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JEE MAIN & ADVANCE 12th PCM Physics -Electrostatics module-2 Demo Videos
Welcome back, students.
So, students, till now we have been dealing with examples of field and cases of field due to discrete charges. There was a charge or two point charges or multiple charges placed at different, different points. Now, let us discuss the case of continuous charge systems and to start with we will be finding electric field to a line charge at axial position. Now, students, we will be following a certain strategy to solve these questions or these cases. Number one, we need to choose the element whose field is already known to us. For example if you talk of this line charge. Now, this line charge, we can, we can imagine this line charge is made of small, small point charges and since we know the field to point charge. Therefore we can consider a very small length which will be just like a point charge. Similarly, for this ring, if we closely observe and take a very small part on this particular ring, it will also be like a point charge whose field is known to us. And so if we talk of this disc and this disc, if we closely analyse, it will be made up of small, small rings and if we know the field to a ring then we should be taking our element as the ring. So, first step is to choose an element. Step 2 is to find field to this element and next step is to integrate. So, let’s apply that in this particular scenario wherein we have to find electric field due to a finite line charge, at any point along the line charge. So, what do we have? We have a line charge and we need to find field at point P and what is given to us? It is given all these conditions that is length, distance of this point from one of the edges and we are given that this line charge has a linear charge density Lamda and length L. So, step one will be to choose a very small element whose field is known to us. So, let’s consider a small element whose field is known to us. Now, what will that element be? Remember, for line charges the elements will be a very small length which will be like a point charge. Now, if we closely observe this line charge is made up of small charge plus small charge plus small charge and so on and out of these small charges if we closely observe and then take a very small portion and say it is at a distance of x from O and has a thickness of dx then this element will be just like a point charge and so it qualifies to be an element. So our element is a point charge at distance x from origin having length dx and having a charge dq which will be equal to Lamda dx. Why Lamda dx, because linear charge density Lambda implies charge on 1 unit length is Lambda. So, charge on dx unit length will be Lambda dx. So, we know charge on element is Lambda dx. Now, what is our step 2? Step 2 is to find field due to the small element that is find dE vector. Now, what is dE in magnitude? Remember field to a point charge is k charge upon distance square. So, the charge of this element is dq and what is the distance, distance will be total length, I mean, L plus a minus x. So, hence magnitude of the field will be kdq upon L plus a minus x square. Now, as far as direction is concerned. Since this dq charge is positive. So, field due to dq charge will be in this direction that is this d vector will be kdq in this direction. We can rewrite it as kdq upon L plus a minus x square I caron. So, now we know field due to our element. We have to find field at P, what is our step 3? Step 3 is to integrate. Now, before we integrate remember, students, our all the terms should be in same variables. Now, since dq is Lambda dx, we can rewrite this equation as dE vector is equal to k Lambda dx upon a plus L minus x square I caron. Now, we can integrate now. So, integrating we need to apply limits. Now, what will be the limits be? If we closely observe limits of x will be from 0 to x equal to L, because if we closely observe line charge starts from x is equal to 0 and goes till x is equal to L. So, hence this limit will become 0 to L. Now, what is the integration of this expression? Remember, integration of this expression L plus a minus x raise to power n dx is as shown. Provided n is not equal to 1. That is integral is L plus a minus x raise to power n plus 1 upon n plus 1 and there is a minus sign outside it because in this term there is coefficient of x is minus 1. So, let’s integrate we get E vector as minus k Lambda into a plus L minus x raise to power minus 1 upon minus 1 into minus 1 I caron, and if we apply limits we will get this result. I hope you know how to apply limits. Value of expression at x is equal to upper limit minus value of expression at x is equal to lower limit. So, in this case we will get this expression. On further solving we will get the result to be k lambda L upon a plus L into a I caron.
Students, if you feel that I have been fast here please pause re-watch this particular thing because you need repetitions to understand it nicely. So, we will get back in next module till then students, thank you.
