A body of area `0.5m^(2)` makes an angle of `60^(@)` with the uniform magnetic field of `4Wb m^(-2)`. The magnetic flux passing through the body will be

Options

A. 2Wb
B. `sqrt3Wb`
C. 0.5 Wb
D. 1Wb

Explanation

To calculate the magnetic flux (Φ) passing through a body, we can use the formula:

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

Where:
- \( \Phi \) is the magnetic flux,
- \( B \) is the magnetic field strength (in Webers per square meter, Wb/m²),
- \( A \) is the area of the body (in square meters, m²),
- \( \theta \) is the angle between the magnetic field and the normal (perpendicular) to the surface of the area.

Given:
- Area \( A = 0.5 \, m^2 \)
- Magnetic field \( B = 4 \, Wb/m^2 \)
- Angle \( \theta = 60^\circ \)

Now, we can substitute the values into the formula step by step.

### Step 1: Calculate the cosine of the angle
\[ \cos(60^\circ) = \frac{1}{2} \]

### Step 2: Substitute the values into the magnetic flux formula
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
\[ \Phi = 4 \, Wb/m^2 \cdot 0.5 \, m^2 \cdot \frac{1}{2} \]

### Step 3: Perform the multiplication
\[ \Phi = 4 \cdot 0.5 \cdot \frac{1}{2} \]
\[ \Phi = 2 \cdot \frac{1}{2} \]
\[ \Phi = 1 \, Wb \]

### Final Answer
The magnetic flux passing through the body is \( 1 \, Wb \).

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