JEE PYQ: Units & Measurements - Question ID 8380d1509075 (JEE Main 2022)

Core Formula & Concept:

Mutual inductance ($M$) is a measure of the ability of one coil to induce an electromotive force (emf) in another coil due to a changing current. The fundamental physics governing mutual inductance is Faraday’s Law of Induction, which states that the induced emf ($\mathcal{E}$) in a coil is proportional to the rate of change of magnetic flux ($\Phi_B$) through it:

$\mathcal{E} = -N \frac{d\Phi_B}{dt}$ where $N$ is the number of turns in the coil. For two coils, the induced emf in the second coil ($\mathcal{E}_2$) due to a changing current ($I_1$) in the first coil is given by: $\mathcal{E}_2 = -M \frac{dI_1}{dt}$ Here, $M$ is the mutual inductance between the two coils. The negative sign indicates the direction of the induced emf (Lenz’s Law), but for dimensional analysis, we focus on the magnitude. The key formula for mutual inductance can also be expressed in terms of magnetic flux linkage: $M = \frac{N_2 \Phi_{21}}{I_1}$ where: - $N_2$ is the number of turns in the second coil, - $\Phi_{21}$ is the magnetic flux through the second coil due to the current in the first coil, - $I_1$ is the current in the first coil.

Step-by-Step Derivation:

To find the dimensions of mutual inductance, we analyze the defining equation: $\mathcal{E}_2 = -M \frac{dI_1}{dt}$ Rearranging for $M$: $M = \frac{\mathcal{E}_2}{dI_1/dt}$ Now, let’s determine the dimensions of each term:

1. Dimensions of emf ($\mathcal{E}$): Emf is work done per unit charge. Work has dimensions of energy, which is $[ML^2T^{-2}]$, and charge ($Q$) has dimensions of $[AT]$ (since $I = Q/t$). Thus: $[\mathcal{E}] = \frac{[ML^2T^{-2}]}{[AT]} = [ML^2T^{-3}A^{-1}]$
2. Dimensions of $dI/dt$: Current ($I$) has dimensions of $[A]$, and time ($t$) has dimensions of $[T]$. Thus: $\left[\frac{dI}{dt}\right] = \frac{[A]}{[T]} = [AT^{-1}]$
3. Dimensions of $M$: Substitute the dimensions of $\mathcal{E}$ and $dI/dt$ into the equation for $M$: $[M] = \frac{[\mathcal{E}]}{[dI/dt]} = \frac{[ML^2T^{-3}A^{-1}]}{[AT^{-1}]} = [ML^2T^{-2}A^{-2}]$

Thus, the dimension of mutual inductance is $[ML^2T^{-2}A^{-2}]$, which corresponds to Option C.

Common Traps & Exam Tip:

Students often make the following mistakes in this question:

1. Confusing mutual inductance with self-inductance: The dimension of self-inductance ($L$) is the same as mutual inductance ($M$), but students sometimes incorrectly associate it with resistance or capacitance dimensions. Remember that both $L$ and $M$ have the same dimensions.
2. Incorrectly handling the current term: Some students forget that $dI/dt$ has dimensions of $[AT^{-1}]$ and mistakenly treat it as $[A]$, leading to an incorrect dimension for $M$. Always verify the dimensions of derivatives (e.g., $dI/dt$ is current per unit time).
3. Ignoring the negative sign in Faraday’s Law: The negative sign in $\mathcal{E} = -M dI/dt$ is for direction (Lenz’s Law) and does not affect the dimensions. Students sometimes overcomplicate the derivation by including it unnecessarily.
4. Miscounting the exponent of $A$: A common error is to write $[ML^2T^{-2}A^{-1}]$ (Option A) instead of $[ML^2T^{-2}A^{-2}]$ (Option C). This happens when students forget that the denominator in $M = \mathcal{E} / (dI/dt)$ introduces an additional $[A^{-1}]$ term.

Exam Tip: Always cross-verify dimensions using an alternative formula. For example, using $M = N_2 \Phi_{21} / I_1$: - Magnetic flux ($\Phi$) has dimensions of $[ML^2T^{-2}A^{-1}]$ (since $\Phi = BA$ and $B = F/(IL)$). - Thus, $[M] = \frac{[N_2][\Phi_{21}]}{[I_1]} = \frac{[1][ML^2T^{-2}A^{-1}]}{[A]} = [ML^2T^{-2}A^{-2}]$. This confirms the result.