A flux of 1m Wb passes through a strip having an area `A=0.02 m^(2)`. The plane of the strip is at an angle of `60^(@)` to the direction of a uniform field B. The value of B is-

Options

A. `0.1T`
B. `0.058T`
C. `4.0mT`
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 normal to the surface. The formula for magnetic flux is given by:

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

### Step 1: Identify the given values
- Magnetic flux \( \Phi = 1 \, \text{mWb} = 1 \times 10^{-3} \, \text{Wb} \)
- Area \( A = 0.02 \, \text{m}^2 \)
- Angle \( \theta = 60^\circ \)

### Step 2: Determine the angle to use in the formula
The angle \( \theta \) given in the problem is the angle between the magnetic field \( B \) and the plane of the strip. However, in the formula for magnetic flux, we need the angle between the magnetic field and the normal to the surface. The normal vector makes an angle of \( 90^\circ - \theta \) with the magnetic field. Therefore:

\[ \theta' = 90^\circ - 60^\circ = 30^\circ \]

### Step 3: Calculate \( \cos(\theta') \)
Now we calculate \( \cos(30^\circ) \):

\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]

### Step 4: Substitute values into the flux formula
Now we can substitute the values into the magnetic flux formula:

\[ \Phi = B \cdot A \cdot \cos(30^\circ) \]

Substituting the known values:

\[ 1 \times 10^{-3} = B \cdot 0.02 \cdot \frac{\sqrt{3}}{2} \]

### Step 5: Solve for \( B \)
Rearranging the equation to solve for \( B \):

\[ B = \frac{1 \times 10^{-3}}{0.02 \cdot \frac{\sqrt{3}}{2}} \]

Calculating the denominator:

\[ 0.02 \cdot \frac{\sqrt{3}}{2} = 0.01\sqrt{3} \]

Now substituting back into the equation for \( B \):

\[ B = \frac{1 \times 10^{-3}}{0.01\sqrt{3}} = \frac{1 \times 10^{-1}}{\sqrt{3}} = \frac{0.1}{\sqrt{3}} \]

### Step 6: Final calculation
Calculating the numerical value:

\[ B \approx \frac{0.1}{1.732} \approx 0.0577 \, \text{T} \approx 0.058 \, \text{T} \]

### Conclusion
Thus, the value of \( B \) is approximately \( 0.058 \, \text{T} \).

---