Question

Prove theoretically the relation between e.m.f. induced in a coil and rate of change of magnetic flux in electromagnetic induction. A parallel plate air condenser has a capacity of 20 $\mu F$. What will be the new capacity if:
(a) the distance between the two plates is doubled?
(b) a marble slab of dielectric constant 8 is introduced between the two plates?

Answer 1

We have the equation $e=-\frac{d\varphi }{dt}.$ 'e' is emf and $\varphi$ is flux.

This we'll prove now.

The diagram shows a situation in which a closed coil is placed in a magnetic field. The $\frac{{}_X}{}$ sign shows that the magnetic field is perpendicular to the plane of the screen and directed into the screen.

We are now trying to pull out the coil towards the right side.

Flux= magnetic field intensity$\times$area of coil. $\varphi =BA$

From diagram $A=lx$. So, $\frac{d\varphi }{dt}=Bl\frac{dx}{dt}=Blv$. Here $v$ is speed ------------------------------(1)

Lenz's law states that when the magnetic flux associated with a coil is changed, the emf induced in the coil is in a direction which tries to oppose the change of flux.

When we pull out the coil, the flux in it reduces. Hence the current generated will be such that it'll try to increase the flux. In order to increase the flux, it has to produce flux in the same direction as the field. By right hand rule, this direction should be in the clockwise sense. Hence the induced current is as shown in the diagram by the arrows.

Flemming's left hand rule states that when the thumb, the index-finger and the middle-finger of your left hand are held mutually perpendicular to each other, then the thumb shows the force acting in a straight conductor carrying current in the direction of the middle-finger placed in a magnetic field in the direction of the index-finger. You can remember this as $FBI.$ Thumb $F,$ index-finger $B$ and middle-finger $I.$

As per this rule, forces $f,$ $f_1$ and $f_2$ will act on the coil in directions as shown in the diagram. The $f_1$ and $f_2$ will cancel each other and the only unbalanced force is $f$ acting towards left. When we pull the coil towards right, we have to do work against this force.

If we apply equal and opposite force, then the work done by the force acting on the coil is negative. Let $dx$ be a small displacement in the right side and $dw$ be the work done. $dw=-fdx$

By Flemming's left hand rule, $f=Bil.$ The $i$ is current. So $dw=-Bildx.$ The mechanical work done gets converted into electric energy. So the mechanical power$\frac{dw}{dt}$ gets converted to electrical power $ei.$

But $\frac{dw}{dt}=-f\frac{dx}{dt}=-fv=-Bilv$

$\therefore ei=-Bilv$ $\therefore e=-Blv$----------------------(2)

From equations 1 and 2 we have $e=-\frac{d\varphi }{dt}$

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$C_1=20\mu F$ $C_1=\frac{A{\epsilon }_{{}_0}}{d}$ A is are of plate, d is distance between them.

$C_2=\frac{A{\epsilon }_{{}_0}}{2d}$ $\therefore C_2=\frac{C_1}{2}=10\mu F$

Ans:- $C_2=10\mu F$

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$C_1=\frac{A{\epsilon }_{{}_0}}{d}$

$C_3=\frac{A{\epsilon }_{{}_0}k}{d}$ $k$ is the dielectric constant.

$C_3=\frac{A{\epsilon }_{{}_0}\times 8}{d}=C_1\times 8=160\mu F$

Ans:- $C_3=160\mu F$