A straight wire carring current I is turned into a circular loop. If the magnitude of magnetic moment associated with it in M.K.S. unit is M, the length of wire will be

Options

A. a)`(4piI)/(M)`
B. b) `sqrt((4pi M)/(I))`
C. c) `sqrt((4pi I)/(M))`
D. d) `(M pi)/(4I)`

Answer

To solve the problem, we need to find the length of a wire that has been bent into a circular loop, given its magnetic moment \( M \) and the current \( I \) flowing through it. Here’s a step-by-step solution:

### Step 1: Understand the magnetic moment of a circular loop
The magnetic moment \( M \) of a circular loop is given by the formula:
\[ M = I \times A \]
where \( I \) is the current and \( A \) is the area of the loop.

### Step 2: Calculate the area of the circular loop
The area \( A \) of a circular loop with radius \( R \) is given by:
\[ A = \pi R^2 \]

### Step 3: Relate the radius to the length of the wire
The length \( L \) of the wire when it is formed into a circular loop is equal to the circumference of the circle:
\[ L = 2\pi R \]
From this, we can express the radius \( R \) in terms of the length \( L \):
\[ R = \frac{L}{2\pi} \]

### Step 4: Substitute the radius into the area formula
Substituting \( R \) into the area formula gives:
\[ A = \pi \left(\frac{L}{2\pi}\right)^2 = \pi \frac{L^2}{4\pi^2} = \frac{L^2}{4\pi} \]

### Step 5: Substitute the area back into the magnetic moment formula
Now substituting the area \( A \) back into the magnetic moment formula:
\[ M = I \times A = I \times \frac{L^2}{4\pi} \]

### Step 6: Rearrange to find the length \( L \)
Rearranging the equation to solve for \( L^2 \):
\[ L^2 = \frac{4\pi M}{I} \]
Taking the square root to find \( L \):
\[ L = \sqrt{\frac{4\pi M}{I}} \]

### Final Answer
Thus, the length of the wire \( L \) is:
\[ L = \sqrt{\frac{4\pi M}{I}} \]