Question

Select the dimensional formula of $${\displaystyle \frac{B^2}{2\mu }}$$
$$\begin{array}{rl}& A.{\text{M}}^1{\text{L}}^1{\text{T}}^2\\ & B.{\text{M}}^{-1}{\text{L}}^1{\text{T}}^2\\ & C.{\text{M}}^{-1}{\text{L}}^{-1}{\text{T}}^{-2}\\ & D.{\text{M}}^1{\text{L}}^{-1}{\text{T}}^{-2}\end{array}$$

Answer 1

Hint: The magnetic density is forced by the product of the charge and velocity. The permeability is Henry per meter. Using these equations, we will compute the dimensional formula of magnetic density and permeability separately. Then, finally, we will divide these units.
 Formula used:
$$\begin{array}{rl}& B={\displaystyle \frac{F}{qv}}\\ & \mu ={\displaystyle \frac{H}{m}}\end{array}$$

Complete step by step answer:
The magnetic density is force by the product of the charge and velocity. The mathematical representation of the same is given as follows.
$$B={\displaystyle \frac{F}{qv}}$$
The permeability is Henry per meter. The mathematical representation of the same is given as follows.
$$\mu ={\displaystyle \frac{H}{m}}$$
Using these equations, we will compute the dimensional formula of magnetic density and permeability separately.
Consider the formula for computing the magnetic density.
$$B={\displaystyle \frac{F}{qv}}$$
Substitute the dimensional formula of the parameters such as force, charge and velocity.
$$B={\displaystyle \frac{[ML{T}^{-2}]}{[AT][L{T}^{-1}]}}$$
Continue further computation.
$$B=[M^1{L}^0{T}^{-2}{A}^{-1}]$$
Consider the formula for computing the permeability.
$$\mu ={\displaystyle \frac{H}{m}}$$
Substitute the dimensional formula of the parameters such as force, charge and velocity.
$$\mu ={\displaystyle \frac{[ML{T}^{-2}]}{[A^2]}}$$
Continue further computation.
$$\mu =[ML{T}^{-2}{A}^{-2}]$$
Now, we will divide the dimensional formula of the magnetic density by the dimensional formula of the permeability.
Consider the formula.
$${\displaystyle \frac{B^2}{2\mu }}$$
Substitute the dimensional formula of the magnetic density and permeability.
$${\displaystyle \frac{B^2}{2\mu }}={\displaystyle \frac{{[M{L}^0{T}^{-2}{A}^{-1}]}^2}{[ML{T}^{-2}{A}^{-2}]}}$$
Continue further computation.
$$\begin{array}{rl}& {\displaystyle \frac{B^2}{2\mu }}={\displaystyle \frac{{[M{L}^0{T}^{-2}{A}^{-1}]}^2}{[ML{T}^{-2}{A}^{-2}]}}\\ & \Rightarrow {\displaystyle \frac{B^2}{2\mu }}={\displaystyle \frac{[M^2{L}^0{T}^{-4}{A}^{-2}]}{[ML{T}^{-2}{A}^{-2}]}}\\ & \therefore {\displaystyle \frac{B^2}{2\mu }}=[M{L}^{-1}{T}^{-2}]\end{array}$$
$$\therefore$$ The dimensional formula of $${\displaystyle \frac{B^2}{2\mu }}$$ is $$[M^1{L}^{-1}{T}^{-2}]$$

So, the correct answer is “Option D”.

Note: There are many methods of finding the dimensional formula of equations, such as, by using the direct units of the parameters used in the equations, by further substitution of parameters within the parameters. In this case, we have used both methods, the first method is used for permeability and the second method is used for magnetic density.