Animated Solution for Physics - Physics and Measurement: The dimensions of magnetic field in $M,L,T$ and $C$ (coulomb) is given as

When faced with finding the dimensions of a complex physical quantity like the magnetic field ($B$), the most reliable strategy is to anchor yourself to a fundamental formula. Dimensional analysis is like balancing a cosmic scale—whatever physical dimensions exist on the left side of an equation must perfectly match the dimensions on the right side.

To find the dimensions of $B$, we need an equation where $B$ is a key player, and where we already know the dimensions of all the other characters.

Lorentz Force

The simplest and most elegant formula connecting the magnetic field to fundamental quantities is the Lorentz force law. Imagine a charged particle moving through space. If it enters a magnetic field, it experiences a force. For a charge $q$ moving with velocity $v$ perpendicular to a magnetic field $B$, the magnitude of the force is given by:

This equation is our golden key. We know the dimensions of force, charge, and velocity. By substituting these knowns, we can easily unlock the unknown dimension of $B$.

Let's list the dimensional formulas for the quantities we know.

1. Force ($F$): From Newton's second law ($F=ma$), force is mass times acceleration. Therefore, its dimensional formula is:

2. Charge ($q$): The problem specifically asks us to express the answer using $C$ for coulombs (charge). So, we don't need to break it down into current and time. We simply write:

3. Velocity ($v$): Velocity is displacement over time, giving us:

Now, we substitute these dimensional formulas back into our master equation:

To isolate $[B]$, we divide the force dimension by the dimensions of charge and velocity:

Now comes the satisfying part—cancellation and simplification. - The $L$ in the numerator and the $L$ in the denominator perfectly cancel each other out ($L^{1-1}=L^0$). - For time ($T$), we have $T^{-2}$ in the numerator and $T^{-1}$ in the denominator. Using exponent rules, $T^{-2-(-1)}=T^{-1}$. - The charge ($C$) moves from the denominator to the numerator, becoming $C^{-1}$.

Putting it all together, we arrive at our final answer:

This matches option (c).

Could we have used a different formula? Absolutely! You could use the formula for the magnetic force on a current-carrying wire: $F=BIl$.

In this case, you would need to remember that current ($I$) is the rate of flow of charge, so its dimension would be $[C{T}^{-1}]$. Substituting $[F]=[ML{T}^{-2}]$, $[I]=[C{T}^{-1}]$, and $[l]=[L]$ into $B=\frac{F}{Il}$ will lead you to the exact same result: $[M{T}^{-1}{C}^{-1}]$. It is always empowering to have multiple mathematical tools at your disposal!