Which of the following has the dimensions of electrical resistance (${{\varepsilon }_{0}}$ is the permittivity of vacuum and ${{\mu }_{0}}$ is the permeability of vacuum)
$\begin{align}
& \text{A}\text{. }\sqrt{\dfrac{{{\varepsilon }_{0}}}{{{\mu }_{0}}}} \\
& B.\text{ }\dfrac{{{\mu }_{0}}}{{{\varepsilon }_{0}}} \\
& C.\text{ }\sqrt{\dfrac{{{\mu }_{0}}}{{{\varepsilon }_{0}}}} \\
& D.\text{ }\dfrac{{{\varepsilon }_{0}}}{{{\mu }_{0}}} \\
\end{align}$
Answer
645.3k+ views
Hint: In order to find out which of these is the correct option, we need to know the dimensional formula for the resistance, which is the ratio of electric voltage to the electric current. After that, we determine the dimensional formula for ${{\varepsilon }_{0}}\text{ and }{{\mu }_{0}}$ from Coulomb's law and Biot-Sawart’s Law respectively.
Complete answer:
We know that the electrical resistance is defined as the ratio of the electric voltage to the electric current. So, we can write,
$R=\dfrac{\text{Voltage}}{\text{Current}}$ …. Equation (1)
Electric voltage can be defined as the work done divided by an electric charge, and the charge can be expressed as the product of electric current and time.
$V=\dfrac{\text{Work}}{\text{Charge}}=\dfrac{\text{Work}}{\left( \text{Current} \right)\text{ }\!\!\times\!\!\text{ Time}}$… Equation (2)
Substituting Equation (2) in Equation (1), we get,
$R=\dfrac{\text{Work}}{{{\left( \text{Current} \right)}^{2}}\text{ }\!\!\times\!\!\text{ Time}}$
Substituting the dimensional formula for work, electric current and time, we get
$R=\dfrac{\left[ M{{L}^{2}}{{T}^{-2}} \right]}{\left[ {{A}^{2}} \right]\times \left[ T \right]}$
$R=\left[ M{{L}^{2}}{{A}^{-2}}{{T}^{-3}} \right]$ … Equation (3)
From Coulomb's law, we know that the force acting between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. So, we can write,
$F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$
So, the permittivity of vacuum can be written as,
${{\varepsilon }_{0}}=\dfrac{{{q}_{1}}{{q}_{2}}}{F\times {{r}^{2}}}$
Writing down the dimensional formula for various quantities in the above Equation, we get,
${{\varepsilon }_{0}}=\dfrac{{{\left[ AT \right]}^{2}}}{\left[ ML{{T}^{-2}} \right]\times \left[ {{L}^{2}} \right]}$
$\therefore {{\varepsilon }_{0}}=\left[ {{M}^{-1}}{{L}^{-3}}{{A}^{2}}{{T}^{4}} \right]$ … Equation (4)
From Biot-Sawart’s law, we know that the magnetic field produced by a current-carrying conductor at a point is directly proportional to the current flowing through the conductor and inversely proportional to the square of the distance between the point and the conductor. So, we can write,
$B=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{Idl}{{{r}^{2}}}$
So, the permittivity of vacuum can be written as,
${{\mu }_{0}}=\dfrac{B\times {{r}^{2}}}{Idl}$
Writing down the dimensional formula for various quantities in the above Equation, we get,
${{\mu }_{0}}=\dfrac{\left[ M{{T}^{-2}}{{A}^{-1}} \right]\times {{\left[ L \right]}^{2}}}{\left[ AT \right]\times \left[ L \right]}$
$\therefore {{\mu }_{0}}=\left[ {{M}^{1}}L{{A}^{-2}}{{T}^{-3}} \right]$ … Equation (5)
Dividing equation (5) by Equation (4), we get,
$\dfrac{{{\mu }_{0}}}{{{\varepsilon }_{0}}}=\dfrac{\left[ {{M}^{1}}L{{A}^{-2}}{{T}^{-3}} \right]}{\left[ {{M}^{-1}}{{L}^{-3}}{{A}^{2}}{{T}^{4}} \right]}$
$\dfrac{{{\mu }_{0}}}{{{\varepsilon }_{0}}}=\left[ {{M}^{2}}{{L}^{4}}{{A}^{-4}}{{T}^{-6}} \right]$
This dimensional formula is just the square of the dimensional formula of the electric resistance,
$\sqrt{\dfrac{{{\mu }_{0}}}{{{\varepsilon }_{0}}}}=\left[ {{M}^{1}}{{L}^{2}}{{A}^{-2}}{{T}^{-3}} \right]$
So, $\sqrt{\dfrac{{{\mu }_{0}}}{{{\varepsilon }_{0}}}}$ has the same dimension as resistance.
