Animated Solution for Physics - Physics and Measurement: In the formula $X=3Y{Z}^2$, $X$ and $Z$ have dimensions of capacitance and magnetic induction, respectively. What are the dimensions of $Y$ in MKSQ system?

Imagine you are an explorer in the vast universe of physics, and dimensional analysis is your trusty compass. It allows you to break down the most complex physical quantities into their fundamental building blocks. In this problem, we are tasked with finding the dimensions of an unknown quantity $Y$ from a given equation. Let's embark on this dimensional journey!

We are given the formula:

Our goal is to find the dimensions of $Y$. To do this, we first need to isolate $Y$. Rearranging the equation gives us:

In the realm of dimensional analysis, pure numbers like $3$ are completely dimensionless. They don't carry any physical units, so we can safely ignore them when equating dimensions. Therefore, the dimensional equation becomes:

A Quick Refresher Before we dive into the calculations, let's address a crucial detail: the MKSQ system. You might be more familiar with the standard SI system, where the fundamental units are Mass ($M$), Length ($L$), Time ($T$), and Current ($A$). However, in the MKSQ system, Charge ($Q$) is treated as a fundamental quantity instead of Current. This means we don't need to convert charge into current and time ($Q=AT$); we simply leave it as $Q$.

Let's derive its dimensions from scratch to avoid any memorization errors. We know that capacitance $C$ is defined as the charge stored per unit voltage:

But what is voltage ($V$)? Voltage, or electric potential, is the work done per unit charge:

Substituting this back into our capacitance formula, we get:

Now, we substitute the well-known dimensions of work ($W=\text{Force}\times \text{Displacement}=[M{L}^2{T}^{-2}]$):

Bringing the terms from the denominator to the numerator, we find the dimensions of $X$:

A reliable way to find this is by using the Lorentz force formula for a moving charge:

Rearranging for $B$, we get:

Let's plug in the fundamental dimensions for force ($[ML{T}^{-2}]$), charge ($[Q]$), and velocity ($[L{T}^{-1}]$):

Notice how the length dimension ($L$) beautifully cancels out:

Bringing It All Together Now that we have the dimensions of both $X$ and $Z$, we can substitute them back into our rearranged master equation:

First, let's square the dimensions of $Z$:

Now, substitute everything into the equation for $[Y]$:

When we bring the terms from the denominator up to the numerator, their powers change sign:

Now, we simply add the exponents for each fundamental dimension: - For $M$: $-1+(-2)=-3$ - For $L$: $-2$ (remains unchanged) - For $T$: $2+2=4$ - For $Q$: $2+2=4$

Putting it all together, we arrive at our final, elegant result:

$$[Y]=[M^{-3}{L}^{-2}{T}^4{Q}^4]$$

This perfectly matches option (b). By breaking down complex quantities into their fundamental definitions, we've successfully navigated through the dimensional maze!