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: Start from the force on a moving charge in a magnetic field.
\[ F = qvB \Rightarrow B = \frac{F}{qv} \]
Step 2: Write the dimensions of each quantity.
Force has dimensions \( [M L T^{-2}] \), charge has dimensions \( [A T] \) (current times time), and velocity has dimensions \( [L T^{-1}] \).
Step 3: Substitute and simplify.
\[ [B] = \frac{[M L T^{-2}]}{[A T][L T^{-1}]} = \frac{[M L T^{-2}]}{[A L T^{0}]} = [M L^0 T^{-2} A^{-1}] \]
The length dimension cancels out, leaving mass, inverse time squared, and inverse current.
\[ \boxed{[M^1 L^0 T^{-2} A^{-1}]} \]