Question

The inner and outer coils of the two long co-axial solenoids are the same length and have radii of ${\mathrm{r}}_1\mathrm{and}{\mathrm{r}}_2$, as well as different numbers of turns per unit length, ${\mathrm{n}}_1\mathrm{and}{\mathrm{n}}_2$, respectively. The inner-self-inductance coil's and mutual inductance are compared as follows:

• A. $\begin{array}{l}\frac{n_2}{n_1}:\frac{r_2^2}{r_1^2}\end{array}$
• B. $\begin{array}{l}\frac{n_2}{n_1}:\frac{r_1}{r_2}\end{array}$
• C. $\begin{array}{l}\frac{n_1}{n_2}\end{array}$
• D. $\begin{array}{l}\frac{n_2}{n_1}\end{array}$

Answer 1

The correct option is D

$\begin{array}{l}\frac{n_2}{n_1}\end{array}$

The explanation for the correct option:

D: $\begin{array}{l}\frac{n_2}{n_1}\end{array}$

Calculation for the Ratio of Mutual Inductance to the Self Inductance:

• The expression used for calculating Mutual Inductance:$\begin{array}{l}={\mathrm{\mu }}_0{\mathrm{n}}_1{\mathrm{n}}_2{\mathrm{R}}_2^2\mathrm{l}\end{array}$
• An expression used for Self Inductance:$\begin{array}{l}={\mathrm{\mu }}_0{\mathrm{n}}_1^2{\mathrm{R}}_2^2\mathrm{l}\end{array}$
• $\begin{array}{rcl}\frac{\mathrm{Mutual}\mathrm{Inductance}}{\mathrm{Self}\mathrm{Inductance}}& =& \frac{\begin{array}{l}\begin{array}{l}{\mathrm{\mu }}_0{\mathrm{n}}_1{\mathrm{n}}_2{\mathrm{R}}_2^2\mathrm{l}\end{array}\end{array}}{{\mathrm{\mu }}_0{\mathrm{n}}_1^2{\mathrm{R}}_2^2\mathrm{l}}\\ & =& \frac{{\mathrm{n}}_2}{{\mathrm{n}}_1}\end{array}$

Final Answer: Therefore, option D is the correct option.