A circular coil of radius `10 cm "and" 100` turns carries a current 1A. What is the magnetic moment of the coil?

Options

A. `3.142xx10^(4) A-m^(2)`
B. `10^(4)A-m^(2)`
C. `3,142 A-m^(2)`
D. `3 A-m^(2)`

Answer

To find the magnetic moment of a circular coil, we can use the formula:

\[ \text{Magnetic Moment} (M) = n \cdot I \cdot A \]

where:
- \( n \) = number of turns in the coil
- \( I \) = current flowing through the coil
- \( A \) = area of the coil

### Step 1: Identify the given values
- Radius of the coil, \( r = 10 \, \text{cm} = 0.1 \, \text{m} \) (convert cm to m)
- Number of turns, \( n = 100 \)
- Current, \( I = 1 \, \text{A} \)

### Step 2: Calculate the area of the coil
The area \( A \) of a circular coil is given by the formula:

\[ A = \pi r^2 \]

Substituting the radius in meters:

\[ A = \pi (0.1)^2 = \pi (0.01) = 0.0314 \, \text{m}^2 \]

### Step 3: Substitute the values into the magnetic moment formula
Now we can substitute the values of \( n \), \( I \), and \( A \) into the magnetic moment formula:

\[ M = n \cdot I \cdot A \]

\[ M = 100 \cdot 1 \cdot 0.0314 \]

\[ M = 3.14 \, \text{A m}^2 \]

### Step 4: Final result
Thus, the magnetic moment of the coil is:

\[ M \approx 3.142 \, \text{A m}^2 \]

### Conclusion
The correct answer is approximately \( 3.142 \, \text{A m}^2 \).

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