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JEE MAIN & ADVANCE 12th PCM Physics -Electrostatics Module-3 Demo Videos
Welcome back, students. So, students in previous module we had just mentioned potential due to various continuous charges. We learnt potential due to ring of charge, due to an arc of charge. More over we also talked about potential due to a line of charge. Now, let us extend the discussion and study potential due to spherical charge configurations.
So, let us start with hollow spherical charge. Let us say we have a hollow sphere whose charge is capital Q and radius is capital R. Now, what will be potential at any point inside, say at a distance r from the centre? Students, potential at any point inside as a matter of fact is k capital Q by capital R. It is constant, it is independent of small r, right, surprised. So remember students, potential inside a hollow spherical charge distribution is constant, it is same as that of potential at surface. So, if we draw the graph of potential versus R for inside the region it will be a constant. So, till r small r does not become capital R, potential will be a constant, as shown. Now, what will be potential at any point on the outside? So, at any point on outside potential will be equal to k capital Q by small r. So, its graph will be a hyperbolic graph as shown which will be inversely proportional to small r. Students, I hope you remember what was electric field due to hollow spherical charge distributions. Electric field outside was kQ by small r square and inside it was zero. So, we had learned that for outside points the sphere of charge behaves like a point charge placed at centre. The story remains the same even for potentials. But for inside electric field inside was zero and here electric inside it is k capital Q by capital R that is constant.
Now, let’s talk about solid spherical charge distributions. Again let us consider we have solid sphere having charge capital Q and radius r. Now, what will be potential at a small r distance inside the solid sphere, will it be constant like the case of hollow spheres? No, students, it will not be constant. It will be equal to some (2:39) some tough complicated formula but quite easy to remember. k capital Q by 2 capital R into 3 minus small r square by capital R square. It is very big complicated formula but still not that tough to remember. Remember, repeat it 3 times with me, potential inside a solid spherical charge distribution is k capital Q by 2 capital R into 3 minus small r square by capital R square. Two more times, potential inside due to solid spherical charge distribution is k capital Q by 2 capital R into 3 minus small r square by capital R square. Third time, potential inside due to solid spherical charge distribution is k capital Q by 2 capital R into 3 minus small r square by capital R square. Students, you please repeat it, I will have to move on but you can repeat it as many times as you want. Now, if you have to draw the graph of this potential. You can see in this case, potential is a quadratic function, it is proportional to minus r square. So, its graph will be something like this. Potential at o will be basically potential at centre having small r is equal to zero. So, it will be 3kQ by 2R and potential at surface will be at small r is equal to capital R it will be equal to k capital Q by R. So, potential inside is, we are aware of, what will be potential outside. Now, at outside point potential just like hollow spherical charges, potential is constant, is like potential at centre. Its capital KQ by small r. So, at outside points potential due to the solid sphere is same as that of potential due to a point charged placed at centre. It will be equal to k capital Q by small r. So, if you have to draw the graph it will again be a hyperbolic function. So, students, I hope you remember potential due to hollow sphere inside is constant, due to solid sphere inside is this, tough, this big formula. Potential outside due to hollow sphere and solid sphere is same.
We will get back in next module. You please repeat this module as these particular formulas are important. Till then thank you, students.
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JEE MAIN & ADVANCE 12th PCM Physics -Electromagnetic Induction Module-3 Demo Videos
Hello, students, now let’s try to find the induced EMF in a rod which is placed in a time bearing magnetic field. So see, here’s the situation, there is a rod of length L which is kept at a distance a from the centre. And this is a cylindrical magnetic field region. Now what will happen, since the magnetic field is changing and electric field we induced in the region and because of this electric field basically an EMF will be developed across the ends of the rod. We know the formula for induced electric field it is half rdB by dt. This is the induced electric field inside the cylindrical region. Now what will we do, we will analyze the situation, see this is a rod, electric field is at an angle theta with the rod. So how will we calculate the potential difference? So dv is equal to E.dl, or magnitude of dv is magnitude of E.dl. Now what we can do E dl, that means E.dl is E cos theta into dl because l is along the length, dl is along the length and E is at an angle theta. So d is equal to E cos theta into dl. And the value of E electric field is half rdB by dt so we replace here half r dB by dt cos theta dl.
