JEE PYQ: Units & Measurements - Question ID 6dbc1fe748e1 (JEE Main 2026)

Core Formula & Concept:

In electromagnetism, the relationship between electric and magnetic fields can be derived from Maxwell’s equations. One of the key results is that a time-varying electric field generates a magnetic field. The relevant physical law here is the Ampère-Maxwell Law, which in differential form is:

$\nabla \times \vec{H} = \vec{J} + \frac{\partial \vec{D}}{\partial t}$

where:

• $\vec{H}$ is the magnetic field (A/m),
• $\vec{J}$ is the current density (A/m²),
• $\vec{D}$ is the electric displacement field (C/m²), related to the electric field $\vec{E}$ by $\vec{D} = \epsilon \vec{E}$, where $\epsilon$ is the permittivity of the medium (F/m).

In free space (where $\vec{J} = 0$), this simplifies to: $\nabla \times \vec{H} = \epsilon \frac{\partial \vec{E}}{\partial t}$

To find the dimensions of $\vec{H}$ in terms of $\epsilon$, $\vec{E}$, $x$, and $t$, we use dimensional analysis. The goal is to express $H$ in the form: $H = \frac{x^p \epsilon^q E^r}{t^s}$ and determine the exponents $p, q, r, s$ such that the dimensions on both sides match.

Step-by-Step Derivation:

Step 1: Write down the dimensions of each quantity.

We use the following dimensional formulas (in terms of mass $M$, length $L$, time $T$, and current $I$):

• Magnetic field $H$: From $\nabla \times \vec{H} = \vec{J}$, we know $\nabla \times \vec{H}$ has dimensions of current density, $[J] = I L^{-2}$. Since $\nabla$ has dimension $L^{-1}$, we get: $[H] = I L^{-1}$
• Electric field $E$: From $\vec{F} = q \vec{E}$, force $F = M L T^{-2}$, and charge $q = I T$, so: $[E] = \frac{[F]}{[q]} = \frac{M L T^{-2}}{I T} = M L T^{-3} I^{-1}$
• Permittivity $\epsilon$: From $\vec{D} = \epsilon \vec{E}$, and $[D] = I T L^{-2}$ (since $D$ is charge per unit area), so: $[\epsilon] = \frac{[D]}{[E]} = \frac{I T L^{-2}}{M L T^{-3} I^{-1}} = M^{-1} L^{-3} T^4 I^2$
• Distance $x$: $[x] = L$
• Time $t$: $[t] = T$

Step 2: Substitute dimensions into the given equation.

The equation is: $H = \frac{x^p \epsilon^q E^r}{t^s}$ Substitute the dimensions: $I L^{-1} = \frac{L^p \cdot (M^{-1} L^{-3} T^4 I^2)^q \cdot (M L T^{-3} I^{-1})^r}{T^s}$

Step 3: Expand the right-hand side.

Multiply the dimensions: $= L^p \cdot M^{-q} L^{-3q} T^{4q} I^{2q} \cdot M^r L^r T^{-3r} I^{-r} \cdot T^{-s}$ Combine like terms:

• Mass: $M^{-q + r}$
• Length: $L^{p - 3q + r}$
• Time: $T^{4q - 3r - s}$
• Current: $I^{2q - r}$

So, the right-hand side becomes: $M^{-q + r} L^{p - 3q + r} T^{4q - 3r - s} I^{2q - r}$

Step 4: Equate dimensions on both sides.

Left-hand side: $I L^{-1} = M^0 L^{-1} T^0 I^1$
Right-hand side: $M^{-q + r} L^{p - 3q + r} T^{4q - 3r - s} I^{2q - r}$

Equate the exponents of $M, L, T, I$:

1. $M$: $-q + r = 0$ → $r = q$
2. $L$: $p - 3q + r = -1$
3. $T$: $4q - 3r - s = 0$
4. $I$: $2q - r = 1$

Step 5: Solve the system of equations.

From equation (1): $r = q$
Substitute $r = q$ into equation (4): $2q - q = 1 → q = 1$ Then $r = q = 1$
Substitute $q = 1$ and $r = 1$ into equation (2): $p - 3(1) + 1 = -1 → p - 2 = -1 → p = 1$ Substitute $q = 1$, $r = 1$ into equation (3): $4(1) - 3(1) - s = 0 → 4 - 3 - s = 0 → s = 1$

Step 6: Write the final exponents.

We find: $p = 1, \quad q = 1, \quad r = 1, \quad s = 1$ Thus, the correct option is A: $1, 1, 1, 1$.

Common Traps & Exam Tip:

Students often make the following mistakes:

• Incorrect dimensional formulas: Confusing the dimensions of $H$, $E$, and $\epsilon$. For example, some assume $[H] = M T^{-2} I^{-1}$ (like magnetic flux density $B$), but $H$ has units A/m, so $[H] = I L^{-1}$.
• Sign errors in exponents: Forgetting that $\epsilon$ has negative mass and length exponents, leading to incorrect balancing.
• Overcomplicating the problem: Trying to use vector calculus or integral forms instead of simple dimensional analysis. This question only requires dimensional consistency.
• Ignoring current dimension: Many forget that current $I$ is a fundamental dimension and must be balanced, leading to wrong values for $q$ and $r$.

Exam Tip: Always write down the dimensions of each quantity clearly before substituting. Double-check the signs and powers, especially for derived quantities like $\epsilon$. In dimensional analysis, consistency across all fundamental dimensions (M, L, T, I) is crucial.