Dimensional Analysis of Physical Quantities

Dimensional analysis is a fundamental concept in physics that helps us understand the nature of physical quantities and verify the consistency of equations. Every physical quantity can be expressed in terms of a combination of fundamental dimensions such as Mass (M), Length (L), Time (T), and Electric Current (A). The question asks us to identify which physical quantity among the given options has the dimension $\text{MLT}^{-2}\text{A}^{-2}$. Let's systematically determine the dimensions of each option.

1. Magnetic Field (B)

• The magnetic field $B$ is a measure of the strength of a magnetic field. One common way to define it is through the force experienced by a current-carrying conductor. The force $F$ on a wire of length $L$ carrying a current $I$ in a magnetic field $B$ is given by the formula: F = BIL (assuming the wire is perpendicular to the magnetic field).
• From this formula, we can express the magnetic field $B$ as: B = \frac{F}{IL}
• Now, let's find the dimensions for each term:
  • Dimension of Force $[F]$ = $\text{MLT}^{-2}$ (Mass × Acceleration)
  • Dimension of Current $[I]$ = $\text{A}$
  • Dimension of Length $[L]$ = $\text{L}$
• Substituting these into the equation for $[B]$: [B] = \frac{\text{MLT}^{-2}}{\text{A} \cdot \text{L}} = \text{MT}^{-2}\text{A}^{-1}

2. Torsion Constant

• The torsion constant (often denoted as $C$ or $k$) is a property of a material's resistance to twisting. It relates the applied torque $\tau$ to the angle of twist $\theta$ produced: \tau = C\theta
• From this, the torsion constant $C$ can be written as: C = \frac{\tau}{\theta}
• Let's determine the dimensions:
  • Torque $\tau$ has the dimensions of Force multiplied by distance: $[\tau] = [F][L] = \text{MLT}^{-2} \cdot \text{L} = \text{ML}^2\text{T}^{-2}$.
  • Angle $\theta$ is a dimensionless quantity (it's a ratio of arc length to radius).
• Therefore, the dimension of the torsion constant $[C]$ is: [C] = \frac{\text{ML}^2\text{T}^{-2}}{\text{1}} = \text{ML}^2\text{T}^{-2}

3. Magnetic Moment

• The magnetic moment (usually denoted by $M$ or $\mu$) of a current loop is a measure of its strength as a magnetic dipole. For a simple current loop, it is defined as the product of the current $I$ flowing through the loop and the area $A$ enclosed by the loop: M = IA
• Let's find the dimensions:
  • Dimension of Current $[I]$ = $\text{A}$
  • Dimension of Area $[A]$ = $\text{L}^2$ (Length × Length)
• Therefore, the dimension of the magnetic moment $[M]$ is: [M] = \text{A} \cdot \text{L}^2 = \text{L}^2\text{A}

4. Permeability of Free Space ($\mu_0$)

• The permeability of free space $\mu_0$ is a fundamental physical constant that describes how a magnetic field passes through a vacuum. It appears in many equations in electromagnetism.
• One common formula involving $\mu_0$ is the force per unit length between two long, parallel current-carrying wires: \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r} where $F/L$ is the force per unit length, $I_1$ and $I_2$ are the currents in the wires, and $r$ is the distance between them.
• Rearranging this formula to find $\mu_0$: \mu_0 = \frac{2\pi r (F/L)}{I_1 I_2}
• Let's determine the dimensions for each term:
  • Dimension of Force $[F]$ = $\text{MLT}^{-2}$
  • Dimension of Length $[L]$ = $\text{L}$ (for force per unit length denominator)
  • Dimension of distance $[r]$ = $\text{L}$
  • Dimension of Current $[I_1]$ and $[I_2]$ = $\text{A}$
  • The constant $2\pi$ is dimensionless.
• Substituting these dimensions into the expression for $[\mu_0]$: [\mu_0] = \frac{\text{L} \cdot (\text{MLT}^{-2}\text{L}^{-1})}{\text{A} \cdot \text{A}} = \frac{\text{MLT}^{-2}}{\text{A}^2} = \text{MLT}^{-2}\text{A}^{-2}
• This derived dimension of $\text{MLT}^{-2}\text{A}^{-2}$ for the permeability of free space matches the dimension given in the question.

Conclusion

By analyzing the dimensional formulas of each physical quantity, we have determined that the Permeability of free space has the dimension $\text{MLT}^{-2}\text{A}^{-2}$.