Now in this formula what is r, r is the distance of the point from the centre. So in this diagram r will be basically a divided by cos theta. So we can replace r which is a variable actually by a cos theta. Now when we do that, r is a by cos theta and we had cos theta dl. So cos theta gets cancelled and we have a very simple expression a by 2 dB by dt integral dl. Now integral dl will be l so basically we have induced EMF across the ends as aL by 2 dB by dt. Now also note we have taken dB by dt outside the integration. That means we have assumed dB by dt is constant that means rate at which magnetic field is changing is constant, then only the formula is aL by 2 dB by dt. What is a, a is the distance of the rod from the centre. L is the length of the rod. So we have to remember this formula aL by 2 dB by dt is the induced EMF across the ends of the rod placed in a cylindrical time bearing magnetic field.
Now what about the direction, the direction basically can be given by the direction of induced electric field. So suppose here induced electric field is clockwise that means it will try to make the current flow in clockwise direction. So the direction of induced EMF will be such that A is at high potential, B is at low potential.
Now let’s take a few basic examples. See in this situation, a rod of length L is at a distance a. Now what about the direction, B will be at high potential or Q will be at high potential. See B is decreasing in this region, so the cause of induced electric field is decreasing B inside the plane. That means the induced electric field should be clockwise since that the B due to is also inside the plane, so the induced electric field is clockwise and induced EMF will be such that it will try to make the current flow in clockwise direction. So we know the formula is aL by 2 dB by dt, and t will be at high potential that means VP minus VQ is half aL dB by dt. a is the distance of the rod from the centre and L is the length of the rod.
Let’s take another example, very similar situation, P and Q length a, distance is a. Now P is decreasing, again if B is decreasing the direction of induced electric field will be clockwise. So that means the formula is aL by 2 dB by dt so VP minus VQ is half aL dB by dt, again the same result because the distance of the rod is the same and the length is L, so again half aL dB by dt and P will be at high potential such that it will also try to make the current flow in clockwise direction.
Now let’s take a different situation. Here one end of the rod is at the centre, other end is P, we have to find the induced EMF across the ends. So let’s first take the direction of the induced electric field, Now B is increasing in this situation so direction of induced electric field will be anti-clockwise, such that B due to is outside so it is opposing the Len’s Law. So basically induced electric field is in anti clockwise direction. Now the formula was aL by 2 dB by dt. Now what was a? a was the distance of the rod from the centre. Now this rod itself is passing through the centre. So basically distance of rod from centre is zero that means induced EMF in this case will be 0. So if a rod is passing through the centre, induced EMF in that rod will be 0.
Now what we can do, we can analyse it in a different way. We have shown the induced electric field, now if the rod is passing through the centre, the induced electric field will be perpendicular to length, now if induced electric field is perpendicular to length, we know dv is E.dL, dot product that means cos theta and cos 90 is 0 that means E.dl is 0. So we can look at it in this way because electric field is perpendicular to the length, induced EMF will be 0.
Now let’s take another example, here we have two rods of length a and it is symmetrically placed above the centre. And here dB by dt is k, k is greater than 0. That means magnetic field inside the plane is increasing. So we have the formula aL by 2 dB by dt we can analyze the situation, a is the distance of rod from the centre. So here how do we find the distance of rod from centre? See because it is symmetric angles will be 60 degree, half angle will be 30 degree. The length is a by 2 half the length of the rod so basically the perpendicular distance here will be a by 2 tan 30. Using trigonometry we can see a by 2 tan 30 is the perpendicular distance. So a by 2 root 3 is the perpendicular distance now just use the formula aL by 2 dB by dt so half as it is, a is the perpendicular distance which will be a by 2 root 3, l is the length of the rod, here it is a, so a and into dB by dt which is equal to k in this case. So we have induced EMF is ka square divide by 4 root 3. Now the direction, see, because field is increasing, direction of induced electric field will be anti clockwise so induced EMF will be such that it will try to make the current flow in anti clockwise direction. So here we can draw the equivalent batteries like this in the left side and in the right side, such that if it is connected it will try to make the current flow in anti clockwise direction. Now VP minus VQ in the diagram we can see VP minus VQ will be 2 times E, that means VP minus VQ is ka square divide by 2 root 3.
