Electromagnetic Induction And Alternating Current Ques 5
5 There are two long coaxial 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. The ratio of mutual inductance to the self-inductance of the inner coil is
(Main 2019, 11 Jan I)
(a) $\frac{n_2}{n_1} \cdot \frac{r_1}{r_2}$
(b) $\frac{n_2}{n_1} \cdot \frac{r_2^2}{r_1^2}$
(c) $\frac{n_2}{n_1}$
(d) $\frac{n_1}{n_2}$
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Answer:
Correct Answer: 5.( c )
Solution:
Formula:
- Mutual inductance for a coaxial solenoid of radius $r_1$ and $r_2$ and number of turns $n_1$ and $n_2$, respectively is given as, $M=\mu_0 n_1 n_2 \pi r_1^2 l$ (for internal coil of radius $r_1$ ) Self inductance for the internal coil,
$ \begin{aligned} L & =\mu_0 n_1^2 \pi r_1^2 l \\ \therefore \quad \frac{M}{L} & =\frac{n_1 n_2}{n_1^2}=\frac{n_2}{n_1} \end{aligned} $
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