Question

Two coils are wrapped around a cylindrical form in such a way that the same flux passes through every turn of both coils. (In practice this is achieved by inserting an iron core through the cylinder; this has the effect of concentrating the flux.) The primary coil has${\mathit{N}}_{\mathbf{1}}$turns and the secondary has${\mathit{N}}_{\mathbf{2}}$(Fig. 7.57). If the current in the primary is changing, show that the emf in the secondary is given by

$\frac{{\mathbf{\epsilon }}_{\mathbf{2}}}{{\mathbf{\epsilon }}_{\mathbf{1}}}\mathbf{=}\frac{{\mathbf{N}}_{\mathbf{2}}}{{\mathbf{N}}_{\mathbf{1}}}$(7.67)

where${\mathbf{\epsilon }}_{\mathbf{1}}$is the (back) emf of the primary. [This is a primitive transformer-a device for raising or lowering the emf of an alternating current source. By choosing the appropriate number of turns, any desired secondary emf can be obtained. If you think this violates the conservation of energy, study Prob. 7.58.]

Short answer

The expression$\frac{{\epsilon }_2}{{\epsilon }_1}=\frac{N_2}{N_1}$is verified.

Step-by-step solution

Given information

The number of turns in the primary coils are,$N_1$.

The number of turns in the secondary coils are,$N_2$.

The current passing through the primary coil is,I .

Magnetic flux on coils

When the magnitude of the current flowing in the coil varies, a magnetic flux is generated and a certain amount of emf is also induced in the coil.

If the number of turns in the coil are increased then the magnetic flux produced in the coil also increases.

Determine the magnetic flux on both coils

When the current passes through a single loop of primary coil, the magnetic field is generated and the change in current value creates the magnetic flux$\Phi$.

Then, the formula for the magnetic flux on the primary coil is given by,

localid="1658291056948" ${\mathrm{\Phi }}_1={\mathrm{N}}_1\mathrm{\Phi }$

And, the formula for the magnetic flux on the secondary coil is given by,

localid="1658291081661" ${\mathrm{\Phi }}_2={\mathrm{N}}_2\mathrm{\Phi }$

Determine emf induced in the primary coil

The formula for the emf induced in the primary coil is given by,

$$\begin{gathered} {\epsilon }_1=-\frac{d{\Phi }_1}{dt} \\ {\epsilon }_1=-\frac{d\left(N_1\Phi \right)}{dt} \\ {\epsilon }_1=-N_1\frac{d\Phi }{dt}........\left(1\right) \end{gathered}$$

Determine emf induced in the secondary coil

The formula for the emf induced in the secondary coil is given by,

$$\begin{gathered} {\epsilon }_2=-\frac{d{\Phi }_2}{dt} \\ {\epsilon }_2=-\frac{d\left(N_2\Phi \right)}{dt} \\ {\epsilon }_2=-N_2\frac{d\Phi }{dt}.......\left(2\right) \end{gathered}$$

By dividing equation (2) by equation (1),

$$\begin{gathered} \frac{{\epsilon }_2}{{\epsilon }_1}=\frac{-N_2\frac{d\Phi }{dt}}{-N_1\frac{d\Phi }{dt}} \\ \frac{{\epsilon }_2}{{\epsilon }_1}=\frac{N_2}{N_1} \end{gathered}$$

Hence proved.