The dimensions of the quantity \( \epsilon_0 \frac{d\Phi_E}{dt} \) are to be determined.- \( \epsilon_0 \) (permittivity of free space) has dimensions: \[ [\epsilon_0] = \frac{\text{C}^2}{\text{Nm}^2} = \frac{\text{A}^2 \cdot \text{s}^4}{\text{kg} \cdot \text{m}^3} \]- \( \Phi_E \) (electric flux) is defined as \( \Phi_E = E \cdot A \), where \( E \) is electric field and \( A \) is area. The dimensions of \( E \) are: \[ [E] = \frac{\text{N}}{\text{C}} = \frac{\text{kg} \cdot \text{m/s}^2}{\text{A}} \] The dimensions of area \( A \) are: \[ [A] = \text{m}^2 \] Consequently, the dimensions of \( \Phi_E \) 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, the dimensions of \( \epsilon_0 \frac{d\Phi_E}{dt} \) are calculated:- The dimensions of the rate of change of electric flux, \( \frac{d\Phi_E}{dt} \), are: \[ \left[\frac{d\Phi_E}{dt}\right] = \frac{\text{kg} \cdot \text{m}^3}{\text{A} \cdot \text{s}^3} \]Therefore, the dimensions of \( \epsilon_0 \frac{d\Phi_E}{dt} \) are:\[[\epsilon_0 \frac{d\Phi_E}{dt}] = [\epsilon_0] \cdot \left[\frac{d\Phi_E}{dt}\right] = \frac{\text{A}^2 \cdot \text{s}^4}{\text{kg} \cdot \text{m}^3} \cdot \frac{\text{kg} \cdot \text{m}^3}{\text{A} \cdot \text{s}^3} = \text{A}^1 \cdot \text{s}^1 = \text{A}\]This result corresponds to the dimension of electric current, which is \( \text{A} \).Thus, the correct answer is (B) Electric current.
