Animated Solution for Physics - Physics and Measurement: The dimension of $\frac{B^2}{2{\mu }_0}$, where $B$ is magnetic field and ${\mu }_0$ is the magnetic permeability of vacuum, is

When you first look at the expression $\frac{B^2}{2{\mu }_0}$, your immediate instinct as a physics student might be to roll up your sleeves and start deriving the dimensional formulas for each individual term. This is a very natural reaction, especially when you have spent hours practicing dimensional analysis by breaking down complex formulas into their fundamental base quantities.

You might recall that the magnetic field $B$ can be derived from the Lorentz force equation, which is given by $F=qvB$. By rearranging this equation to solve for $B$, you get $B=\frac{F}{qv}$. Substituting the known dimensions for force ($[ML{T}^{-2}]$), charge ($[AT]$), and velocity ($[L{T}^{-1}]$), you would eventually arrive at the dimensional formula for the magnetic field, which is $[M{T}^{-2}{A}^{-1}]$.

Next, you would have to tackle the magnetic permeability of free space, ${\mu }_0$. You might use Ampere's Law or the formula for the magnetic field at the center of a current-carrying circular loop, $B=\frac{{\mu }_{0}I}{2r}$. Rearranging for ${\mu }_0$ gives ${\mu }_0=\frac{2rB}{I}$. Substituting the dimensions of radius ($[L]$), magnetic field ($[M{T}^{-2}{A}^{-1}]$), and current ($[A]$), you would find that the dimensional formula for ${\mu }_0$ is $[ML{T}^{-2}{A}^{-2}]$.

Finally, you would take these two derived dimensional formulas, square the dimension of $B$, divide it by the dimension of ${\mu }_0$, and carefully cancel out the mass, length, time, and ampere terms to reach your final answer.

While this brute-force method is mathematically sound and will absolutely lead you to the correct answer, it is a massive time sink. In highly competitive and time-constrained exams like JEE Main and JEE Advanced, time is your most precious resource. Falling into this trap not only eats up valuable minutes that you could spend on much harder, high-weightage problems, but it also opens the door wide open to silly algebraic mistakes. A single misplaced negative sign in your exponents could ruin the entire calculation.

This is exactly where a deep, conceptual understanding of physics becomes your greatest weapon. The examiners who design these questions don't actually want you to do all that tedious, error-prone algebra. Instead, they are testing your ability to recognize standard physical quantities that are hidden in plain sight.

If you have thoroughly studied the chapter on Electromagnetic Induction and understand the fundamental properties of magnetic fields, you should immediately recognize the expression $\frac{B^2}{2{\mu }_0}$.

But what exactly does that mean? Whenever a magnetic field is established in a region of space—whether it is inside a current-carrying solenoid, around a permanent bar magnet, or propagating through the vacuum of space as part of an electromagnetic wave—it stores energy. The magnetic field itself is a reservoir of energy.

The amount of this magnetic energy stored per unit volume of space is precisely what the expression $\frac{B^2}{2{\mu }_0}$ calculates. By recognizing this profound physical truth, we completely bypass the need to find the individual dimensions of the magnetic field $B$ and the permeability ${\mu }_0$. We have effectively transformed a complex, multi-step derivation into a trivial, one-step division!

Since we now confidently know that the expression $\frac{B^2}{2{\mu }_0}$ is simply the magnetic energy density, we can write its dimensional formula as the dimension of energy divided by the dimension of volume. This is a much safer and faster path to the solution.

Let's break this down into its fundamental components.

The dimensional formula for energy is one of the most common, fundamental, and important formulas in all of physics. Since energy is dimensionally equivalent to work done, and work is defined as force multiplied by displacement, we can easily construct its dimensions. Force is mass times acceleration ($[ML{T}^{-2}]$), and displacement is length ($[L]$). Therefore, the dimensional formula for energy is $[M{L}^2{T}^{-2}]$.

Next, we consider the denominator of our conceptual fraction: volume. Volume is simply a three-dimensional measure of space. For any regular geometric shape, it is calculated as length multiplied by width multiplied by height. Therefore, its dimensional formula is simply $[L^3]$.

Now, we substitute these fundamental dimensions into our energy density relation:

When we divide these terms, the mass $M$ and time $T^{-2}$ components remain completely unchanged, as there are no corresponding terms in the denominator to interact with them.

For the length dimension, we use the fundamental laws of exponents. We have $L^2$ in the numerator and $L^3$ in the denominator. Dividing them means we subtract the exponent of the denominator from the exponent of the numerator: $L^{2-3}$, which simplifies perfectly to $L^{-1}$.

And just like that, in a matter of seconds and with virtually zero risk of algebraic error, we have arrived at the correct answer!

This problem serves as a beautiful and powerful reminder of why physics is not just about memorizing a long list of disconnected formulas. It is about deeply understanding what those formulas represent in the real, physical world.

When you see a combination of constants and variables, ask yourself: "Does this represent a fundamental property of nature?" More often than not, in exams like the JEE, it does.

As a bonus tip for your future studies, remember that this concept applies symmetrically to electric fields as well. The electric energy density—the energy stored per unit volume in an electric field—is given by the expression $\frac{1}{2}{ϵ}_0{E}^2$.

Because it is also an energy density, representing the exact same physical concept of energy per unit volume, it shares the exact same dimensional formula: $[M{L}^{-1}{T}^{-2}]$.

Mastering these conceptual shortcuts and recognizing the underlying symmetry of nature will not only drastically boost your problem-solving speed but also deepen your overall appreciation for the elegant and interconnected structure of physics!