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← JEE Advanced 2022 · Paper 2
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JEE Advanced 2022 · Paper 2, Physics question 4

In a particular system of units, a physical quantity can be expressed in terms of the electric charge ee, electron mass mem_e, Planck's constant hh, and Coulomb's constant k=14πϵ0k = \dfrac{1}{4\pi\epsilon_0}, where ϵ0\epsilon_0 is the permittivity of vacuum. In terms of these physical constants, the dimension of the magnetic field is [B]=[e]α[me]β[h]γ[k]δ[B] = [e]^\alpha[m_e]^\beta[h]^\gamma[k]^\delta. The value of α+β+γ+δ\alpha + \beta + \gamma + \delta is _____.

Answer and worked solution

Official answer: 4

Dimensions (writing Q for charge):

  • From F=qvBF=qvB: [B]=MT−1Q−1[B]=MT^{-1}Q^{-1}.
  • [h]=ML2T−1[h]=ML^2T^{-1}.
  • [k]=ML3T−2Q−2[k]=ML^3T^{-2}Q^{-2}.

Match the exponents in [e]α[me]β[h]γ[k]δ[e]^\alpha[m_e]^\beta[h]^\gamma[k]^\delta:

  • MM: β+γ+δ=1\beta+\gamma+\delta=1.
  • LL: 2γ+3δ=02\gamma+3\delta=0.
  • TT: γ+2δ=1\gamma+2\delta=1.
  • QQ: α−2δ=−1\alpha-2\delta=-1.

From LL and TT: δ=2\delta=2 and γ=−3\gamma=-3. Then β=2\beta=2 and α=3\alpha=3.

Answer. α+β+γ+δ=3+2−3+2=4\alpha+\beta+\gamma+\delta=3+2-3+2=4.

Topics:Dimensional analysis
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