Two solenoids of equal number of turns have their lengths and the radii in the same ratio 1:4. The ratio of their self inductances will be

Options

A. `1:2`
B. `2:1`
C. `1:1`
D. `1:4`

Explanation

To solve the problem of finding the ratio of self-inductances of two solenoids with equal number of turns and given dimensions, we can follow these steps:

### Step 1: Understand the given ratios
We are given that the lengths and radii of the two solenoids are in the ratio of 1:4. Let:
- Length of solenoid A = L
- Length of solenoid B = 4L
- Radius of solenoid A = R
- Radius of solenoid B = 4R

### Step 2: Write the formula for self-inductance
The self-inductance \( L \) of a solenoid is given by the formula:
\[ L = \frac{\mu_0 N^2 A}{l} \]
where:
- \( \mu_0 \) = permeability of free space
- \( N \) = number of turns
- \( A \) = cross-sectional area of the solenoid
- \( l \) = length of the solenoid

### Step 3: Calculate the self-inductance for both solenoids
For solenoid A:
- Cross-sectional area \( A_A = \pi R^2 \)
- Length \( l_A = L \)

Thus, the self-inductance \( L_1 \) for solenoid A is:
\[ L_1 = \frac{\mu_0 N^2 (\pi R^2)}{L} \]

For solenoid B:
- Cross-sectional area \( A_B = \pi (4R)^2 = 16\pi R^2 \)
- Length \( l_B = 4L \)

Thus, the self-inductance \( L_2 \) for solenoid B is:
\[ L_2 = \frac{\mu_0 N^2 (16\pi R^2)}{4L} = \frac{16\mu_0 N^2 \pi R^2}{4L} = \frac{4\mu_0 N^2 \pi R^2}{L} \]

### Step 4: Find the ratio of self-inductances
Now, we can find the ratio \( \frac{L_1}{L_2} \):
\[ \frac{L_1}{L_2} = \frac{\frac{\mu_0 N^2 \pi R^2}{L}}{\frac{4\mu_0 N^2 \pi R^2}{L}} = \frac{1}{4} \]

### Conclusion
The ratio of the self-inductances of the two solenoids is:
\[ L_1 : L_2 = 1 : 4 \]

### Final Answer
The correct option is D: \( 1 : 4 \).

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