So, the answer to the question is option (C).
Note: Ohm’s Law: It states that the current flowing through a conductor is directly proportional to the voltage across the ends of the conductor. $V=IR$, where V is the potential difference, I is the electric current and R is the electric resistance.
Electrical Resistance is an opposition to the flow of electric current. Its SI unit is Ohms. The resistance of a material depends on the resistivity of that particular material, the length of material and the surface area of the material.
Complete answer:
We know that the electrical resistance is defined as the ratio of the electric voltage to the electric current. So, we can write,
$R=\dfrac{\text{Voltage}}{\text{Current}}$ …. Equation (1)
Electric voltage can be defined as the work done divided by an electric charge, and the charge can be expressed as the product of electric current and time.
$V=\dfrac{\text{Work}}{\text{Charge}}=\dfrac{\text{Work}}{\left( \text{Current} \right)\text{ }\!\!\times\!\!\text{ Time}}$… Equation (2)
Substituting Equation (2) in Equation (1), we get,
$R=\dfrac{\text{Work}}{{{\left( \text{Current} \right)}^{2}}\text{ }\!\!\times\!\!\text{ Time}}$
Substituting the dimensional formula for work, electric current and time, we get
$R=\dfrac{\left[ M{{L}^{2}}{{T}^{-2}} \right]}{\left[ {{A}^{2}} \right]\times \left[ T \right]}$
$R=\left[ M{{L}^{2}}{{A}^{-2}}{{T}^{-3}} \right]$ … Equation (3)
From Coulomb's law, we know that the force acting between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. So, we can write,
$F=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}$
So, the permittivity of vacuum can be written as,
${{\varepsilon }_{0}}=\dfrac{{{q}_{1}}{{q}_{2}}}{F\times {{r}^{2}}}$
Writing down the dimensional formula for various quantities in the above Equation, we get,
${{\varepsilon }_{0}}=\dfrac{{{\left[ AT \right]}^{2}}}{\left[ ML{{T}^{-2}} \right]\times \left[ {{L}^{2}} \right]}$
$\therefore {{\varepsilon }_{0}}=\left[ {{M}^{-1}}{{L}^{-3}}{{A}^{2}}{{T}^{4}} \right]$ … Equation (4)
From Biot-Sawart’s law, we know that the magnetic field produced by a current-carrying conductor at a point is directly proportional to the current flowing through the conductor and inversely proportional to the square of the distance between the point and the conductor. So, we can write,
$B=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{Idl}{{{r}^{2}}}$
So, the permittivity of vacuum can be written as,
${{\mu }_{0}}=\dfrac{B\times {{r}^{2}}}{Idl}$
Writing down the dimensional formula for various quantities in the above Equation, we get,
${{\mu }_{0}}=\dfrac{\left[ M{{T}^{-2}}{{A}^{-1}} \right]\times {{\left[ L \right]}^{2}}}{\left[ AT \right]\times \left[ L \right]}$
$\therefore {{\mu }_{0}}=\left[ {{M}^{1}}L{{A}^{-2}}{{T}^{-3}} \right]$ … Equation (5)
Dividing equation (5) by Equation (4), we get,
$\dfrac{{{\mu }_{0}}}{{{\varepsilon }_{0}}}=\dfrac{\left[ {{M}^{1}}L{{A}^{-2}}{{T}^{-3}} \right]}{\left[ {{M}^{-1}}{{L}^{-3}}{{A}^{2}}{{T}^{4}} \right]}$
$\dfrac{{{\mu }_{0}}}{{{\varepsilon }_{0}}}=\left[ {{M}^{2}}{{L}^{4}}{{A}^{-4}}{{T}^{-6}} \right]$
This dimensional formula is just the square of the dimensional formula of the electric resistance,
$\sqrt{\dfrac{{{\mu }_{0}}}{{{\varepsilon }_{0}}}}=\left[ {{M}^{1}}{{L}^{2}}{{A}^{-2}}{{T}^{-3}} \right]$
So, $\sqrt{\dfrac{{{\mu }_{0}}}{{{\varepsilon }_{0}}}}$ has the same dimension as resistance.
So, the answer to the question is option (C).
Note: Ohm’s Law: It states that the current flowing through a conductor is directly proportional to the voltage across the ends of the conductor. $V=IR$, where V is the potential difference, I is the electric current and R is the electric resistance.
Electrical Resistance is an opposition to the flow of electric current. Its SI unit is Ohms. The resistance of a material depends on the resistivity of that particular material, the length of material and the surface area of the material.
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