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AIIMS 2003
Exam
AIIMS
Year
2003
Subject
Physics
Chapter
Units and Dimensions

AIIMS 2003 · Physics · Units and Dimensions

Using mass (M) , length (L) , time (T) and current (A) as...

Using mass (M)(M), length (L)(L), time (T)(T) and current (A)(A) as fundamental quantities, the dimension of permeability is

Options

  1. AM−1LT−2 A\mathrm{M}^{-1} \mathrm{LT}^{-2} \mathrm{~A}
  2. BML2 T−2 A−1\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-1}
  3. CMLT−2 A−2\mathrm{MLT}^{-2} \mathrm{~A}^{-2}
  4. DMLT−1 A−1\mathrm{MLT}^{-1} \mathrm{~A}^{-1}.

Correct answer

C. MLT−2 A−2\mathrm{MLT}^{-2} \mathrm{~A}^{-2}

Step-by-step solution

We know that the force per unit length of a wire carrying current due to another parallel wire carrying current is given by dFdl=μ0i1i22πd⇒μ0=2πdi1i2dFdl∴[μ0]=[L][A2]⋅[MLT−2][L] or  [μ0]=[MLT−2][A−2]=[MLT−2A−2].\begin{gathered}\frac{d F}{d l}=\frac{\mu_0 i_1 i_2}{2 \pi d} \Rightarrow \mu_0=\frac{2 \pi d}{i_1 i_2} \frac{d F}{d l}\\ \therefore \quad\left[\mu_0\right]=\frac{[\mathrm{L}]}{\left[\mathrm{A}^2\right]} \cdot \frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{L}]}\\ \text{\text{ or } }\left[\mu_0\right]=\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{A}^{-2}\right]=\left[\mathrm{MLT}^{-2} \mathrm{ A}^{-2}\right] . \end{gathered}
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