If M is the magnetic moment, `mu` is the magnetic permeability and H is the intensity of magnetising field, then the dimensional formula for `mu` MH is

Options

A. `ML^(2)T^(-2)`
B. `ML^(2)T^(-2)I^(2)`
C. `ML^(2)T^(-3)`
D. `ML T^(-2)I`

Answer

To find the dimensional formula for the product of magnetic permeability (μ), magnetic moment (M), and intensity of magnetizing field (H), we will follow these steps:

### Step 1: Understand the Definitions
- **Magnetic Moment (M)**: It is defined as the product of the current (I) flowing through a loop and the area (A) of the loop.
- **Magnetic Permeability (μ)**: It relates the magnetic induction (B) to the intensity of the magnetizing field (H) through the equation B = μH.
- **Intensity of Magnetizing Field (H)**: It is a measure of the strength of the magnetic field.

### Step 2: Find the Dimensional Formula for Magnetic Moment (M)
The magnetic moment (M) can be expressed as:
\[ M = I \cdot A \]
Where:
- \( I \) (current) has the dimension of [A] (Ampere).
- \( A \) (area) has the dimension of [L²] (Length squared).

Thus, the dimensional formula for magnetic moment (M) is:
\[ [M] = [I] \cdot [A] = [A] \cdot [L^2] = A \cdot L^2 \]

### Step 3: Find the Dimensional Formula for Magnetic Induction (B)
From the equation \( B = μH \), we can express μ as:
\[ μ = \frac{B}{H} \]
To find the dimensional formula for B, we can use the relationship with force (F), charge (q), and velocity (v):
\[ F = q(v \times B) \]
The dimension of force (F) is:
\[ [F] = [M][L][T^{-2}] \]
The dimension of charge (q) is:
\[ [q] = [A][T] \]
The dimension of velocity (v) is:
\[ [v] = [L][T^{-1}] \]

Thus, the dimension of B can be derived as:
\[ [B] = \frac{[F]}{[q][v]} = \frac{[M][L][T^{-2}]}{[A][T][L][T^{-1}]} = \frac{[M]}{[A][T]} \]

This simplifies to:
\[ [B] = [M][A^{-1}][T^{-2}] \]

### Step 4: Find the Dimensional Formula for Magnetic Permeability (μ)
Now substituting the dimensions of B and H into the equation for μ:
\[ μ = \frac{B}{H} \]
Where H has the dimension of [A][T^{-1}][L^{-1}]. Therefore:
\[ [H] = [A][L^{-1}][T^{-1}] \]

So, we have:
\[ [μ] = \frac{[M][A^{-1}][T^{-2}]}{[A][L^{-1}][T^{-1}]} = [M][A^{-2}][L][T^{-1}] \]

### Step 5: Combine the Dimensional Formulas
Now we need to find the dimensional formula for \( μMH \):
\[ [μMH] = [μ] \cdot [M] \cdot [H] \]
Substituting the dimensions we found:
\[ [μMH] = ([M][A^{-2}][L][T^{-1}]) \cdot ([A][L^2]) \cdot ([A][L^{-1}][T^{-1}]) \]

### Step 6: Simplify the Expression
Now simplifying:
\[ [μMH] = [M][A^{-2}][L][T^{-1}] \cdot [A][L^2] \cdot [A][L^{-1}][T^{-1}] \]
\[ = [M][A^{-2}][L^2][T^{-2}] \]
\[ = [M][L^2][T^{-2}] \]

### Final Result
Thus, the dimensional formula for \( μMH \) is:
\[ [μMH] = [M][L^2][T^{-2}] \]