A square-shaped coil of side 10 cm is placed in a region of magnetic field 0.2 T such that it makes an angle of `30^(@)` with the magnetic field. Calculate the magnetic flux linked with the coil.

Answer

To calculate the magnetic flux linked with a square-shaped coil placed in a magnetic field, we can follow these steps:

### Step 1: Understand the formula for magnetic flux
The magnetic flux (Φ) linked with a coil is given by the formula:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
where:
- \(B\) is the magnetic field strength (in Tesla),
- \(A\) is the area of the coil (in square meters),
- \(\theta\) is the angle between the magnetic field and the normal (perpendicular) to the surface of the coil.

### Step 2: Convert the side length of the coil to meters
The side length of the square coil is given as 10 cm. We need to convert this to meters:
\[ \text{Side length} = 10 \, \text{cm} = 10 \times 10^{-2} \, \text{m} = 0.1 \, \text{m} \]

### Step 3: Calculate the area of the coil
The area \(A\) of a square coil is given by:
\[ A = \text{side}^2 = (0.1 \, \text{m})^2 = 0.01 \, \text{m}^2 \]

### Step 4: Identify the magnetic field strength and angle
The magnetic field strength \(B\) is given as 0.2 T, and the angle \(\theta\) is given as 30°. However, we need to find the angle between the magnetic field and the area vector. The area vector is perpendicular to the surface of the coil, so:
\[ \text{Angle between area vector and magnetic field} = 90° - 30° = 60° \]

### Step 5: Calculate the magnetic flux
Now we can substitute the values into the magnetic flux formula:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
Substituting the values:
\[ \Phi = 0.2 \, \text{T} \cdot 0.01 \, \text{m}^2 \cdot \cos(60°) \]
We know that \(\cos(60°) = \frac{1}{2}\), so:
\[ \Phi = 0.2 \cdot 0.01 \cdot \frac{1}{2} \]
\[ \Phi = 0.2 \cdot 0.01 \cdot 0.5 = 0.001 \, \text{Wb} \]
Thus, the magnetic flux linked with the coil is:
\[ \Phi = 10^{-3} \, \text{Wb} \]

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

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