The dimension of magnetic field in M, L, T and C (Coulomb) is given as :

Options

A. `MLT^(-1)C^(-1)`
B. `MT^(-2)C^(-2)`
C. `MT^(-1)C^(-1)`
D. `MT^(-2)C^(-1)`

Answer

To find the dimensions of the magnetic field \( B \) in terms of mass \( M \), length \( L \), time \( T \), and charge \( C \), we can start by using the formula for the magnetic force acting on a current-carrying wire.

### Step-by-Step Solution:

1. **Understanding the Magnetic Force**:
The magnetic force \( F \) on a current-carrying wire of length \( L \) in a magnetic field \( B \) is given by the equation:
\[ F = BIL \sin(\theta) \]
For simplicity, we can consider the case where the wire is perpendicular to the magnetic field, so \( \sin(\theta) = 1 \). Thus, we have:
\[ F = BIL \]

2. **Rearranging the Formula**:
We can rearrange the formula to express the magnetic field \( B \):
\[ B = \frac{F}{IL} \]

3. **Finding Dimensions of Each Quantity**:
- The dimension of force \( F \) is given by:
\[ [F] = [M][L][T^{-2}] = MLT^{-2} \]
- The dimension of current \( I \) is represented as charge per unit time:
\[ [I] = \frac{[Q]}{[T]} = CT^{-1} \]
- The dimension of length \( L \) is simply:
\[ [L] = L \]

4. **Substituting Dimensions into the Formula**:
Now we can substitute the dimensions of force, current, and length into the equation for \( B \):
\[ [B] = \frac{[F]}{[I][L]} = \frac{MLT^{-2}}{(CT^{-1})(L)} \]

5. **Simplifying the Expression**:
When we simplify this expression, we get:
\[ [B] = \frac{MLT^{-2}}{CTL} = \frac{M}{C} T^{-1} \]

6. **Final Result**:
Therefore, the dimensions of the magnetic field \( B \) can be expressed as:
\[ [B] = ML^{-1}T^{-2}C^{-1} \]

### Conclusion:
The dimensions of the magnetic field \( B \) in terms of mass \( M \), length \( L \), time \( T \), and charge \( C \) are:
\[ [B] = M L^{-1} T^{-2} C^{-1} \]