The dimensions of `(B^(2)R^(2)C^(2))/(2mu_(0))` (Where B is magnetic filed, and `mu_(0)` is permeability of free space , R resistance and C is capacitance) is

Options

A. `[ML^(-1)]`
B. `[MLT^(-1)]`
C. `[ML^(2)T^(-1)]`
D. `[MLT^(2)]`

Answer

To find the dimensions of the expression \((B^2 R^2 C^2)/(2\mu_0)\), where \(B\) is the magnetic field, \(R\) is resistance, \(C\) is capacitance, and \(\mu_0\) is the permeability of free space, we will follow these steps:

### Step-by-Step Solution:

1. **Determine the dimensions of \(B\)**:
- The magnetic field \(B\) can be derived from the Lorentz force law, where the force \(F = QVB\). Rearranging gives us \(B = \frac{F}{QV}\).
- The dimensions of force \(F\) are \([M L T^{-2}]\).
- The dimensions of charge \(Q\) are \([A T]\) (where \(A\) is current and \(T\) is time).
- The dimensions of velocity \(V\) are \([L T^{-1}]\).
- Therefore, the dimensions of \(B\) are:
\[ [B] = \frac{[F]}{[Q][V]} = \frac{[M L T^{-2}]}{[A T][L T^{-1}]} = \frac{[M L T^{-2}]}{[A L T^{-1}]} = [M A^{-1} T^{-2}] \]

2. **Determine the dimensions of \(R\)**:
- Resistance \(R\) is defined by Ohm's law \(V = IR\), or \(R = \frac{V}{I}\).
- The dimensions of voltage \(V\) are \([M L^2 T^{-3} A^{-1}]\) (derived from \(V = IR\)).
- The dimensions of current \(I\) are \([A]\).
- Thus, the dimensions of \(R\) are:
\[ [R] = \frac{[V]}{[I]} = \frac{[M L^2 T^{-3} A^{-1}]}{[A]} = [M L^2 T^{-3} A^{-2}] \]

3. **Determine the dimensions of \(C\)**:
- Capacitance \(C\) is defined as \(C = \frac{Q}{V}\).
- The dimensions of charge \(Q\) are \([A T]\) and voltage \(V\) has dimensions \([M L^2 T^{-3} A^{-1}]\).
- Therefore, the dimensions of \(C\) are:
\[ [C] = \frac{[Q]}{[V]} = \frac{[A T]}{[M L^2 T^{-3} A^{-1}]} = [M^{-1} L^{-2} T^4 A^2] \]

4. **Combine the dimensions**:
- Now we can find the dimensions of \(B^2 R^2 C^2\):
\[ [B^2] = [M^2 A^{-2} T^{-4}] \]
\[ [R^2] = [M^2 L^4 T^{-6} A^{-4}] \]
\[ [C^2] = [M^{-2} L^{-4} T^8 A^4] \]
- Therefore, the combined dimensions are:
\[ [B^2 R^2 C^2] = [M^2 A^{-2} T^{-4}] \cdot [M^2 L^4 T^{-6} A^{-4}] \cdot [M^{-2} L^{-4} T^8 A^4] \]
- Simplifying this gives:
\[ [B^2 R^2 C^2] = [M^{2+2-2} L^{4-4} T^{-4-6+8} A^{-2-4+4}] = [M^2 L^0 T^{-2} A^{-2}] = [M^2 T^{-2} A^{-2}] \]

5. **Determine the dimensions of \(\mu_0\)**:
- The permeability of free space \(\mu_0\) has dimensions:
\[ [\mu_0] = [M L A^{-2} T^{-2}] \]

6. **Combine everything**:
- Now we can find the dimensions of \(\frac{B^2 R^2 C^2}{2\mu_0}\):
\[ \left[\frac{B^2 R^2 C^2}{\mu_0}\right] = \frac{[M^2 T^{-2} A^{-2}]}{[M L A^{-2} T^{-2}]} = [M^{2-1} L^{-1} T^{0} A^{0}] = [M L^{-1}] \]

### Final Answer:
The dimensions of \(\frac{B^2 R^2 C^2}{2\mu_0}\) are \([M L^{-1}]\).

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