A coil of 200 turns `8cm^2` area is placed in external magnetic field of 0.4 Tesla (S.I) in such a way that its area vector makes an angle `60^@` with the field direction. Calculate magnetic flux through the coil (in weber)

Options

A. `5 xx 10^(-3)`
B. `3.2 xx 10^(-2)`
C. `1.2 xx 10^(-3)`
D. `4.3 xx 10^(-2)`

Explanation

To calculate the magnetic flux through the coil, we can use the formula for magnetic flux (Φ):

\[ \Phi = N \cdot B \cdot A \cdot \cos(\theta) \]

Where:
- \( \Phi \) = magnetic flux (in Weber)
- \( N \) = number of turns in the coil
- \( B \) = magnetic field strength (in Tesla)
- \( A \) = area of the coil (in square meters)
- \( \theta \) = angle between the area vector and the magnetic field (in degrees)

### Step-by-Step Solution:

**Step 1: Convert the area from cm² to m².**
- Given area = \( 8 \, \text{cm}^2 \)
- Convert to m²:
\[ A = 8 \, \text{cm}^2 \times \left(\frac{1 \, \text{m}}{100 \, \text{cm}}\right)^2 = 8 \times 10^{-4} \, \text{m}^2 \]

**Step 2: Identify the values given in the problem.**
- Number of turns, \( N = 200 \)
- Magnetic field strength, \( B = 0.4 \, \text{T} \)
- Angle, \( \theta = 60^\circ \)

**Step 3: Calculate \( \cos(60^\circ) \).**
- From trigonometry, we know:
\[ \cos(60^\circ) = \frac{1}{2} \]

**Step 4: Substitute the values into the magnetic flux formula.**
\[ \Phi = N \cdot B \cdot A \cdot \cos(\theta) \]
Substituting the known values:
\[ \Phi = 200 \cdot 0.4 \cdot (8 \times 10^{-4}) \cdot \cos(60^\circ) \]
\[ \Phi = 200 \cdot 0.4 \cdot (8 \times 10^{-4}) \cdot \frac{1}{2} \]

**Step 5: Simplify the expression.**
\[ \Phi = 200 \cdot 0.4 \cdot (8 \times 10^{-4}) \cdot 0.5 \]
\[ \Phi = 200 \cdot 0.2 \cdot (8 \times 10^{-4}) \]
\[ \Phi = 40 \cdot (8 \times 10^{-4}) \]
\[ \Phi = 320 \times 10^{-4} \]
\[ \Phi = 3.2 \times 10^{-2} \, \text{Wb} \]

### Final Answer:
The magnetic flux through the coil is \( \Phi = 3.2 \times 10^{-2} \, \text{Wb} \).