For a given length l of wire carrying a current l, how many circular turns would produce the maximum magnetic moment ?

Options

A. 1
B. 2
C. 4
D. Any value

Answer

To solve the problem of determining how many circular turns of wire carrying a current will produce the maximum magnetic moment, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Magnetic Moment**:
The magnetic moment (μ) produced by a circular loop of wire is given by the formula:
\[ \mu = I \times A \]
where \(I\) is the current flowing through the wire and \(A\) is the area of the loop.

2. **Relating Length of Wire to Number of Turns**:
If we have \(n\) circular turns, the circumference of one loop is \(2\pi r\). Therefore, the total length \(L\) of the wire can be expressed as:
\[ L = n \times (2\pi r) \]
From this, we can solve for the radius \(r\):
\[ r = \frac{L}{2\pi n} \]

3. **Calculating the Area of One Loop**:
The area \(A\) of one circular loop is given by:
\[ A = \pi r^2 \]
Substituting the expression for \(r\):
\[ A = \pi \left(\frac{L}{2\pi n}\right)^2 = \pi \frac{L^2}{4\pi^2 n^2} = \frac{L^2}{4\pi n^2} \]

4. **Finding the Expression for Magnetic Moment**:
Now, substituting the area back into the magnetic moment formula:
\[ \mu = n \times I \times A = n \times I \times \frac{L^2}{4\pi n^2} \]
Simplifying this gives:
\[ \mu = \frac{I L^2}{4\pi n} \]

5. **Maximizing the Magnetic Moment**:
From the expression \(\mu = \frac{I L^2}{4\pi n}\), we can see that the magnetic moment is inversely proportional to the number of turns \(n\). This means that to maximize the magnetic moment, we should minimize \(n\).

6. **Conclusion**:
The minimum number of turns \(n\) that can be made is 1. Therefore, the maximum magnetic moment is produced with:
\[ n = 1 \]

### Final Answer:
The number of circular turns that would produce the maximum magnetic moment is **1**.

---