JEE PYQ: Units & Measurements - Question ID c432df107932 (JEE Main 2020)

ID: c432df107932JEE Main 2020Single Correct MCQ
The dimension of B22μ0{{{B^2}} \over {2{\mu _0}}}, where B is magnetic field and μ0{{\mu _0}} is the magnetic permeability of vacuum, is :

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Step-by-step Explanation

Core Formula & Concept:

In electromagnetism, the expression B22μ0{B^2 \over 2\mu_0} represents the energy density stored in a magnetic field BB in vacuum. The constant μ0\mu_0 is the permeability of free space.

To find its dimensions, we recall:

  • Force between two current-carrying wires gives the dimension of μ0\mu_0: F=μ0I1I22πr[μ0]=[F][r][I]2[].F = {\mu_0 I_1 I_2 \ell \over 2\pi r} \quad\Longrightarrow\quad [\mu_0] = { [F]\,[r] \over [I]^2\,[\ell] }.
  • Magnetic field BB is defined via the Lorentz force F=qvBF = q\,v\,B, so [B]=[F][q][v].[B] = { [F] \over [q]\,[v] }.
Step-by-Step Derivation:

Step 1: Dimensions of μ0\mu_0

From the force formula F=μ0I22πrF = {\mu_0 I^2 \ell \over 2\pi r} we isolate μ0\mu_0: μ0=2πrFI2.\mu_0 = {2\pi r\,F \over I^2\,\ell}. Taking dimensions: [μ0]=[r][F][I]2[]=L  MLT2A2  L=MLT2A2.[\mu_0] = { [r]\,[F] \over [I]^2\,[\ell] } = { \mathrm{L}\;\mathrm{MLT}^{-2} \over \mathrm{A}^2\;\mathrm{L} } = \mathrm{MLT}^{-2}\,\mathrm{A}^{-2}.

Step 2: Dimensions of BB

From F=qvBF = q\,v\,B we get B=Fqv.B = {F \over q\,v}. Hence [B]=[F][q][v]=MLT2AT  LT1=MT2A1.[B] = { [F] \over [q]\,[v] } = { \mathrm{MLT}^{-2} \over \mathrm{AT}\;\mathrm{LT}^{-1} } = \mathrm{MT}^{-2}\,\mathrm{A}^{-1}.

Step 3: Dimensions of B2B^2

Square the dimension of BB: [B2]=(MT2A1)2=M2T4A2.[B^2] = \bigl(\mathrm{MT}^{-2}\,\mathrm{A}^{-1}\bigr)^2 = \mathrm{M}^2\,\mathrm{T}^{-4}\,\mathrm{A}^{-2}.

Step 4: Combine to find B22μ0\displaystyle{B^2 \over 2\mu_0}

Divide [B2][B^2] by [μ0][\mu_0]: [B2μ0]=M2T4A2MLT2A2=M21L1T4+2=ML1T2.\left[{B^2 \over \mu_0}\right] = { \mathrm{M}^2\,\mathrm{T}^{-4}\,\mathrm{A}^{-2} \over \mathrm{MLT}^{-2}\,\mathrm{A}^{-2} } = \mathrm{M}^{2-1}\,\mathrm{L}^{-1}\,\mathrm{T}^{-4+2} = \mathrm{ML}^{-1}\,\mathrm{T}^{-2}. The factor of 22 is dimensionless, so the final dimension is ML1T2.\boxed{\mathrm{ML}^{-1}\,\mathrm{T}^{-2}}.

Common Traps & Exam Tip:

1. Confusing μ0\mu_0 with ϵ0\epsilon_0: Students sometimes use the dimensions of permittivity ϵ0\epsilon_0 instead of permeability μ0\mu_0. 2. Forgetting the square on BB: Omitting the exponent 22 leads to an incorrect power of mass or time. 3. Sign errors in exponents: Careful bookkeeping of negative signs in time and length exponents is essential.

Tip: Always write each step dimensionally and cancel units systematically to avoid arithmetic slips.

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