We are given the quantity \( \epsilon_0 \frac{d\Phi_E}{dt} \), and we need to find its dimensions.
- \(\epsilon_0\) is the permittivity of free space. Its dimensions are:
\[
[\epsilon_0] = \frac{\text{C}^2}{\text{Nm}^2} = \frac{\text{A}^2 \cdot \text{s}^4}{\text{kg} \cdot \text{m}^3}
\]
- \( \Phi_E \) is the electric flux. The electric flux is given by \( \Phi_E = E \cdot A \), where \( E \) is the electric field and \( A \) is the area. The dimensions of \( E \) are:
\[
[E] = \frac{\text{N}}{\text{C}} = \frac{\text{kg} \cdot \text{m/s}^2}{\text{A}}
\]
And the dimensions of area \( A \) are:
\[
[A] = \text{m}^2
\]
Thus, the dimensions of electric flux \( \Phi_E \) are:
\[
[\Phi_E] = [E] \cdot [A] = \frac{\text{kg} \cdot \text{m/s}^2}{\text{A}} \cdot \text{m}^2 = \frac{\text{kg} \cdot \text{m}^3}{\text{A} \cdot \text{s}^2}
\]
Now, we calculate the dimensions of \( \epsilon_0 \frac{d\Phi_E}{dt} \):
- The dimensions of \( \frac{d\Phi_E}{dt} \) are:
\[
\left[\frac{d\Phi_E}{dt}\right] = \frac{\text{kg} \cdot \text{m}^3}{\text{A} \cdot \text{s}^3}
\]
Thus, the dimensions of \( \epsilon_0 \frac{d\Phi_E}{dt} \) are:
\[
[\epsilon_0 \frac{d\Phi_E}{dt}] = [\epsilon_0] \cdot \left[\frac{d\Phi_E}{dt}\right] = \frac{\text{A}^2 \cdot \text{s}^4}{\text{kg} \cdot \text{m}^3} \cdot \frac{\text{kg} \cdot \text{m}^3}{\text{A} \cdot \text{s}^3} = \frac{\text{A}^1 \cdot \text{s}^1}{\text{s}^3} = \text{A} \cdot \text{s}^{-2}
\]
This is the dimension of electric current, as electric current has the dimension of \( \text{A} \).
Therefore, the correct answer is (B) Electric current.
