A circular coil of 35 turns and radius 25 cm carries a current of 11 A. What will be its magnetic dipole moment?

Options

A. `72.2Am^(2)`
B. `70.5Am^(2)`
C. `74.56Am^(2)`
D. `75.56Am^(2)`

Explanation

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

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

where:
- \(\mu\) is the magnetic dipole moment,
- \(n\) is the number of turns in the coil,
- \(I\) is the current flowing through the coil, and
- \(A\) is the area of the coil.

### Step 1: Identify the given values
From the problem, we have:
- Number of turns, \(n = 35\)
- Current, \(I = 11 \, \text{A}\)
- Radius of the coil, \(r = 25 \, \text{cm} = 0.25 \, \text{m}\) (conversion from cm to m)

### 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 into the formula:

\[ A = \pi (0.25 \, \text{m})^2 \]

Calculating \(A\):

\[ A = \pi (0.0625 \, \text{m}^2) \approx 3.14 \times 0.0625 \approx 0.19625 \, \text{m}^2 \]

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

\[ \mu = n \cdot I \cdot A = 35 \cdot 11 \cdot 0.19625 \]

Calculating the product:

\[ \mu = 385 \cdot 0.19625 \approx 75.56 \, \text{A m}^2 \]

### Step 4: Final answer
Thus, the magnetic dipole moment of the circular coil is:

\[ \mu \approx 75.56 \, \text{A m}^2 \]

### Summary
The magnetic dipole moment of the circular coil is approximately \(75.56 \, \text{A m}^2\).

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