A cube of edge a has its edges parallel to X, Y and Z -axis of rectangular coordinate system. A uniform electric held E is parallel to Y-axis and uniform magnetic field B parallel to x axis. The rate at which energy flows through each face of the cube is

Options

A. `(a^(2)EB)/(2mu_(0))` parallel to X-Y plane and zero in others
B. `(a^(2)EB)/(mu_(0))` parallel to Y-Z plane and zero in others
C. `(a^(2)EB)/(2mu_(0))`from all faces
D. `(a^(2)EB)/(2mu_(0))` parallel to X-Z faces and zero in others

Answer

To solve the problem of determining the rate at which energy flows through each face of a cube in the presence of a uniform electric field \( E \) and a uniform magnetic field \( B \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have a cube with edge length \( a \).
- The electric field \( E \) is directed along the Y-axis (let's denote it as \( \hat{j} \)).
- The magnetic field \( B \) is directed along the X-axis (denote it as \( \hat{i} \)).

2. **Using the Poynting Vector**:
- The Poynting vector \( \vec{S} \) represents the rate of energy flow per unit area and is given by the formula:
\[ \vec{S} = \frac{1}{\mu_0} \vec{E} \times \vec{B} \]
- Here, \( \mu_0 \) is the permeability of free space.

3. **Calculating the Cross Product**:
- We need to calculate \( \vec{E} \times \vec{B} \):
\[ \vec{E} = E \hat{j}, \quad \vec{B} = B \hat{i} \]
\[ \vec{E} \times \vec{B} = (E \hat{j}) \times (B \hat{i}) = E B (\hat{j} \times \hat{i}) \]
- The cross product \( \hat{j} \times \hat{i} = -\hat{k} \) (using the right-hand rule).

4. **Substituting into the Poynting Vector**:
- Therefore, we have:
\[ \vec{S} = \frac{1}{\mu_0} E B (-\hat{k}) = -\frac{E B}{\mu_0} \hat{k} \]
- This indicates that the energy flows in the negative Z-direction.

5. **Calculating the Rate of Energy Flow**:
- The magnitude of the Poynting vector is:
\[ |\vec{S}| = \frac{E B}{\mu_0} \]
- The area of each face of the cube is \( A = a^2 \).

6. **Finding the Total Energy Flow Through Each Face**:
- The rate of energy flow through the face of the cube perpendicular to the Z-axis (which is the XY-plane) is:
\[ \text{Rate of energy flow} = |\vec{S}| \cdot A = \frac{E B}{\mu_0} \cdot a^2 \]

7. **Final Result**:
- Thus, the rate at which energy flows through the face of the cube that is parallel to the XY-plane is:
\[ \text{Rate of energy flow} = \frac{E B a^2}{\mu_0} \]
- For the other faces (YZ-plane and XZ-plane), since the Poynting vector is directed along the Z-axis, the energy flow through those faces is zero.

### Summary:
- The rate at which energy flows through the faces of the cube is:
- \( \frac{E B a^2}{\mu_0} \) through the XY-plane,
- \( 0 \) through the YZ-plane and XZ-plane.