Question

State the principle on which transformer works. Explain its working with construction. Derive an expression for ratio of e.m.f.s and currents in terms of number of turns in primary and secondary coil.
A conductor of any shape, having area $40c{m}^2$ placed in air is uniformly charged with a charge $0.2\mu C$. Determine the electric intensity at a point just outside its surface. Also, find the mechanical force per unit area of the charged conductor.
$[{ϵ}_0=8\cdot 85\times {10}^{-12}S.I.units]$

Answer 1

Principle Of working of a transformer :

A transformer works on the principle that Whenever the magnetic flux linked with a coil changes, an emf is induced in the neighbouring coil.

Construction : It consists of two coils primary (P) and secondary (S), insulated from each other and Wound on a soft iron core as shown in the figure below.

The primary coil is called the input coil and the secondary coil is called the output coil.

Working : When.an alternating voltage is applied to the primary coil, the current through the coil goes, on changing. Hence, the magnetic flux through the core also changes- As this changing magnetic flux is linked with both coils, an emf is induced in each of them. The amount of magnetic flux linked with the coil depends on the number of twins of the coil.

Derivation : Lett be the magnetic flux,linked per turn with both coils at a certain instant of time 'f'. Let the number. f turns of the primary and secondary cOils be $N_p$ and $N_s$ respectively. Therefore, the, total magnetic flux linked with the primary coil at certain instant of time 't' is $N_p\varphi$. Similarly, the total magnetic flux linked with.the secondary coil at certain instant of time 't' is $N_s\varphi$.

Now, the induced emf in a coil is

$e=\frac{d\varphi }{dt}$

Therefore, the induced emf in the primary coil is

$e_p=-\frac{d\varphi p}{dt}=-\frac{dNp\varphi }{dt}=-N_p\frac{d\varphi }{dt}$ ...(1)

Similarly, the induced emf in the secondary coil is

$e_s=-\frac{d\varphi s}{dt}=-\frac{d{N}_s\varphi }{dt}=-N_s\frac{d\varphi }{dt}$ .....(2)

Dividing equations (1) and (2), we get

$\frac{e_s}{s_p}=\frac{-N_s\frac{d\varphi }{dt}}{-N\frac{d\varphi }{dt}}=\frac{N_s}{N_p}$ ...(3)

The above equation is called the equation of the transfer and the ratio $\frac{N_s}{N_p}$ is known as the turns ratio of the transformer.

Now, for an ideal transformer, we know that the input power is equal to the output .

power.

$\frac{e_s}{e_p}=\frac{N_s}{N_p}$

$\therefore \frac{e_s}{e_p}=\frac{N_s}{N_p}=\frac{i_p}{i_s}$

Numerical:

Given: $Q=0.2\mu C=0.2\times {10}^{-6}C$

$A=40c{m}^2=4\times {10}^{-4}{m}^2$

${\epsilon }_0=8.85\times {10}^{-12}$SI units

The electric field intensity just outside the surface of a charged conductor of any shape is

$E=\frac{\sigma }{{\epsilon }_0}=\frac{Q}{A{\epsilon }_0}$

$\therefore E=\frac{0.2\times {10}^{-6}}{40\times {10}^{-4}\times 8.85\times {10}^{-12}}$

$\therefore E=5.65\times {10}^{6}N/C$

Now, the mechanical force per unit area of a conductor is

$f=\frac{1}{2}{\epsilon }_0{E}^2=\frac{1}{2}\times 8.85\times {10}^{-12}\times (5.65\times {10}^6{)}^2$

$\therefore f=12141.25N/m^2$