Question

The quantities $x=\frac{1}{\sqrt{{\mu }_{\circ }{\epsilon }_{\circ }}},y=\frac{E}{B}$ and $z=\frac{L}{CR}$are defined where $C$-capacitance, $R$- Resistance,$L$ - length, $E$- Electric field, $B$ - magnetic field and ${\epsilon }_0,{\mu }_0$ - free space permittivity and permeability respectively. Then:

• A. Only $y$ and $z$ have the same dimension
• B. $x,y$ and $z$ have the same dimension
• C. Only $x$ and $y$ have the same dimension
• D. Only $x$ and $z$ have the same dimension

Answer 1

The correct option is B

$x,y$ and $z$ have the same dimension

Step 1: Given data:

The given physical quantities-

$$\begin{gathered} x=\frac{1}{\sqrt{{\mu }_{\circ }{\epsilon }_{\circ }}} \\ y=\frac{E}{B} \end{gathered}$$

$z=\frac{L}{CR}$

where $C$-capacitance, $R$- Resistance,$L$ - length, $E$- Electric field, $B$ - magnetic field and ${\epsilon }_0,{\mu }_0$ - free space permittivity and permeability respectively.

Step 2: Formula used:

We know that the speed of wave is given by-

$c=\frac{1}{\sqrt{{\mu }_o{\epsilon }_o}}$ where $c$ is speed of light

Step 3: Used dimensions

Length's dimension$\left(L\right)=\left[L\right]$

Dimension of speed using formula $speed=\frac{dis\tan ce}{time}$, $\left(v\right)=\left[L{T}^{-1}\right]........\left(1\right)$

Dimension of capacitance is calculated using the formula, $C=\frac{q}{V}$ Where, $q$is charge, $V$ is voltage

The dimension of charge $q=I\times t=\left[I^1{T}^1\right]$

The dimension of voltage is calculated using the formula, $V=electricfield\times d$

The dimension of electric field is calculated using the formula, $E=\frac{F}{q}$

The dimensional formula of the electric field will be- $\frac{F}{q}=\left[M{L}^1{T}^{-2}\right]\left[I^{-1}{T}^{-1}\right]=\left[M{L}^1{I}^{-1}{T}^{-3}\right]........\left(2\right)$

The dimensional formula for voltage will be- $V=E\times d=\left[M{L}^2{I}^{-1}{T}^{-3}\right]$

Thus the dimensional formula for capacitance is calculated as-

$C=\frac{\left[IT\right]}{\left[M{L}^2{I}^{-1}{T}^{-3}\right]}=\left[M^{-1}{L}^{-2}{I}^2{T}^4\right]$

$\left(C\right)=\left[M^{-1}{L}^{-2}{T}^4{I}^2\right]...........\left(3\right)$

The dimension of resistance is calculated using the ohm's law $V=IR$

Resistance's dimension$\left(R\right)=\left[M{L}^2{T}^{-3}{I}^{-2}\right].........\left(4\right)$

The magnetic field can be calculated using the formula, $F=qvB$ where $v$ is velocity

Thus, the dimensional formula for magnetic field is as follows-

$B=\frac{F}{qv}=\frac{\left[ML{T}^{-2}\right]}{\left[IT\right]\left[L{T}^{-1}\right]}$

$\left(B\right)=\left[M{T}^{-2}{I}^{-1}\right]..........\left(5\right)$

Step 4: Find the dimension of $x$

As we know that,

$speed\left(v\right)=\frac{1}{\sqrt{{\mu }_{\circ }{\epsilon }_{\circ }}}........\left(6\right)$

Using equation (6)-

$$\begin{gathered} x=\frac{1}{\sqrt{{\mu }_{\circ }{\epsilon }_{\circ }}} \\ x=V............\left(7\right) \end{gathered}$$

Now substituting, the dimension of speed in equation (7)

Thus, the dimension of $x$ will be-

$\left[x\right]=\left[L{T}^{-1}\right]$

Step 4: Compute the dimension of $y$

It is understood that $y=\frac{E}{B}$. So,

Substitute the known dimensions of electric field and magnetic field from equations (2) and (5) in the relation,

$$\begin{gathered} \left[y\right]=\frac{\left[ML{I}^{-1}{T}^{-3}\right]}{\left[M{T}^{-2}{I}^{-1}\right]} \\ \left[y\right]=\left[L{T}^{-1}\right] \end{gathered}$$

Step 5: Compute the dimension of $z$

We know that $z=\frac{L}{CR}$.

To find the dimension of $z$, substitute the known dimension from equation (3) and (4) in the relation,

$$\begin{gathered} \left[z\right]=\frac{\left[L\right]}{\left[M^{-1}{L}^{-2}{T}^4{I}^2\right]\left[M{L}^2{T}^{-3}{I}^{-2}\right]} \\ \left[z\right]=\left[L{T}^{-1}\right] \end{gathered}$$

Thus, $x,y,z$ have the same dimensions.

Hence, option B is the correct answer.