The dimensional formula for the physical quantity `(E^(2)mu_(0)epsilon_(0))/(B^(2))` is
(E = electric field and B = magnetic field)

Options

A. `L^(0)M^(0)T^(0)`
B. `L^(1)M^(0)T^(-1)`
C. `L^(-1)M^(0)T^(1)`
D. `L^(1//2)M^(0)T^(-1//2)`

Answer

To find the dimensional formula for the physical quantity \((E^{2} \mu_{0} \epsilon_{0})/(B^{2})\), we will follow these steps:

### Step 1: Write the known quantities
We know:
- \(E\) is the electric field.
- \(B\) is the magnetic field.
- \(\mu_{0}\) is the permeability of free space.
- \(\epsilon_{0}\) is the permittivity of free space.

### Step 2: Determine the dimensional formulas
1. **Electric Field \(E\)**:
The electric field \(E\) has the dimensional formula:
\[ [E] = \frac{[F]}{[q]} = \frac{[M L T^{-2}]}{[I T]} = [M L T^{-3} I^{-1}] \]

2. **Magnetic Field \(B\)**:
The magnetic field \(B\) has the dimensional formula:
\[ [B] = \frac{[F]}{[q v]} = \frac{[M L T^{-2}]}{[I T L]} = [M T^{-2} I^{-1}] \]

3. **Permeability of Free Space \(\mu_{0}\)**:
The permeability \(\mu_{0}\) has the dimensional formula:
\[ [\mu_{0}] = \frac{[F]}{[I^{2}]} = \frac{[M L T^{-2}]}{[I^{2}]} = [M L T^{-2} I^{-2}] \]

4. **Permittivity of Free Space \(\epsilon_{0}\)**:
The permittivity \(\epsilon_{0}\) has the dimensional formula:
\[ [\epsilon_{0}] = \frac{[q^{2}]}{[F L]} = \frac{[I^{2} T^{2}]}{[M L T^{-2}]} = [M^{-1} L^{-1} T^{4} I^{2}] \]

### Step 3: Substitute the dimensional formulas into the expression
Now we substitute these dimensional formulas into the expression \((E^{2} \mu_{0} \epsilon_{0})/(B^{2})\):
\[ \frac{[E^{2} \mu_{0} \epsilon_{0}]}{[B^{2}]} = \frac{([M L T^{-3} I^{-1}])^{2} \cdot [M L T^{-2} I^{-2}] \cdot [M^{-1} L^{-1} T^{4} I^{2}]}{([M T^{-2} I^{-1}])^{2}} \]

### Step 4: Simplify the expression
Calculating the numerator:
\[ [E^{2} \mu_{0} \epsilon_{0}] = [M^{2} L^{2} T^{-6} I^{-2}] \cdot [M L T^{-2} I^{-2}] \cdot [M^{-1} L^{-1} T^{4} I^{2}] \]
Combining these:
\[ = [M^{2} L^{2} T^{-6} I^{-2} \cdot M L T^{-2} I^{-2} \cdot M^{-1} L^{-1} T^{4} I^{2}] \]
\[ = [M^{2 + 1 - 1} L^{2 + 1 - 1} T^{-6 - 2 + 4} I^{-2 - 2 + 2}] \]
\[ = [M^{2} L^{2} T^{-4} I^{-2}] \]

Calculating the denominator:
\[ [B^{2}] = ([M T^{-2} I^{-1}])^{2} = [M^{2} T^{-4} I^{-2}] \]

### Step 5: Final simplification
Now we can simplify:
\[ \frac{[M^{2} L^{2} T^{-4} I^{-2}]}{[M^{2} T^{-4} I^{-2}]} = [L^{2}] \]

### Conclusion
Thus, the dimensional formula for the physical quantity \((E^{2} \mu_{0} \epsilon_{0})/(B^{2})\) is:
\[ [L^{2}] \]