambe I think the answer is (b) 20N. Assuming I'm correct you can do the question by taking moments about the centre of mass of the rod or about the pivot. I think taking moments about the pivot is slightly simpler though there isn't a lot in it.
I'm out for a couple of hours now but will be back around 16:00 UK time or tomorrow morning.
And also one more question ,the space between two metallic coaxial cylinders are filled with material of resistivity rho .What is resistnace between the cylinders
well how will measure it?
we used travelling microscope to measure ri of glass slab.
@starunique2016 you get a range of fluids with different refractive indexes and put the diamond in each fluid in turn. When the RI of the diamond and fluid are the same the diamond will disappear.
@JohnRennie Suspensions are propagated from the main site account to chat, so this user's main chat account is, in fact, also currently suspended. It is technically already a breach of the "don't use sock to do something you couldn't without them" for them to chat in any room at all.
Since I don't see any reason to forbid them chatting in rooms not associated with physics.SE, I'm not going to look for that, but this room is parented to physics.SE.
You mean reflect in the convex mirror to produce a virtual image then reflect that in the concave mirror to produce a real image at the position of the object?
@IceInkberry We are on refraction? Are you ahead? Is optics easier or harder than Electrodynamics and mechanics. Arrange in increasing order of hardness: Mechanics, Electro-StatDynamics, Optics.
In this case the object is the image formed by reflection in the first mirror (the mirror on the right) so it isn't a physical thing, it's just a place where light rays converge.
Suppose the image is formed to the right of the convex mirror i.e. somewhere in between the two mirrors.
Suppose we have light rays coming in from the right. If the mirror weren't there they would converge at a point to the left of the mirror as shown by the dashed lines, so that's where the image would form.
What actually happens is that the mirror intercepts the light rays before they have converged to form an image and reflects them back, so now they converge somewhere on the right of the mirror and form the image there.
Anonymous
@Abcd We almost completed the chapter. It is easy, atleast geometrical optics is easy. Don't know about wave optics. Geometrical optics is almost the same what we had learnt in lower classes except that there are variations in questions. There are fixed type of questions here. If they want to make it difficult, they mix kinematics/mechanics with it.
But in fact the mirror behaves just as if there were an (imaginary) object to the left. That is, if you put $u$ in as a negative number then it will give the correct (positive) result for $v$.
Consider a simple situation:
Of an infinite wire and a loop kept in front of it. We are to find the mutual inductance of the system.
$\phi = Mi$, where M denotes mutual inductance.
The method we adopt to solve such questions is:
Make current flow through the object you like and calculate t...