Thank you.
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JEE MAIN & ADVANCE 12th PCM Physics -Electromagnetic Induction Module-1 Demo Video
Hello, students, now let’s study magnetic flux before we move onto electromagnetic induction. Now same as electric flux, magnetic flux is also defined as the surface integral of the normal component of the magnetic field passing through that surface. And here the SI unit of magnetic flux is weber. So the mathematical formula is flux is integral B.ds. Like in electric field we had seen flux was E.ds. Here it is integral B.ds. So if this is a surface, the normal component of B.product takes care of that. So flux the mathematical formula is integral B.ds. Now B is the magnetic field vector and ds is basically the area vector. Now if we have an open surface the area vector can be taken either in the upward direction or in the downward direction, so there are two possibilities. But if we have a closed surface we always take the area vector as the outward normal. So if we have an open surface you have two choices, if we have a closed surface only take the outward normal as the area. Now see, if the magnitude of magnetic field that is B is constant and also the angle between B and area is constant, then we can simplify our equation. See flux is integral, B can be taken out of the integral. So B.integral ds. Now if we integrate ds, ds is the area vector. So if you integrate ds we will get the area vector so flux is B.A. And since the angle is constant we can also write flux is BA cos theta. So this is also the formula for flux but this is valid only when B is constant and the angle between B and area is constant over the entire surface. Then we can use this simple formula otherwise we have to use flux is integral B.ds.
Now let’s calculate magnetic flux in some simple situations. So in this figure you see there is an area and magnetic field lines are passing through that area. What we can do is first of all draw the area vector so area vector we had drawn perpendicular to the surface. We can also take the area vector towards the left, it is upto you. Now sometimes the diagram can be drawn like this, this is the side view. If you see this diagram from the front we can draw like this. If you see the same diagram from the left hand side or basically call it the front view, then see the magnetic field lines are crossing the area perpendicularly. So this is also a way of drawing. Now flux has two formulas, integral B.ds or BA cos theta. Now how do we decide what formula to use and how to calculate flux? So basically follow three simple steps. First check the magnetic field is constant over the entire surface. So in this question, see B is constant over the entire surface, okay, fine. Now let’s move on to checking the angle between B and area vector. So if you see the angle between B vector and area vector is 0 degree. So angle is 0 which is also constant, then find the flux. So flux is BA cos 0 that means flux is equal to B into A.
Let’s take another situation, here the area is tilted a bit, B is horizontal and which is constant. So again if you draw the area vector we can draw perpendicular to the surface. The other possibility is also outward but any one you can choose. Now if we mark the angle suppose the angle of the area vector was theta from the vertical. So just calculate the angle between area vector and B vector. So we can see the angle between area and B is again theta. So B is constant over the entire surface, first step is valid. Angle between B and A is also constant theta. So we can write flux is equal to BA cos theta.
So basically this thing we have studied in electrostatics, so in magnetism also or EMI also we need the concept of flux. We can also see, if we divide the area or basically take the component one will be A cos theta and other will be A sin theta. So the component of area along the field will not contribute in flux, because the normal will be perpendicular to B. Only the component of area which is perpendicular to B contributes in the flux, so we can also write flux is equal to B into A perpendicular. In this case A perpendicular is A cos theta. So flux will be BA cos theta.
Now let’s take another situation, let’s take it one by one. See this is an area vector and D is as shown in the figure. Now directly you can see, D is constant over the surface, now angle between D and A, see the area vector is inside the plane or outside the plane. So you can take any of the two options, let’s take the area vector out of the plane. So angle between B and A is 90 degrees, you can see in the diagram the angle between B and A is 90 degrees. So we can write flux is BA cos theta that means BA cos 90 which is equal to 0.
