A circular coil of `300` turns and diameter `14cm` carries a current of `15A`. What is the magnitude of magnetic moment linked with the loop?

Options

A. `51.7JT^(-1)`
B. `69.2JT^(-1)`
C. `38.6JT^(-1)`
D. `19.5JT^(-1)`

Explanation

To find the magnitude of the magnetic moment linked with the circular coil, we can follow these steps:

### Step 1: Identify the given values
- Number of turns (N) = 300
- Diameter of the coil = 14 cm
- Current (I) = 15 A

### Step 2: Convert diameter to radius
The radius (r) is half of the diameter. Therefore:
\[ \text{Radius} = \frac{\text{Diameter}}{2} = \frac{14 \text{ cm}}{2} = 7 \text{ cm} \]
Convert this to meters:
\[ r = 7 \text{ cm} = 7 \times 10^{-2} \text{ m} \]

### Step 3: Calculate the area of the circular coil
The area (A) of a circle is given by the formula:
\[ A = \pi r^2 \]
Substituting the value of r:
\[ A = \pi (7 \times 10^{-2})^2 = \pi (49 \times 10^{-4}) = 49\pi \times 10^{-4} \text{ m}^2 \]

### Step 4: Calculate the magnetic moment (M)
The magnetic moment (M) for a coil is given by the formula:
\[ M = N \cdot I \cdot A \]
Substituting the values:
\[ M = 300 \cdot 15 \cdot (49\pi \times 10^{-4}) \]
Calculating the numerical values:
\[ M = 300 \cdot 15 \cdot 49 \cdot 3.14 \times 10^{-4} \]
\[ M = 300 \cdot 15 \cdot 153.86 \times 10^{-4} \]
\[ M = 300 \cdot 2307.9 \times 10^{-4} \]
\[ M = 692370 \times 10^{-4} = 69.237 \text{ J/T} \approx 69.2 \text{ J/T} \]

### Final Answer
The magnitude of the magnetic moment linked with the loop is approximately **69.2 J/T**.

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