1. Write dimensions of electric flux

Electric flux is $\Phi_E = \vec E \cdot \vec A$ So, $[\Phi_E] = [E][L^2]$

Now electric field $E = \frac{F}{q}$ with $[F] = MLT^{-2}, \qquad [q] = AT$ Thus, $[E] = \frac{MLT^{-2}}{AT} = MLA^{-1}T^{-3}$

Therefore, $[\Phi_E] = MLA^{-1}T^{-3} \cdot L^2 = ML^3A^{-1}T^{-3}$

2. Write dimensions of magnetic flux

Magnetic flux is $\Phi_B = B \cdot A$ So, $[\Phi_B] = [B][L^2]$

Magnetic field can be obtained from $F = qvB$ Hence, $[B] = \frac{F}{qv}$ Now, $[v] = LT^{-1}$ So, $[B] = \frac{MLT^{-2}}{(AT)(LT^{-1})} = MT^{-2}A^{-1} \cdot T = MA^{-1}T^{-2}$ Actually simplifying carefully, $(AT)(LT^{-1}) = AL$ Thus, $[B] = \frac{MLT^{-2}}{AL} = MA^{-1}T^{-2}$

Therefore, $[\Phi_B] = MA^{-1}T^{-2} \cdot L^2 = ML^2A^{-1}T^{-2}$

3. Find ratio of electric flux to magnetic flux

$\frac{\Phi_E}{\Phi_B} = \frac{ML^3A^{-1}T^{-3}}{ML^2A^{-1}T^{-2}}$

Subtracting powers, $= L^{3-2}T^{-3-(-2)} = LT^{-1}$

So the required dimensions are $M^0L^1T^{-1}A^0$

Hence, $Q=1, \qquad R=-1$

4. Check options

• A: $(3,-5)$ — incorrect
• B: $(-2,1)$ — incorrect
• C: $(-2,2)$ — incorrect
• D: $(1,-1)$ — correct

5. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So the derived answer agrees with the stored answer.