In electromagnetic theory, electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related. In the questions below, $[E]$ and $[B]$ stand for dimensions of electric and magnetic fields respectively, while $[\varepsilon_0]$ and $[\mu_0]$ stand for dimensions of the permittivity and permeability of free space respectively. $[L]$ and $[T]$ are dimensions of length and time respectively. All quantities are in $SI$ units.
$(1)$ The relation between $[E]$ and $[B]$ is:
$(A)$ $[E] = [B][L][T]$
$(B)$ $[E] = [B][L]^{-1}[T]$
$(C)$ $[E] = [B][L][T]^{-1}$
$(D)$ $[E] = [B][L]^{-1}[T]^{-1}$
$(2)$ The relation between $[\varepsilon_0]$ and $[\mu_0]$ is:
$(A)$ $[\mu_0] = [\varepsilon_0][L]^2[T]^{-2}$
$(B)$ $[\mu_0] = [\varepsilon_0][L]^{-2}[T]^2$
$(C)$ $[\mu_0] = [\varepsilon_0]^{-1}[L]^2[T]^{-2}$
$(D)$ $[\mu_0] = [\varepsilon_0]^{-1}[L]^{-2}[T]^2$
Give the answers for questions $(1)$ and $(2)$.
$(1)$ The relation between $[E]$ and $[B]$ is:
$(A)$ $[E] = [B][L][T]$
$(B)$ $[E] = [B][L]^{-1}[T]$
$(C)$ $[E] = [B][L][T]^{-1}$
$(D)$ $[E] = [B][L]^{-1}[T]^{-1}$
$(2)$ The relation between $[\varepsilon_0]$ and $[\mu_0]$ is:
$(A)$ $[\mu_0] = [\varepsilon_0][L]^2[T]^{-2}$
$(B)$ $[\mu_0] = [\varepsilon_0][L]^{-2}[T]^2$
$(C)$ $[\mu_0] = [\varepsilon_0]^{-1}[L]^2[T]^{-2}$
$(D)$ $[\mu_0] = [\varepsilon_0]^{-1}[L]^{-2}[T]^2$
Give the answers for questions $(1)$ and $(2)$.
- A$C, D$
- B$C, A$
- C$C, B$
- D$B, C, D$
Explore More
Similar Questions
Write the dimensions of $a / b$ in the relation $P = \frac{a - t^2}{bx}$; where $P$ is the pressure,$x$ is the distance,and $t$ is the time.
Difficult
View Solution$A$ length-scale $(l)$ depends on the permittivity $(\varepsilon)$ of a dielectric material,the Boltzmann constant $(k_B)$,the absolute temperature $(T)$,the number density $(n)$ of certain charged particles,and the charge $(q)$ carried by each of the particles. Which of the following expression$(s)$ for $l$ is(are) dimensionally correct?
$(A)$ $l=\sqrt{\left(\frac{n q^2}{\varepsilon k_B T}\right)}$
$(B)$ $l=\sqrt{\left(\frac{\varepsilon k_B T}{n q^2}\right)}$
$(C)$ $l=\sqrt{\left(\frac{q^2}{\varepsilon n^{2 / 3} k_B T}\right)}$
$(D)$ $l=\sqrt{\left(\frac{q^2}{\varepsilon n^{1 / 3} k_B T}\right)}$
$(A)$ $l=\sqrt{\left(\frac{n q^2}{\varepsilon k_B T}\right)}$
$(B)$ $l=\sqrt{\left(\frac{\varepsilon k_B T}{n q^2}\right)}$
$(C)$ $l=\sqrt{\left(\frac{q^2}{\varepsilon n^{2 / 3} k_B T}\right)}$
$(D)$ $l=\sqrt{\left(\frac{q^2}{\varepsilon n^{1 / 3} k_B T}\right)}$
In terms of potential difference $V$,electric current $I$,permittivity $\varepsilon_0$,permeability $\mu_0$,and speed of light $c$,the dimensionally correct equation$(s)$ is(are):
$(A)$ $\mu_0 I^2 = \varepsilon_0 V^2$
$(B)$ $\varepsilon_0 I = \mu_0 V$
$(C)$ $I = \varepsilon_0 cV$
$(D)$ $\mu_0 cI = V$
$(A)$ $\mu_0 I^2 = \varepsilon_0 V^2$
$(B)$ $\varepsilon_0 I = \mu_0 V$
$(C)$ $I = \varepsilon_0 cV$
$(D)$ $\mu_0 cI = V$
MediumIIT 2015
View SolutionIf mass is written as $m=kc^{p} G^{-1 / 2} \,h^{1 / 2}$ then the value of $P$ will be : (Constants have their usual meaning with $k$ a dimensionless constant)
DifficultJEE MAIN 2024
View SolutionThe frequency $(v)$ of an oscillating liquid drop may depend upon the radius $(r)$ of the drop,the density $(\rho)$ of the liquid,and the surface tension $(s)$ of the liquid as: $v = r^{a} \rho^{b} s^{c}$. The values of $a, b,$ and $c$ respectively are:
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo