A rectangular frame of area `10 m^2` is placed in a uniform electric field of `20 NC^(-1)`, with normal drawn on the surface of the frame making `60^@` angle with the direction of field. What will be the electric flux through the frame ?

Options

A. 100 Vm
B. 200 Vm
C. 50 Vm
D. 150 Vm

Answer

To find the electric flux through the rectangular frame, we can use the formula for electric flux, which is given by:

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

where:
- \(\Phi\) is the electric flux,
- \(E\) is the magnitude of the electric field,
- \(A\) is the area of the surface,
- \(\theta\) is the angle between the electric field and the normal (perpendicular) to the surface.

### Step 1: Identify the given values
- Area of the frame, \(A = 10 \, m^2\)
- Electric field strength, \(E = 20 \, N/C\)
- Angle with the normal, \(\theta = 60^\circ\)

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

### Step 3: Substitute the values into the flux formula
Now, we can substitute the values into the electric flux formula:
\[ \Phi = E \cdot A \cdot \cos(\theta) \]
\[ \Phi = 20 \, N/C \cdot 10 \, m^2 \cdot \frac{1}{2} \]

### Step 4: Perform the calculation
\[ \Phi = 20 \cdot 10 \cdot \frac{1}{2} = 200 \cdot \frac{1}{2} = 100 \, Nm^2/C \]

### Conclusion
The electric flux through the frame is:
\[ \Phi = 100 \, Nm^2/C \]