Two coils are made of copper wires of same length in the first coil the number turns is 3n and radius is r. In the second coil number of turns is n and radius is 3r the ratio of self inductances of the coils will be :-

Options

A. `9:1`
B. `3:1`
C. `1:3`
D. `1:9`

Answer

To find the ratio of the self-inductances of the two coils, we can use the formula for the self-inductance of a coil:

\[ L = \frac{\mu_0 N^2 A}{l} \]

where:
- \( L \) is the self-inductance,
- \( \mu_0 \) is the permeability of free space,
- \( N \) is the number of turns,
- \( A \) is the cross-sectional area of the coil,
- \( l \) is the length of the coil.

### Step 1: Calculate the self-inductance of the first coil (L1)

For the first coil:
- Number of turns, \( N_1 = 3n \)
- Radius, \( r_1 = r \)
- Length of the wire is the same for both coils.

The cross-sectional area \( A_1 \) of the first coil is given by:
\[ A_1 = \pi r^2 \]

Now, substituting the values into the self-inductance formula:
\[ L_1 = \frac{\mu_0 (3n)^2 A_1}{l} = \frac{\mu_0 (3n)^2 \pi r^2}{l} = \frac{9 \mu_0 n^2 \pi r^2}{l} \]

### Step 2: Calculate the self-inductance of the second coil (L2)

For the second coil:
- Number of turns, \( N_2 = n \)
- Radius, \( r_2 = 3r \)

The cross-sectional area \( A_2 \) of the second coil is given by:
\[ A_2 = \pi (3r)^2 = 9\pi r^2 \]

Now, substituting the values into the self-inductance formula:
\[ L_2 = \frac{\mu_0 (n)^2 A_2}{l} = \frac{\mu_0 n^2 (9\pi r^2)}{l} = \frac{9 \mu_0 n^2 \pi r^2}{l} \]

### Step 3: Find the ratio of self-inductances (L1/L2)

Now, we can find the ratio of the self-inductances:
\[ \frac{L_1}{L_2} = \frac{\frac{9 \mu_0 n^2 \pi r^2}{l}}{\frac{9 \mu_0 n^2 \pi (9 r^2)}{l}} \]

This simplifies to:
\[ \frac{L_1}{L_2} = \frac{9 \mu_0 n^2 \pi r^2}{9 \mu_0 n^2 \pi (9 r^2)} \]

Canceling out the common terms:
\[ \frac{L_1}{L_2} = \frac{1}{9} \]

Thus, the ratio of the self-inductances of the two coils is:
\[ L_1 : L_2 = 1 : 9 \]

### Final Answer
The ratio of self-inductances of the coils will be \( 1 : 3 \).

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