A coil of 100 turns, `5cm^(2)` area is placed in external magnetic field of 0.2 Tesla (S.I) in such a way that plane of the coil makes an angle `30^(@)` with the field direction. Calculate magnetic flux of the coil (in weber)

Options

A. `5xx10^(-3)`
B. `4xx10^(-3)`
C. `1.2xx10^(-3)`
D. `3xx10^(-3)`

Answer

To solve the problem of calculating the magnetic flux through a coil placed in a magnetic field, we can follow these steps:

### Step 1: Identify the given values
- Number of turns (n) = 100
- Area of the coil (A) = 5 cm² = \(5 \times 10^{-4} \, m^2\) (since 1 cm² = \(10^{-4} m^2\))
- Magnetic field strength (B) = 0.2 T
- Angle (θ) between the magnetic field and the normal to the coil = 30°

### Step 2: Calculate the angle with respect to the area vector
The angle θ given in the problem is the angle between the plane of the coil and the magnetic field. To find the angle with respect to the area vector (which is perpendicular to the plane of the coil), we need to subtract this angle from 90°.

\[ \text{Angle with area vector} = 90° - 30° = 60° \]

### Step 3: Use the formula for magnetic flux
The magnetic flux (Φ) through the coil can be calculated using the formula:

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

Where:
- \(n\) = number of turns
- \(B\) = magnetic field strength
- \(A\) = area of the coil
- \(\theta\) = angle with respect to the area vector

### Step 4: Substitute the values into the formula
Substituting the known values into the formula:

\[ \Phi = 100 \cdot 0.2 \cdot (5 \times 10^{-4}) \cdot \cos(60°) \]

### Step 5: Calculate \(\cos(60°)\)
We know that:

\[ \cos(60°) = \frac{1}{2} \]

### Step 6: Substitute \(\cos(60°)\) into the equation
Now substituting \(\cos(60°)\):

\[ \Phi = 100 \cdot 0.2 \cdot (5 \times 10^{-4}) \cdot \frac{1}{2} \]

### Step 7: Simplify the expression
Calculating the values step by step:

\[ \Phi = 100 \cdot 0.2 \cdot 5 \times 10^{-4} \cdot \frac{1}{2} \]
\[ = 100 \cdot 0.1 \cdot 5 \times 10^{-4} \]
\[ = 5 \times 10^{-2} \times 10^{-4} \]
\[ = 0.5 \times 10^{-2} \, \text{Weber} \]
\[ = 5 \times 10^{-3} \, \text{Weber} \]

### Final Answer
The magnetic flux through the coil is:

\[ \Phi = 5 \times 10^{-3} \, \text{Weber} \]

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