A flux of `10^(-3)Wb` passes through a strip having an area `A = 0.02m^(2)`. The plane of the strip is at an angle of `60^(@)` to be direction of a uniform field B. The value of B is

Options

A. 0.1 T
B. 0.058 T
C. 4.0 mT
D. None of the above

Answer

To solve the problem, we need to find the value of the magnetic field \( B \) given the magnetic flux \( \Phi \), the area \( A \), and the angle \( \theta \) between the magnetic field and the area vector.

### Step-by-Step Solution:

1. **Understand the Formula for Magnetic Flux**:
The magnetic flux \( \Phi \) through a surface is given by the formula:
\[ \Phi = B \cdot A = B \cdot A \cdot \cos(\theta) \]
where:
- \( B \) is the magnetic field,
- \( A \) is the area,
- \( \theta \) is the angle between the magnetic field and the normal (perpendicular) to the surface.

2. **Identify Given Values**:
From the problem, we have:
- \( \Phi = 10^{-3} \, \text{Wb} \)
- \( A = 0.02 \, \text{m}^2 \)
- The angle between the plane of the strip and the magnetic field is \( 60^\circ \).

3. **Determine the Angle for the Formula**:
The angle \( \theta \) used in the flux formula is the angle between the magnetic field \( B \) and the area vector \( A \). Since the area vector is perpendicular to the surface, we need to find the angle between \( B \) and this area vector.
- If the plane of the strip is at \( 60^\circ \) to the magnetic field, then the angle \( \theta \) between the magnetic field and the area vector is:
\[ \theta = 90^\circ - 60^\circ = 30^\circ \]

4. **Substitute Values into the Flux Formula**:
Now we can substitute the values into the flux formula:
\[ 10^{-3} = B \cdot 0.02 \cdot \cos(30^\circ) \]

5. **Calculate \( \cos(30^\circ) \)**:
The value of \( \cos(30^\circ) \) is:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]

6. **Rearranging the Equation to Solve for \( B \)**:
Substitute \( \cos(30^\circ) \) into the equation:
\[ 10^{-3} = B \cdot 0.02 \cdot \frac{\sqrt{3}}{2} \]
Rearranging gives:
\[ B = \frac{10^{-3}}{0.02 \cdot \frac{\sqrt{3}}{2}} \]

7. **Simplify the Expression**:
Simplifying further:
\[ B = \frac{10^{-3}}{0.01 \sqrt{3}} = \frac{10^{-1}}{\sqrt{3}} \]

8. **Calculate the Final Value of \( B \)**:
Now, calculating \( B \):
\[ B = \frac{0.1}{\sqrt{3}} \approx \frac{0.1}{1.732} \approx 0.0577 \, \text{Tesla} \]
Rounding this gives:
\[ B \approx 0.058 \, \text{Tesla} \]

### Final Answer:
The value of the magnetic field \( B \) is approximately \( 0.058 \, \text{Tesla} \).