A coil of 100 turns and area 5 square centimetre is placed in a magnetic field B = 0.2 T . The normal to the plane of the coil makes an angle of `60^(@)` with the direction of the magnetic field. The magnetic flux linked with the coil is

Options

A. `5xx10^(-3)Wb`
B. `5xx10^(-5)Wb`
C. `5xx10^(-2)Wb`
D. `10^(-4)Wb`

Answer

To find the magnetic flux linked with the coil, we can follow these steps:

### Step 1: Identify the given values
- Number of turns in the coil (N) = 100 turns
- Area of the coil (A) = 5 cm² = 5 × 10⁻⁴ m² (since 1 cm² = 10⁻⁴ m²)
- Magnetic field (B) = 0.2 T
- Angle (θ) = 60° (the angle between the normal to the coil and the magnetic field)

### Step 2: Use the formula for magnetic flux
The magnetic flux (Φ) linked with the coil can be calculated using the formula:
\[ Φ = N \cdot B \cdot A \cdot \cos(θ) \]
where:
- Φ is the magnetic flux,
- N is the number of turns,
- B is the magnetic field strength,
- A is the area of the coil,
- θ is the angle between the magnetic field and the normal to the coil.

### Step 3: Substitute the values into the formula
Substituting the values we have:
\[ Φ = 100 \cdot 0.2 \cdot (5 \times 10^{-4}) \cdot \cos(60°) \]

### Step 4: Calculate cos(60°)
We know that:
\[ \cos(60°) = \frac{1}{2} \]

### Step 5: Substitute cos(60°) into the equation
Now substituting cos(60°) into the equation:
\[ Φ = 100 \cdot 0.2 \cdot (5 \times 10^{-4}) \cdot \frac{1}{2} \]

### Step 6: Perform the calculations
Calculating step by step:
1. Calculate \(0.2 \cdot (5 \times 10^{-4})\):
\[ 0.2 \cdot 5 = 1 \quad \text{and} \quad 1 \times 10^{-4} = 1 \times 10^{-4} \]

2. Now multiply by 100:
\[ 100 \cdot (1 \times 10^{-4}) = 1 \times 10^{-2} \]

3. Finally, multiply by \(\frac{1}{2}\):
\[ 1 \times 10^{-2} \cdot \frac{1}{2} = 0.5 \times 10^{-2} = 5 \times 10^{-3} \]

### Step 7: Conclusion
Thus, the magnetic flux linked with the coil is:
\[ Φ = 5 \times 10^{-3} \, \text{Wb} \]

### Final Answer
The magnetic flux linked with the coil is \(5 \times 10^{-3} \, \text{Wb}\).

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