Select the dimensional formula of `B^2/(2mu0)`

Options

A. `[M^1L^1T^2]`
B. `[M^-1L^1T^2]`
C. `[M^-1L^-1T^-2]`
D. `[M^1L^-1T^-2]`

Answer

To find the dimensional formula of \( \frac{B^2}{2\mu_0} \), we will break down the components involved: the magnetic field \( B \) and the permeability of free space \( \mu_0 \).

### Step 1: Determine the dimensional formula of the magnetic field \( B \)

The magnetic field \( B \) can be expressed in terms of force, charge, and velocity. The formula for magnetic field intensity is given by:

\[ B = \frac{F}{Q \cdot v} \]

Where:
- \( F \) is the force,
- \( Q \) is the charge,
- \( v \) is the velocity.

#### Step 1.1: Find the dimensions of force \( F \)

The dimensional formula for force \( F \) is:

\[ [F] = M L T^{-2} \]

#### Step 1.2: Find the dimensions of charge \( Q \)

Charge \( Q \) can be expressed in terms of current \( I \) and time \( t \):

\[ [Q] = I \cdot T \]

Where \( I \) has the dimensional formula of \( A \) (Ampere). Thus,

\[ [Q] = A \cdot T \]

#### Step 1.3: Find the dimensions of velocity \( v \)

The dimensional formula for velocity \( v \) is:

\[ [v] = L T^{-1} \]

#### Step 1.4: Substitute into the formula for \( B \)

Now substituting the dimensions into the formula for \( B \):

\[ [B] = \frac{[F]}{[Q] \cdot [v]} = \frac{M L T^{-2}}{(A T)(L T^{-1})} \]

This simplifies to:

\[ [B] = \frac{M L T^{-2}}{A L T} = \frac{M}{A T^2} \]

Thus, the dimensional formula for \( B \) is:

\[ [B] = M L^0 T^{-2} A^{-1} \]

### Step 2: Determine the dimensional formula of \( \mu_0 \)

The permeability of free space \( \mu_0 \) has the unit of Newton per Ampere squared:

\[ [\mu_0] = \frac{[F]}{[Q]^2} = \frac{M L T^{-2}}{(A T)^2} \]

This simplifies to:

\[ [\mu_0] = \frac{M L T^{-2}}{A^2 T^2} = M L T^{-4} A^{-2} \]

### Step 3: Calculate the dimensional formula of \( \frac{B^2}{2\mu_0} \)

Now we can find the dimensional formula of \( \frac{B^2}{2\mu_0} \):

\[ \frac{B^2}{\mu_0} = \frac{(M L^0 T^{-2} A^{-1})^2}{M L T^{-4} A^{-2}} \]

Calculating \( B^2 \):

\[ B^2 = M^2 L^0 T^{-4} A^{-2} \]

Now substituting into the formula:

\[ \frac{B^2}{\mu_0} = \frac{M^2 L^0 T^{-4} A^{-2}}{M L T^{-4} A^{-2}} = \frac{M^2}{M} \cdot \frac{L^0}{L} \cdot \frac{T^{-4}}{T^{-4}} \cdot \frac{A^{-2}}{A^{-2}} = M^{2-1} L^{0-1} T^{0} A^{0} \]

This simplifies to:

\[ \frac{B^2}{\mu_0} = M^1 L^{-1} T^0 A^0 = M L^{-1} \]

### Final Answer

Thus, the dimensional formula of \( \frac{B^2}{2\mu_0} \) is:

\[ [M L^{-1}] \]