Question

The dimensional formula of magnetic field B is:

• A. $[M^1 L^0 T^{-2} A^{-1}]$
• B. $[M^1 L^2 T^{-2} A^{-1}]$
• C. $[M^1 T^{-2} A^{-1}]$
• D. $[M^1 L^0 T^{-1} A^{-1}]$

Hint:

Alternatively, use the formula $F = I L B \sin\theta$.
Then, $[B] = [F] / ([I][L]) = [M L T^{-2}] / ([A][L]) = [M T^{-2} A^{-1}]$, which is equivalent to $[M^1 L^0 T^{-2} A^{-1}]$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the dimensional formula of the magnetic field vector ($B$).
This is solved by using a standard physical equation that relates the magnetic field to other quantities with known dimensions.

Step 2: Key Formula or Approach:
We use the magnetic force equation acting on a moving charge (Lorentz Force):
\[ F = q v B \sin\theta \]
Rearranging this to solve for $B$:
\[ B = \frac{F}{q v \sin\theta} \]
We substitute the fundamental dimensions of Force ($F$), Charge ($q$), and Velocity ($v$).

Step 3: Detailed Explanation:

• Let us list the dimensional formulas of the individual variables:
1. Force ($F$):
\[ [F] = [M^1 L^1 T^{-2}] \]
2. Charge ($q$): Since current $I = q/t$, we have $q = I \cdot t$.
\[ [q] = [A^1 T^1] \]
3. Velocity ($v$):
\[ [v] = [L^1 T^{-1}] \]
4. The term $\sin\theta$ is a dimensionless ratio:
\[ [\sin\theta] = [M^0 L^0 T^0] \]

• Now, substitute these dimensions into the expression for $B$:
\[ [B] = \frac{[M^1 L^1 T^{-2}]}{[A^1 T^1] \cdot [L^1 T^{-1}]} \]

• Simplify the denominator first:
\[ [A^1 T^1] \cdot [L^1 T^{-1}] = [A^1 L^1 T^0] \]

• Now divide the numerator by the simplified denominator:
\[ [B] = \frac{[M^1 L^1 T^{-2}]}{[A^1 L^1 T^0]} = [M^1 L^0 T^{-2} A^{-1}] \]

• Therefore, the dimensional formula for the magnetic field is $[M^1 L^0 T^{-2} A^{-1}]$.


Step 4: Final Answer:
The correct dimensional representation of the magnetic field $B$ is $[M^1 L^0 T^{-2} A^{-1}]$.