Question

Find the rate of mutual inductance to the self – inductance of the inner coil if there are two long co – axial solenoids of same length $l$ , the inner and outer coils have radii ${r_1}$ and ${r_2}$ and number of turns per unit length ${n_1}$ and ${n_2}$ respectively.
$\dfrac{{{n_2}}}{{{n_1}}}.\dfrac{{r_2^2}}{{r_1^2}}$
$\dfrac{{{n_2}}}{{{n_1}}}.\dfrac{{{r_1}}}{{{r_2}}}$
$\dfrac{{{n_1}}}{{{n_2}}}$
$\dfrac{{{n_2}}}{{{n_1}}}$

Answer 1

Hint: Derive the expressions for the self – inductance and mutual inductance for solenoid by using the formula for magnetic flux. Then, calculate the of mutual inductance to the self – inductance by dividing mutual inductance by self – inductance.

Complete step by step answer:
Before deriving the expressions of self – inductance and mutual inductance you should know about them.
1. Self – inductance: This is the phenomenon of induction of voltage in current carrying wire. This is the property of current carrying coils which oppose the change of current flowing through it. The S.I unit for this inductance is Henry.
2. Mutual inductance: This is the phenomenon in which the current flowing in one coil generates the voltage in secondary coil.

Now, we have to get the expressions for self – inductance and mutual inductance. So, let there be two coils $P$ and $S$. The area of cross – section be $A$. The length of solenoids, radii and number of turns per unit length are already given in the question.
We know that,
Magnetic flux, $\phi = $ Magnetic field $ \times $ effective area
Putting the values, we get –
$\phi = \dfrac{{{\mu _0}{n_1}I}}{l} \times {n_1}A$
Now, self – inductance of the primary coil is –
\[L = \dfrac{\phi }{I} = \dfrac{{{\mu _0}n_1^2A}}{l} \cdots \left( 1 \right)\]
Similarly, the self – inductance of the secondary coil is –
$ \Rightarrow L = \dfrac{{{\mu _0}n_2^2A}}{l} \cdots \left( 2 \right)$
Now, we have to calculate the mutual induction of both the coils –
Therefore, when current flows from $P$ then, the flux linked with $S$ is –
${\phi _2} = \dfrac{{{\mu _0}{n_1}I}}{l} \times {n_2}A$
So, the mutual inductance of two coils is, $M = \dfrac{{{\phi _2}}}{I}$
Putting the value of flux in mutual induction –
$\therefore M = \dfrac{{{\mu _0}{n_1}{n_2}A}}{l} \cdots \left( 3 \right)$
As we have to calculate the ratio of mutual inductance to the self – inductance for the inner coil. Therefore, -
Dividing the equation $\left( 3 \right)$ by equation $\left( 1 \right)$, we get –
$\dfrac{M}{L} = \dfrac{{{\mu _0}{n_1}{n_2}A}}{{{\mu _0}n_1^2A}}$
Cancelling the same quantities in numerator and denominator, we get –
$\therefore \dfrac{M}{L} = \dfrac{{{n_2}}}{{{n_1}}}$
Hence, the correct option is (D).

Note:The formula for self – inductance is, $E = - L\dfrac{{di}}{{dt}}$ but for the solenoid the self – inductance is different. So, don’t get confused. By using the self – inductance we can get the mutual inductance for the second coil with respect to the first coil.