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.