A circular coil of radius `4 cm` and of `20` turns carries a current of `3` amperes. It is placed in a magnetic field of intensity of `0.5 weber//m^(2)`. The magnetic dipole moment of the coil is

Options

A. `0.15 ampere-m^(2)`
B. `0.3 ampere-m^(2)`
C. `0.45 ampere-m^(2)`
D. `0.6 ampere-m^(2)`

Explanation

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

\[ \mu = NIA \]

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

### Step 1: Identify the given values
- Number of turns, \(N = 20\)
- Current, \(I = 3 \, \text{A}\)
- Radius of the coil, \(r = 4 \, \text{cm} = 0.04 \, \text{m}\)

### 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 value of the radius:

\[ A = \pi (0.04 \, \text{m})^2 \]
\[ A = \pi (0.0016 \, \text{m}^2) \]
\[ A \approx 0.0050265 \, \text{m}^2 \quad (\text{using } \pi \approx 3.14159) \]

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

\[ \mu = NIA = 20 \times 3 \times 0.0050265 \]
\[ \mu = 60 \times 0.0050265 \]
\[ \mu \approx 0.30159 \, \text{A m}^2 \]

### Step 4: Round the answer
Rounding to two decimal places, we get:

\[ \mu \approx 0.30 \, \text{A m}^2 \]

### Final Answer
The magnetic dipole moment of the coil is approximately \(0.30 \, \text{A m}^2\).

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