The dimensional formula of magnetic induction B is

Options

A. `[M^0ALT^0]`
B. `[M^0AL^(-1)T^0]`
C. `[M^0AL^2T^0]`
D. `[M^0A^(-1)T^(-2)]`

Answer

To find the dimensional formula of magnetic induction (B), we can start from the Lorentz force equation, which states that the force (F) experienced by a charge (Q) moving with a velocity (V) in a magnetic field is given by:

\[ F = Q \cdot V \cdot B \]

From this equation, we can rearrange it to express B:

\[ B = \frac{F}{Q \cdot V} \]

Now, we need to find the dimensions of each of the quantities involved in this equation.

1. **Force (F)**:
The dimensional formula for force is given by Newton's second law, which states that force is mass times acceleration. Therefore, we can express the dimensional formula of force as:
\[ [F] = [M][L][T^{-2}] \]

2. **Charge (Q)**:
The charge can be expressed in terms of current (I) and time (T). The dimensional formula for charge is:
\[ [Q] = [I][T] \]
where [I] is the dimension of electric current.

3. **Velocity (V)**:
Velocity is defined as displacement over time. Thus, its dimensional formula is:
\[ [V] = [L][T^{-1}] \]

Now we can substitute these dimensional formulas back into the equation for B:

\[ [B] = \frac{[F]}{[Q][V]} = \frac{[M][L][T^{-2}]}{([I][T])([L][T^{-1}])} \]

Now, simplifying the denominator:

\[ [B] = \frac{[M][L][T^{-2}]}{[I][T][L][T^{-1}]} = \frac{[M][L][T^{-2}]}{[I][L][T]} = \frac{[M][L][T^{-2}]}{[I][L][T]} = \frac{[M]}{[I][T]}[T^{-1}] \]

Now, cancelling out [L] in the numerator and denominator:

\[ [B] = [M][I^{-1}][T^{-1}] \]

Thus, the dimensional formula for magnetic induction (B) is:

\[ [B] = [M^1][I^{-1}][T^{-1}] \]

### Final Answer:
The dimensional formula of magnetic induction \( B \) is \( [M^1 I^{-1} T^{-1}] \).