Question

Find the dimensions of
(a) electric field E.
(b) magnetic field B and
(c) magnetic permeability μ0.
The relevant equation are
F=qE, F=qvB, and B=μ0I2 π a;
where F is force, q is charge, v is speed, I is current, and a is distance.

Answer 1

The dimension of magnetic permeability is $\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2} \mathrm{I}^{-2}\right]$.

Explanation:

Given :

where F is force, q is charge, v is speed, I is current, and a is distance.

(a) Electric field E :

We know that, E $=\frac{F}{q}$

                        E = electric field strength

                        q = amount of charge  

                         f = force on charge  

                           $=\frac{M L T^{-2}}{[I T]}$

                           $=\left[\mathrm{MLT}^{-3} \mathrm{I}^{-1}\right]$

(b) Magnetic field B  :

We know that B  $=\frac{F}{q v}$

                        V = velocity of particle

                         f = force on charge  

                        q = amount of charge  

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

                           $=\left[\mathrm{M} \mathrm{T}^{-2} \mathrm{I}^{-1}\right]$

(c) Magnetic permeability μ0 :

We know that μ0  $=\frac{B 2 \pi a}{I}$

                             $=\frac{M T^{-2} I^{-1}[L]}{I}$

                             $=\left[M L T^{-2} I^-^2\right]$

Therefore, the dimension is $\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2} \mathrm{I}^{-2}\right]$.

Answer 2

$\huge{\boxed{\mathcal\pink{\fcolorbox{red}{yellow}{Answer}}}}$

$MLT^{ - 2} I^{ - 2}$