The magnetic force on a point moving charge is `F=q(vxxB)`.
Here `q=` electric charge
`v=` velocity of the point charge
`B=` magnetic field
The dimensions of `B` is

Options

A. `["MLT"^(-1)"A"]`
B. `["M"^(2)"LT"^(-2)"A"^(-1)]`
C. `["MT"^(-2)"A"^(-1)]`
D. None of these

Explanation

To find the dimensions of the magnetic field \( B \) using the equation for magnetic force on a moving charge, we start with the equation:

\[ F = q(v \times B) \]

Where:
- \( F \) is the magnetic force,
- \( q \) is the electric charge,
- \( v \) is the velocity of the charge,
- \( B \) is the magnetic field.

### Step 1: Identify the dimensional formulas of the quantities involved

1. **Force \( F \)**:
The dimensional formula of force is given by:
\[ [F] = M^1 L^1 T^{-2} \]

2. **Charge \( q \)**:
The dimensional formula of electric charge is:
\[ [q] = A^1 T^1 \]

3. **Velocity \( v \)**:
The dimensional formula of velocity is:
\[ [v] = L^1 T^{-1} \]

### Step 2: Rearrange the equation to solve for \( B \)

From the equation \( F = q(v \times B) \), we can rearrange it to find \( B \):
\[ B = \frac{F}{qv} \]

### Step 3: Substitute the dimensional formulas into the equation

Now, substituting the dimensional formulas we found into the equation for \( B \):
\[ [B] = \frac{[F]}{[q][v]} = \frac{M^1 L^1 T^{-2}}{(A^1 T^1)(L^1 T^{-1})} \]

### Step 4: Simplify the expression

Now we simplify the right-hand side:
\[ [B] = \frac{M^1 L^1 T^{-2}}{A^1 L^1 T^1} \]

Canceling \( L^1 \) from the numerator and denominator:
\[ [B] = \frac{M^1 T^{-2}}{A^1 T^1} \]

Now, simplifying further:
\[ [B] = M^1 T^{-2 - 1} A^{-1} = M^1 T^{-3} A^{-1} \]

### Final Answer

Thus, the dimensional formula of the magnetic field \( B \) is:
\[ [B] = M^1 T^{-3} A^{-1} \]

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