The dimension of pole strength of solenoid per unit current through it can be represented as

A. `M^(0) L^(-1) T^(0) A `
B. `M^(0) L^(0) T^(0) A`
C. `M^(0) L^(+1) T^(0) A^(0)`
D. `M^(0) LT^(0) A^(-1)`

Options

A. `M^(0) L^(-1) T^(0) A `
B. `M^(0) L^(0) T^(0) A`
C. `M^(0) L^(+1) T^(0) A^(0)`
D. `M^(0) LT^(0) A^(-1)`

Answer

To find the dimension of pole strength of a solenoid per unit current, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Terms**:
- Let \( M \) represent the magnetic pole strength.
- Let \( I \) represent the current.

2. **Magnetic Dipole Moment**:
- The magnetic dipole moment \( m \) is given by the formula:
\[ m = I \times A \]
where \( A \) is the area.

3. **Pole Strength Relation**:
- The magnetic dipole moment can also be expressed in terms of pole strength \( M \) and the distance \( L_0 \) (the distance between the poles):
\[ m = M \times L_0 \]

4. **Equating the Two Expressions**:
- From the above two equations, we can equate:
\[ I \times A = M \times L_0 \]

5. **Finding the Pole Strength per Unit Current**:
- We want to find the dimension of \( \frac{M}{I} \):
\[ M = \frac{I \times A}{L_0} \]
- Therefore,
\[ \frac{M}{I} = \frac{A}{L_0} \]

6. **Dimensions of Area and Length**:
- The dimension of area \( A \) is \( L^2 \).
- The dimension of length \( L_0 \) is \( L \).

7. **Calculating the Dimension**:
- Thus, the dimension of \( \frac{M}{I} \) becomes:
\[ \frac{M}{I} = \frac{L^2}{L} = L \]

8. **Final Dimension Representation**:
- The dimension of \( \frac{M}{I} \) can be expressed as:
\[ M^{0} L^{1} T^{0} A^{0} \]

9. **Choosing the Correct Option**:
- From the given options, the correct representation is:
- **Option C**: \( M^{0} L^{1} T^{0} A^{0} \)

### Summary of the Solution:
The dimension of pole strength of a solenoid per unit current is \( M^{0} L^{1} T^{0} A^{0} \), which corresponds to **Option C**.