Let $$[{\varepsilon _0}]$$ denotes the dimensional formula of the permittivity of the vacuum and $$[{\mu _0}]$$ that of the permeability of the vacuum. If $$M = {\rm{mass}}$$, $$L = {\rm{length}}$$, $$T = {\rm{Time}}$$ and $$I = {\rm{electric current}}$$, then

Question

Let $$[{\varepsilon _0}]$$ denotes the dimensional formula of the permittivity of the vacuum and $$[{\mu _0}]$$ that of the permeability of the vacuum. If $$M = {\rm{mass}}$$, $$L = {\rm{length}}$$, $$T = {\rm{Time}}$$ and $$I = {\rm{electric current}}$$,  then

Accepted Answer

$$[{\varepsilon _0}] = {M^{ - 1}}{L^{ - 3}}{T^4}{I^2}$$

Answer

$$[{\varepsilon _0}] = {M^{ - 1}}{L^{ - 3}}{T^2}I$$

Answer

$$[{\mu _0}] = ML{T^{ - 2}}{I^{ - 2}}$$

Answer

$$[{\mu _0}] = M{L^2}{T^{ - 1}}I$$

Units and Measurements

Question: Let \[[{\varepsilon _0}]\] denotes the dimensional formula of the permittivity of the vacuum and \[[{\mu _0}]\] that of the permeability of the vacuum. If \[M = {\rm{mass}}\], \[L = {\rm{length}}\], \[T = {\rm{Time}}\] and \[I = {\rm{electric current}}\], then

Units and Measurements

Question: Let \[[{\varepsilon _0}]\] denotes the dimensional formula of the permittivity of the vacuum and \[[{\mu _0}]\] that of the permeability of the vacuum. If \[M = {\rm{mass}}\], \[L = {\rm{length}}\], \[T = {\rm{Time}}\] and \[I = {\rm{electric current}}\], then

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