If `epsilon_(0), B, V` represent permitivity of free space, magnitude of magnetic field and volume of space respectively, then the dimension of `epsilon_(0)B^(2)V` is `[M^(a)L^(b)T^(c)]`. Find `a+b+c`.

Options

A. 8
B. 4
C. 2
D. 1

Answer

To find the dimensions of the expression \( \epsilon_0 B^2 V \), where \( \epsilon_0 \) is the permittivity of free space, \( B \) is the magnetic field, and \( V \) is the volume, we will follow these steps:

### Step 1: Identify the dimensions of each component

1. **Permittivity of free space (\( \epsilon_0 \))**:
The dimension of \( \epsilon_0 \) is given by:
\[ [\epsilon_0] = \frac{[M^{-1} L^{-3} T^{4}]}{[L^2 T^{-2}]} = [M^{-1} L^{-3} T^{4} L^{-2} T^{2}] = [M^{-1} L^{-1} T^{6}] \]

2. **Magnetic field (\( B \))**:
The magnetic field \( B \) has the dimension:
\[ [B] = [M^{1} L^{0} T^{-2} A^{-1}] \]
where \( A \) is the dimension of electric current.

3. **Volume (\( V \))**:
The dimension of volume \( V \) is:
\[ [V] = [L^3] \]

### Step 2: Substitute the dimensions into the expression

Now we will substitute these dimensions into the expression \( \epsilon_0 B^2 V \):
\[ [\epsilon_0 B^2 V] = [\epsilon_0] \cdot [B^2] \cdot [V] \]

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

Now substituting everything into the expression:
\[ [\epsilon_0 B^2 V] = [M^{-1} L^{-1} T^{6}] \cdot [M^{2} L^{0} T^{-4} A^{-2}] \cdot [L^3] \]

### Step 3: Combine the dimensions

Now we combine the dimensions:
\[ [\epsilon_0 B^2 V] = [M^{-1} L^{-1} T^{6}] \cdot [M^{2} L^{0} T^{-4} A^{-2}] \cdot [L^3] \]
\[ = [M^{-1+2} L^{-1+0+3} T^{6-4} A^{-2}] \]
\[ = [M^{1} L^{2} T^{2} A^{-2}] \]

### Step 4: Identify the coefficients \( a, b, c \)

From the final expression, we can identify:
- \( a = 1 \) (coefficient of \( M \))
- \( b = 2 \) (coefficient of \( L \))
- \( c = 2 \) (coefficient of \( T \))

### Step 5: Calculate \( a + b + c \)

Now we calculate:
\[ a + b + c = 1 + 2 + 2 = 5 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{5} \]