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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Introduction to Physics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Introduction to Physics

Determine the dimensions of the magnetic permeability μ_0 . The...

Determine the dimensions of the magnetic permeability μ0\mu_0. The relevant equations are F=qEF = qE, F=qvBF = qvB, and B=μ0I2πaB = \dfrac{\mu_0 I}{2 \pi a}; where FF is force, qq is charge, vv is speed, II is current, and aa is distance.

Options

  1. AMLT−3I−2MLT^{-3} I^{-2}
  2. BMLT−2I−2MLT^{-2} I^{-2}
  3. CML2T−2I−2ML^2T^{-2} I^{-2}
  4. DMLT−2I−1MLT^{-2} I^{-1}

Correct answer

B. MLT−2I−2MLT^{-2} I^{-2}

Step-by-step solution

From the given formula B=μ0I2πaB = \dfrac{\mu_0 I}{2 \pi a}, we get μ0=2πaBI\mu_0 = \dfrac{2 \pi a B}{I}.

From the magnetic force formula F=qvBF = qvB, we get B=FqvB = \dfrac{F}{qv}.

Substituting the expression for BB gives μ0=2πaFqvI\mu_0 = \dfrac{2 \pi a F}{qvI}.

The dimensional formulas for the quantities are:

[F]=MLT−2[F] = MLT^{-2}

[a]=L[a] = L

[q]=IT[q] = IT

[v]=LT−1[v] = LT^{-1}

[I]=I[I] = I

Substituting these dimensions into the expression for μ0\mu_0:

[μ0]=MLT−2L(IT)(LT−1)I[\mu_0] = \dfrac{MLT^{-2} L}{(IT)(LT^{-1})I}

[μ0]=ML2T−2LI2[\mu_0] = \dfrac{ML^2T^{-2}}{L I^2}

[μ0]=MLT−2I−2[\mu_0] = MLT^{-2}I^{-2}
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