Question

The dimensions of magnetic flux are identical to the dimensions of:

• A. EMF $\times$ Time
• B. Electric field $\times$ Velocity
• C. Force/Current
• D. Magnetic field $\times$ Current

Hint:

Whenever an exam problem asks for dimensional matches, always look for a direct historical law equation before breaking terms down into basic $[M][L][T][A]$ base dimensions. Equations like $e = \frac{\Delta \phi}{\Delta t}$ (Faraday's Law) or $V = L\frac{di}{dt}$ let you deduce correct unit combinations in under ten seconds!

Answer 1

The Correct Option is A

Concept: Dimensional analysis can be approached either by calculating the fundamental SI base unit dimensions ($M, L, T, A$) for each term or, more rapidly, by using fundamental governing physical equations. According to Faraday's Law of Electromagnetic Induction, the induced electromotive force ($\text{EMF}$ or $e$) in a circuit is directly proportional to the time rate of change of magnetic flux ($\phi_B$) linking with it: \[ e = -\frac{d\phi_B}{dt} \] By rearranging this formula, we can establish a direct relationship between the dimensional units of magnetic flux, electromotive force, and time.

Step 1: Relating magnetic flux to EMF and time via Faraday's Law.
From the differential relation of Faraday's Law, ignoring the negative sign (which represents direction due to Lenz's law), we write the relationship in terms of dimensional quantities: \[ [\text{EMF}] = \frac{[\text{Magnetic Flux}]}{[\text{Time}]} \] Rearranging the equation to isolate magnetic flux: \[ [\text{Magnetic Flux}] = [\text{EMF}] \times [\text{Time}] \] This directly matches the combination stated in Option (A).

Step 2: Verifying through fundamental dimensional formulas (Optional confirmation).
Let's find the dimensional formula for both sides to verify:
• Magnetic Flux ($\phi_B = B \cdot A$): Since magnetic force is $F = qvB \implies B = \frac{F}{qv}$, its dimensions are $\frac{[MLT^{-2}]}{[AT][LT^{-1}]} = [MT^{-2}A^{-1}]$. Multiplying by area ($A = [L^2]$): \[ [\phi_B] = [ML^2T^{-2}A^{-1}] \]
• EMF $\times$ Time: $\text{EMF}$ is work done per unit charge, so its dimensions are $\frac{[ML^2T^{-2}]}{[AT]} = [ML^2T^{-3}A^{-1}]$. Multiplying this by time ($[T]$): \[ [\text{EMF} \times \text{Time}] = [ML^2T^{-3}A^{-1}] \times [T] = [ML^2T^{-2}A^{-1}] \] Both methods confirm that the dimensions are completely identical.