The areas of cross-section of three magnets of same length are A,2AA and 6A respectively The ratio of their magnetic moments will be

Options

A. `6:2:1`
B. `1:2:6`
C. `1:4:36`
D. `36:4:1`

Explanation

To solve the problem of finding the ratio of the magnetic moments of three magnets with cross-sectional areas A, 2A, and 6A, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Magnetic Moment**: The magnetic moment (M) of a magnet is given by the formula:
\[ M = N \cdot I \cdot A \]
where \(N\) is the number of turns, \(I\) is the current, and \(A\) is the area of cross-section.

2. **Given Areas**: The areas of cross-section for the three magnets are:
- Magnet 1: \(A_1 = A\)
- Magnet 2: \(A_2 = 2A\)
- Magnet 3: \(A_3 = 6A\)

3. **Assuming Constants**: Since the problem states that the lengths of the magnets are the same, we can assume that the number of turns \(N\) and the current \(I\) are constant for all three magnets.

4. **Calculating Magnetic Moments**:
- For Magnet 1:
\[ M_1 = N \cdot I \cdot A \]
- For Magnet 2:
\[ M_2 = N \cdot I \cdot (2A) = 2N \cdot I \cdot A \]
- For Magnet 3:
\[ M_3 = N \cdot I \cdot (6A) = 6N \cdot I \cdot A \]

5. **Finding the Ratios**: Now, we can find the ratio of the magnetic moments:
\[ M_1 : M_2 : M_3 = (N \cdot I \cdot A) : (2N \cdot I \cdot A) : (6N \cdot I \cdot A) \]
Since \(N\), \(I\), and \(A\) are common in all three terms, they cancel out:
\[ = 1 : 2 : 6 \]

6. **Final Answer**: Therefore, the ratio of the magnetic moments of the three magnets is:
\[ \boxed{1 : 2 : 6} \]