"If B is magnetic field and $\mu_0$ is permeability of free space, then the dimensions of (B/$\mu_0$) is"" (1) MT –2 A –1 (2) L –1 A (3) LT –2 A –1 (4) ML 2 T –2 A –1"

1. Magnetic field from the Lorentz force

For a charge $q$ moving with velocity $v$ through a magnetic field $B$, the magnitude of magnetic force is

$F = q\,v\,B\sin\theta$

Ignoring $\sin\theta$ (dimensionless),

$B = \dfrac{F}{q\,v}$

• Force has dimensions $[M\,L\,T^{-2}]$.
• Charge has dimensions $[I\,T]$ (current × time).
• Velocity has dimensions $[L\,T^{-1}]$.

So

$[B] = \dfrac{M\,L\,T^{-2}}{(I\,T)(L\,T^{-1})} = M\,T^{-2}\,I^{-1}$

2. Permeability of free space $\mu_0$

From the constitutive relation in vacuum

$B = \mu_0\,H$

and the fact that magnetic field intensity $H$ has dimensions of current per length,

$[H] = I\,L^{-1}$

Therefore

$[\mu_0] = \dfrac{[B]}{[H]} = \dfrac{M\,T^{-2}\,I^{-1}}{I\,L^{-1}} = M\,L\,T^{-2}\,I^{-2}$

3. Putting it together: $\dfrac{B}{\mu_0}$

$\left[\dfrac{B}{\mu_0}\right] = \dfrac{M\,T^{-2}\,I^{-1}}{M\,L\,T^{-2}\,I^{-2}} = L^{-1}\,I^{+1}$

So the dimensional formula is $L^{-1} A$ → Option (2).

What is the question?

We want to know what basic "building-blocks" (mass, length, time, current) make up the quantity $\dfrac{B}{\mu_0}$.

Baby-step idea 🧒🏻

1. Think of a magnet pulling on a moving electric charge. The pull (force) depends on the magnetic field $B$.
2. The famous rule is
   $F = q \; v \; B$
   (force = charge × speed × magnetic field).
3. By rearranging we can describe $B$ in terms of the already-known force, charge and speed.
4. Then we use another rule $B = \mu_0\,H$ to express $\mu_0$.
5. Finally we divide the two answers and see which basic units remain.

After the dust settles, only length in the denominator and current in the numerator survive, giving the answer $L^{-1}A$ (option 2).

Step-by-Step Calculation

1. Start with the Lorentz force law $F = q\,v\,B$ [F] = $M L T^{-2}$ [q] = $I T$ [v] = $L T^{-1}$

2. Isolate $B$ and write its dimensions

   $B = \dfrac{F}{q v} \quad\Longrightarrow\quad [B] = \dfrac{M L T^{-2}}{(I T)(L T^{-1})} = M\,T^{-2}\,I^{-1}$

3. Use the relation $B = \mu_0 H$ [H] = $I L^{-1}$

   $[\mu_0] = \dfrac{[B]}{[H]} = \dfrac{M T^{-2} I^{-1}}{I L^{-1}} = M L T^{-2} I^{-2}$

4. Compute $[B/\mu_0]$

   $\left[\dfrac{B}{\mu_0}\right] = \dfrac{M T^{-2} I^{-1}}{M L T^{-2} I^{-2}} = L^{-1} I^{+1}$

5. Write in standard dimensional form

   $[B/\mu_0] = L^{-1} A$

6. Match with the given options

   Option (2): $L^{-1} A$ ✅

Final Answer: Option (2)