Let’s take this situation, now see the magnetic field lines are inside the plane. Area vector can be taken outside or inside. See B is constant over the entire surface, the angle between B and area vector is either 180 or 0 degree depends on how you take the area vector. So we have taken the area vector like this, the angle is 180 degrees, so flux is BA cos 180 that means minus BA.
Let’s take this situation, see this is a tricky situation, this 30 degrees is useless because if you see angle between B and area vector is basically 90 degree. B is constant over the entire surface that is very obvious. Now angle between B and A is again 90 degrees because if you draw any line the plane is perpendicular, the area vector and B vector plane is perpendicular. So if you draw any line at 30 degrees, 0 degrees, 90 degrees, it will all be perpendicular to the area vector. So again flux is equal to BA cos 90 that means zero.
Thank you.
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JEE MAIN & ADVANCE 12th PCM Physics -Electromagnetic Induction Module-2 Demo Videos
Hello, students, now let’s analyse the case of Motional EMF when B is not uniform along the length of the wire. See find the induced EMF in the rod. Current in the wire is constant and the rod is moving with constant velocity v in the vertical direction.
Now see due to the current carrying wire there will be a magnetic field but at a distance X we know magnetic field is Mu0i by 2 pi x . So basically magnetic field is varying along the length of the wire. It is constant with respect to time but along the length of the wire it is changing. Now we know induced EMF or motional EMF is integral v cross B.dl. Now B at different parts of rod is different. So what we have to do we have to find the small induced EMF at small element of the rod. So what we can do, we know v, B and dl are mutually perpendicular, so we can use Bv into dl. Net induced EMF will be the integral of v B into dl. Now see velocity of all the parts of the rod is v, what about the magnitude of B, B is equal to Mu0i by 2pi x. So in the element at a distance x whose length is dx, B at a distance x from the wire will be Mu0i by 2 pi x. So integral v instead of B for that element B will be Mu0i by 2 pi x. And what is the length of that element, the length is dx. So e is equal to integral v B and dx. Now what will be the limits of this integration? See we have to find the net induced EMF in the entire rod, so x will begin from a, it will come at the first end of the rod, so initial limit will be a. Now x has to be a plus L to cover the entire length. So initial was a to the limit a plus L. So this will be our limit, from a to a plus L. Now integration of 1 by dx is lnx constant and we can take it out. So vMu0i by 2 pi lnx, a to a plus L. Now lna minus lnb is lna by v so lna plus L minus lna will be lna plus L by a. So induced EMF will be Mu0iv by 2 pi lna plus L by a.
So when v is not constant over the length what we have to do, we have to take an element, find the small induced EMF in that part, integrate from the initial to final to get the total motional EMF in the rod.
Now suppose we have to find the direction, so direction we can use the same method, v cross b. So if we do v cross b we see left side will be the positive, right side will be the negative. Now suppose it is semi circular, now what we will do, integration will be a bit difficult, so we can use the concept of L effective c. v is constant for all the parts. Now B at different points is different but it is constant with respect to time, see suppose we have taken two points, B is constant now as the wire moves up, the distance of the point remains same. That means B at a point remains same. B at different points will be different, but as the wire moves B at one point will remain same. So B remains same so basically B is constant over the length. And v and B is perpendicular. So we can use the concept of L effective. So v cross B into L effective, L effective will be the same as the previous one. So the answer to this question will again be Mu0iv by 2 pi ln a plus L by a. So we have taken this as L effective. But we can only do this if v is constant, B at different points can be different, but B at one point should remain constant and v and B are perpendicular.
Suppose the situation was like this, this wire is moving towards right. Now if you take a point basically B will change, see if you take L effective and you write the answer as E is equal to 0, because L effective and v is along the length, but this will be incorrect. Why? See, because at same point when the wire moves B changes, suppose we take a point, initial the distance was x then when the wire moves like this the distance changes, so B at the same point changes so we cannot use directly the formula of L effective. So in these situations we have to solve it using integration which will be very complex. But we have to remember but we cannot use L effective in these cases. We have to check that v should be constant, B should be constant and they should be perpendicular.
Thank you.
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