JEE PYQ: Units & Measurements - Question ID bb705a2cbe99 (JEE Main 2025)

Core Formula & Concept:

In electromagnetism, two fundamental fluxes are defined:

• Electric flux $\Phi_E$ through a surface is given by $\Phi_E = \int \vec{E} \cdot d\vec{A}$, where $\vec{E}$ is the electric field and $d\vec{A}$ is the area vector. Its SI unit is volt-metre (V·m), which is equivalent to newton-metre-squared per coulomb (N·m²/C).
• Magnetic flux $\Phi_B$ through a surface is given by $\Phi_B = \int \vec{B} \cdot d\vec{A}$, where $\vec{B}$ is the magnetic field. Its SI unit is weber (Wb), which is equivalent to tesla-metre-squared (T·m²).

The question asks for the dimensions of the ratio $\displaystyle \frac{\Phi_E}{\Phi_B}$. To find these dimensions, we express each flux in terms of the fundamental dimensions: mass ($M$), length ($L$), time ($T$), and electric current ($A$).

Step-by-Step Derivation:

Step 1: Express electric flux $\Phi_E$ in fundamental dimensions.

From $\Phi_E = \int \vec{E} \cdot d\vec{A}$, we know:

• Electric field $\vec{E}$ has dimensions of force per unit charge, i.e., $[E] = \frac{[F]}{[Q]} = \frac{MLT^{-2}}{AT} = MLT^{-3}A^{-1}$.
• Area $d\vec{A}$ has dimension $L^2$.

Therefore, the dimension of electric flux is $[\Phi_E] = [E] \cdot [A] = (MLT^{-3}A^{-1}) \cdot L^2 = ML^3T^{-3}A^{-1}$.

Step 2: Express magnetic flux $\Phi_B$ in fundamental dimensions.

From $\Phi_B = \int \vec{B} \cdot d\vec{A}$, we know:

• Magnetic field $\vec{B}$ has dimensions of force per unit current per unit length, i.e., $[B] = \frac{[F]}{[I][L]} = \frac{MLT^{-2}}{A \cdot L} = MT^{-2}A^{-1}$.
• Area $d\vec{A}$ has dimension $L^2$.

Therefore, the dimension of magnetic flux is $[\Phi_B] = [B] \cdot [A] = (MT^{-2}A^{-1}) \cdot L^2 = ML^2T^{-2}A^{-1}$.

Step 3: Compute the dimension of the ratio $\frac{\Phi_E}{\Phi_B}$.

Using the dimensions obtained:

$\left[\frac{\Phi_E}{\Phi_B}\right] = \frac{[\Phi_E]}{[\Phi_B]} = \frac{ML^3T^{-3}A^{-1}}{ML^2T^{-2}A^{-1}} = L^{3-2} T^{-3 - (-2)} A^{-1 - (-1)} = L^1 T^{-1} A^0.$

Simplifying, we get:

$\left[\frac{\Phi_E}{\Phi_B}\right] = M^0 L^1 T^{-1} A^0.$

Comparing with the given form $M^P L^Q T^R A^S$, we identify:

• $P = 0$,
• $Q = 1$,
• $R = -1$,
• $S = 0$.

The question specifically asks for the values of $Q$ and $R$, which are $1$ and $-1$ respectively.

Step 4: Match with the given options.

The pair $(Q, R) = (1, -1)$ corresponds to option D.

Common Traps & Exam Tip:

Students often confuse the dimensions of electric and magnetic fields, leading to incorrect exponents. A frequent mistake is misapplying the relation between force, charge, and current, especially in the expression for $\vec{B}$. Another common error is overlooking the cancellation of the current dimension $A$ in the ratio, which results in $S = 0$. Always double-check the dimensional consistency of each step to avoid such pitfalls.

Final Answer: Option D is correct.