A square coil of `10^(-2) m^(2)` area is placed perpenducular to a uniform magnetic field of `10^(3) Wb//m^(2)`. What is magnetic flux through the coil?

Options

A. 10 weber
B. `10^(-5)`weber
C. `10^(5)`weber
D. 100 weber

Answer

To find the magnetic flux through the coil, we can use the formula for magnetic flux, which is given by:

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

where:
- \(\Phi\) is the magnetic flux,
- \(B\) is the magnetic field strength,
- \(A\) is the area of the coil,
- \(\theta\) is the angle between the magnetic field and the normal (perpendicular) to the surface of the coil.

### Step-by-Step Solution:

1. **Identify the given values**:
- Area of the coil, \(A = 10^{-2} \, m^2\)
- Magnetic field strength, \(B = 10^3 \, Wb/m^2\)
- Since the coil is placed perpendicular to the magnetic field, the angle \(\theta = 0^\circ\).

2. **Calculate \(\cos(\theta)\)**:
- Since \(\theta = 0^\circ\), we have:
\[ \cos(0^\circ) = 1 \]

3. **Substitute the values into the magnetic flux formula**:
\[ \Phi = B \cdot A \cdot \cos(\theta) = (10^3 \, Wb/m^2) \cdot (10^{-2} \, m^2) \cdot 1 \]

4. **Perform the multiplication**:
\[ \Phi = 10^3 \cdot 10^{-2} = 10^{3-2} = 10^1 = 10 \, Wb \]

5. **Conclusion**:
The magnetic flux through the coil is:
\[ \Phi = 10 \, Wb \]

### Final Answer:
The magnetic flux through the coil is \(10 \, Wb\).

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