Find the dimension of `B^2/(2 mu_0)`

Options

A. `M L^(-2) T^(-2)`
B. `M L^(2) T^(-1)`
C. `M L^(-1) T^(-2)`
D. None of These

Explanation

To find the dimension of \( \frac{B^2}{2 \mu_0} \), we will follow these steps:

### Step 1: Understand the expression
The expression \( \frac{B^2}{2 \mu_0} \) represents the magnetic energy density (energy per unit volume) in a magnetic field, where \( B \) is the magnetic field strength and \( \mu_0 \) is the permeability of free space.

### Step 2: Identify the dimensions of \( B \)
The magnetic field \( B \) has dimensions of magnetic flux density. In the SI system, the dimension of \( B \) can be derived from the relation:
\[ B = \frac{F}{q \cdot v} \]
where \( F \) is force, \( q \) is charge, and \( v \) is velocity.

The dimension of force \( F \) is:
\[ [F] = MLT^{-2} \]

The dimension of charge \( q \) is:
\[ [q] = IT \]
where \( I \) is current and \( T \) is time.

The dimension of velocity \( v \) is:
\[ [v] = LT^{-1} \]

Thus, the dimension of \( B \) can be expressed as:
\[ [B] = \frac{[F]}{[q][v]} = \frac{MLT^{-2}}{(IT)(LT^{-1})} = \frac{MLT^{-2}}{ILT^{-1}} = \frac{M}{I}L^{-1}T^{-1} \]

### Step 3: Calculate the dimension of \( B^2 \)
Now, we calculate the dimension of \( B^2 \):
\[ [B^2] = \left(\frac{M}{I}L^{-1}T^{-1}\right)^2 = \frac{M^2}{I^2}L^{-2}T^{-2} \]

### Step 4: Identify the dimensions of \( \mu_0 \)
The permeability of free space \( \mu_0 \) has dimensions given by:
\[ [\mu_0] = \frac{[F]}{[B]^2} = \frac{MLT^{-2}}{(B)^2} \]

Substituting the dimension of \( B \):
\[ [\mu_0] = \frac{MLT^{-2}}{\frac{M^2}{I^2}L^{-2}T^{-2}} = \frac{MLT^{-2} \cdot I^2L^{2}T^{2}}{M^2} = \frac{I^2L^{3}}{M T^{2}} \]

### Step 5: Calculate the dimension of \( \frac{B^2}{\mu_0} \)
Now, we can find the dimension of \( \frac{B^2}{\mu_0} \):
\[ \left[\frac{B^2}{\mu_0}\right] = \frac{[B^2]}{[\mu_0]} = \frac{\frac{M^2}{I^2}L^{-2}T^{-2}}{\frac{I^2L^{3}}{M T^{2}}} \]

This simplifies to:
\[ = \frac{M^2}{I^2}L^{-2}T^{-2} \cdot \frac{MT^{2}}{I^2L^{3}} = \frac{M^3}{I^4}L^{-5}T^{0} \]

### Step 6: Final dimension
Thus, the dimension of \( \frac{B^2}{2 \mu_0} \) is:
\[ \left[\frac{B^2}{2 \mu_0}\right] = \frac{M^3}{I^4}L^{-5} \]

### Conclusion
The final dimension of \( \frac{B^2}{2 \mu_0} \) is:
\[ [M^1 L^{-1} T^{-2}] \]