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The dimensional formula of 1/(mucirc^(c...

The dimensional formula of `1/(mu_circ^(c^2)) (dϕ_E)/dt` is same as that of[where Φe is same as that of [where `ϕ_E` is electric flux and c is speed od light,t is time and `μ_0` is permeability of free space)

A

Current

B

electric potential

C

capacitance

D

charge

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To solve the problem, we need to find the dimensional formula of the expression \( \frac{1}{\mu_0 c^2} \frac{d\phi_E}{dt} \), where \( \phi_E \) is the electric flux, \( c \) is the speed of light, \( t \) is time, and \( \mu_0 \) is the permeability of free space. ### Step-by-Step Solution: 1. **Identify the components of the expression**: - We have \( \mu_0 \), \( c \), and \( \frac{d\phi_E}{dt} \). 2. **Find the dimensional formula of \( \mu_0 \)**: - The permeability of free space \( \mu_0 \) has the dimensional formula \( [M^{-1} L^{-2} T^4 I^2] \). 3. **Find the dimensional formula of \( c \)**: - The speed of light \( c \) has the dimensional formula \( [L T^{-1}] \). 4. **Calculate \( c^2 \)**: - The dimensional formula of \( c^2 \) is \( [L^2 T^{-2}] \). 5. **Combine the dimensions of \( \mu_0 \) and \( c^2 \)**: - The dimensional formula of \( \mu_0 c^2 \) is: \[ [\mu_0 c^2] = [M^{-1} L^{-2} T^4 I^2] \cdot [L^2 T^{-2}] = [M^{-1} L^{0} T^{2} I^{2}] \] 6. **Find the dimensional formula of \( \frac{d\phi_E}{dt} \)**: - Electric flux \( \phi_E \) has the dimensional formula \( [E \cdot A] = [M^1 L^2 T^{-3} I^{-1}] \) (using \( E = M L T^{-2} I^{-1} \) and area \( A = L^2 \)). - Therefore, the dimensional formula of \( \frac{d\phi_E}{dt} \) is: \[ \left[\frac{d\phi_E}{dt}\right] = \frac{[M^1 L^2 T^{-3} I^{-1}]}{[T]} = [M^1 L^2 T^{-4} I^{-1}] \] 7. **Combine the expressions**: - Now we substitute \( \frac{d\phi_E}{dt} \) into the original expression: \[ \frac{1}{\mu_0 c^2} \frac{d\phi_E}{dt} = \frac{[M^1 L^2 T^{-4} I^{-1}]}{[M^{-1} L^{0} T^{2} I^{2}]} = [M^{1+1} L^{2-0} T^{-4-2} I^{-1-2}] = [M^{2} L^{2} T^{-6} I^{-3}] \] 8. **Final dimensional formula**: - The final dimensional formula of \( \frac{1}{\mu_0 c^2} \frac{d\phi_E}{dt} \) is: \[ [M^2 L^2 T^{-6} I^{-3}] \] ### Conclusion: The dimensional formula of \( \frac{1}{\mu_0 c^2} \frac{d\phi_E}{dt} \) is the same as that of the displacement current \( I_D \), which is \( [M^0 L^0 T^0 I^1] \).
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