Question

The magnetic induction field has the dimensions of

• A. Force
• B. Force constant
• C. Surface tension
• D. $\frac{\text{Surface tension}}{\text{Current}}$
• E. Force constant $\times$ current

Hint:

Always reduce complex physical quantities to basic Force or Energy relations to find their dimensions quickly. For $B$, remember $F = BIl$ is also a very helpful shortcut.

Answer 1

The Correct Option is D

Concept:
To find the dimensions of magnetic induction ($B$), we use the Lorentz force formula: $F = qvB \sin\theta$. [itemsep=8pt]
• Dimensions of Force ($F$): $[MLT^{-2}]$
• Dimensions of Charge ($q = It$): $[AT]$
• Dimensions of Velocity ($v$): $[LT^{-1}]$

Step 1: Derive dimensions of Magnetic Induction ($B$).
From $B = \frac{F}{qv}$: \[ [B] = \frac{[MLT^{-2}]}{[AT][LT^{-1}]} = [MT^{-2}A^{-1}] \]

Step 2: Analyze the given options.
[label=\alph*), itemsep=6pt]
• Surface Tension: $\frac{\text{Force}}{\text{Length}} = \frac{[MLT^{-2}]}{[L]} = [MT^{-2}]$
• Current: $[A]$
• Ratio (Option D): $\frac{\text{Surface Tension}}{\text{Current}} = \frac{[MT^{-2}]}{[A]} = [MT^{-2}A^{-1}]$ This matches the dimensions of $B$.