In the formula `X =5YZ^(2)`, X and Z have dimensions of capacitance and magnetic field, respectively. What are the dimensions of Y in SI units?

Options

A. `[M^(-2)L^(-2)T^(6)A^(3)]`
B. `[M^(-2)L^(0) T^(-4)A^(-2)]`
C. `[M^(-3) L^(-2) T^(8) A^(4)]`
D. `[M^(-1) L^(-2) T^(4) A^(2)]`

Explanation

To determine the dimensions of \( Y \) in the formula \( X = 5YZ^2 \), where \( X \) has the dimensions of capacitance and \( Z \) has the dimensions of magnetic field, we will follow these steps:

### Step 1: Identify the dimensions of \( X \) (Capacitance)
The dimensions of capacitance \( C \) can be derived from the formula:
\[ C = \frac{Q}{V} \]
where \( Q \) is charge and \( V \) is voltage.

Voltage \( V \) can be expressed as:
\[ V = \frac{E}{Q} \]
where \( E \) is energy.

Energy \( E \) can be expressed as:
\[ E = F \cdot d \]
where \( F \) is force and \( d \) is distance.

The dimensions of force \( F \) are:
\[ [F] = M L T^{-2} \]
The dimensions of distance \( d \) are:
\[ [d] = L \]
Thus, the dimensions of energy \( E \) are:
\[ [E] = [F][d] = (M L T^{-2})(L) = M L^2 T^{-2} \]

Now substituting back to find the dimensions of voltage:
\[ [V] = \frac{E}{Q} = \frac{M L^2 T^{-2}}{Q} \]
Now substituting \( Q \) in terms of current \( I \) and time \( T \):
\[ [Q] = [I][T] \]
Thus,
\[ [V] = \frac{M L^2 T^{-2}}{I T} = M L^2 T^{-3} I^{-1} \]

Now substituting \( V \) back into the capacitance formula:
\[ [C] = \frac{Q}{V} = \frac{I T}{M L^2 T^{-3} I^{-1}} = \frac{I^2 T^4}{M L^2} \]
Thus, the dimensions of capacitance \( C \) are:
\[ [X] = M^{-1} L^{-2} T^{4} I^{2} \]

### Step 2: Identify the dimensions of \( Z \) (Magnetic Field)
The dimensions of magnetic field \( B \) can be derived from the Lorentz force equation:
\[ F = I L B \]
Rearranging gives:
\[ B = \frac{F}{I L} \]
Substituting the dimensions of force \( F \):
\[ [B] = \frac{M L T^{-2}}{I L} = \frac{M}{I T^2} \]
Thus, the dimensions of magnetic field \( B \) are:
\[ [Z] = M T^{-2} I^{-1} \]

### Step 3: Substitute into the original equation
From the equation \( X = 5YZ^2 \), we can isolate \( Y \):
\[ Y = \frac{X}{Z^2} \]

### Step 4: Calculate the dimensions of \( Y \)
Substituting the dimensions we found:
\[ [Y] = \frac{[X]}{[Z]^2} = \frac{M^{-1} L^{-2} T^{4} I^{2}}{(M T^{-2} I^{-1})^2} \]
Calculating \( [Z]^2 \):
\[ [Z]^2 = M^2 T^{-4} I^{-2} \]
Now substituting back:
\[ [Y] = \frac{M^{-1} L^{-2} T^{4} I^{2}}{M^2 T^{-4} I^{-2}} = M^{-1 - 2} L^{-2} T^{4 + 4} I^{2 + 2} = M^{-3} L^{-2} T^{8} I^{4} \]

### Final Answer
Thus, the dimensions of \( Y \) are:
\[ [Y] = M^{-3} L^{-2} T^{8} I^{4} \]