Understanding Characteristic Impedance Dimensions

This problem asks us to determine the fundamental dimensions of the characteristic impedance of free space, $Z_0$. We are given the formula $Z_0 = \sqrt{\frac{\mu_0}{\epsilon_0}}$. Here, $\mu_0$ represents the magnetic permeability and $\epsilon_0$ represents the electric permittivity of free space. We need to find the dimensions, expressed in terms of Mass (M), Length (L), Time (T), and Electric Current (A).

Deriving Dimensions from Electrical Concepts

A straightforward way to find the dimensions of impedance ($Z$) is by relating it to voltage ($V$) and current ($I$), since impedance is essentially the opposition to current flow, measured in Ohms ($\Omega$). In general, $Z = V/I$. Let's find the dimensions of voltage first.

• Dimensions of Electric Current ($I$): The standard unit for electric current is the Ampere (A). Therefore, its dimensions are simply $[A]$.
• Dimensions of Voltage ($V$): Voltage is defined as energy ($E$) per unit charge ($Q$). So, $V = E/Q$.
  • The dimensions of Energy ($E$) are $[ML^2T^{-2}]$. This comes from the definition of kinetic energy ($\frac{1}{2}mv^2$) or work done (Force × Distance).
  • The dimensions of Electric Charge ($Q$) are $[AT]$, as charge is current multiplied by time ($Q = I \times t$).
  • Therefore, the dimensions of Voltage ($V$) are: $ [V] = \frac{[E]}{[Q]} = \frac{[ML^2T^{-2}]}{[AT]} = [ML^2T^{-3}A^{-1}] $
• Dimensions of Impedance ($Z$): Since impedance $Z$ is the ratio of voltage $V$ to current $I$ ($Z = V/I$), its dimensions are: $ [Z] = \frac{[V]}{[I]} = \frac{[ML^2T^{-3}A^{-1}]}{[A]} = [ML^2T^{-3}A^{-2}] $

This means the dimensions of the characteristic impedance $Z_0$ are $[ML^2T^{-3}A^{-2}]$.

Calculating Dimensions Using Permeability and Permittivity

We can verify this result using the provided formula $Z_0 = \sqrt{\frac{\mu_0}{\epsilon_0}}$. To do this, we need the dimensions of magnetic permeability ($\mu_0$) and electric permittivity ($\epsilon_0$).

Dimensions of Magnetic Permeability ($\mu_0$):

The magnetic permeability $\mu_0$ appears in the formula for the force between two parallel current-carrying wires. A related quantity is the magnetic field $B$. For example, the field around a long straight wire is $B = \frac{\mu_0 I}{2\pi r}$. Force on a charge $q$ moving with velocity $v$ in a magnetic field $B$ is $F = qvB$. Combining these leads to dimensions:

$ [\mu_0] = [MLT^{-2}A^{-2}] $

Dimensions of Electric Permittivity ($\epsilon_0$):

The electric permittivity $\epsilon_0$ appears in Coulomb's Law for the force between two charges: $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$. From this, we can derive the dimensions:

$ [\epsilon_0] = [M^{-1}L^{-3}T^4A^2] $

Now, let's calculate the dimensions of the ratio $\frac{\mu_0}{\epsilon_0}$:

$ \left[\frac{\mu_0}{\epsilon_0}\right] = \frac{[MLT^{-2}A^{-2}]}{[M^{-1}L^{-3}T^4A^2]} = [M^{1-(-1)} L^{1-(-3)} T^{-2-4} A^{-2-2}] = [M^2L^4T^{-6}A^{-4}] $

Finally, we take the square root to find the dimensions of $Z_0$:

$ [Z_0] = \sqrt{[M^2L^4T^{-6}A^{-4}]} = [M^{2/2}L^{4/2}T^{-6/2}A^{-4/2}] = [ML^2T^{-3}A^{-2}] $

Both methods give the same result, confirming our calculation.

Identifying the Correct Option

We need to match our derived dimensions, $[ML^2T^{-3}A^{-2}]$, with the options provided.

Our calculated dimensions $[ML^2T^{-3}A^{-2}]$ match exactly with Option 4.