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PhyMedAdv 2022 • Paper 2

Dimensional Analysis of Magnetic Field using Fundamental Physical Constants

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 = \frac{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 ______.

Official Numerical Answer4

Topics & Concepts

Units and MeasurementsElectric FieldCoulomb's Law

Step-by-Step Solution

To find the value of α+β+γ+δ\alpha + \beta + \gamma + \delta, we express the dimensions of the given physical quantities in terms of fundamental dimensions: Mass (MM), Length (LL), Time (TT), and Charge (QQ).

  1. Electric charge (ee): [e]=Q[e] = Q

  2. Electron mass (mem_e): [me]=M[m_e] = M

  3. Planck's constant (hh): Using E=hνE = h\nu, where energy [E]=ML2T−2[E] = M L^2 T^{-2} and frequency [ν]=T−1[\nu] = T^{-1}: [h]=[E][ν]=ML2T−2T−1=ML2T−1[h] = \frac{[E]}{[\nu]} = \frac{M L^2 T^{-2}}{T^{-1}} = M L^2 T^{-1}

  4. Coulomb's constant (k=14πϵ0k = \frac{1}{4\pi\epsilon_0}): Using Coulomb's law F=ke2r2F = \frac{k e^2}{r^2}, where force [F]=MLT−2[F] = M L T^{-2}: [k]=[F][r]2[e]2=(MLT−2)L2Q2=ML3T−2Q−2[k] = \frac{[F][r]^2}{[e]^2} = \frac{(M L T^{-2}) L^2}{Q^2} = M L^3 T^{-2} Q^{-2}

  5. Magnetic field (BB): Using Lorentz force F=qvBF = q v B, where velocity [v]=LT−1[v] = L T^{-1}: [B]=[F][q][v]=MLT−2Q(LT−1)=MT−1Q−1[B] = \frac{[F]}{[q][v]} = \frac{M L T^{-2}}{Q (L T^{-1})} = M T^{-1} Q^{-1}

Now, setting up the dimensional equation: [B]=[e]α[me]β[h]γ[k]δ[B] = [e]^\alpha [m_e]^\beta [h]^\gamma [k]^\delta

Substitute the dimensional formulas: MT−1Q−1=(Q)α(M)β(ML2T−1)γ(ML3T−2Q−2)δM T^{-1} Q^{-1} = (Q)^\alpha (M)^\beta (M L^2 T^{-1})^\gamma (M L^3 T^{-2} Q^{-2})^\delta

Group the terms corresponding to base dimensions: M1L0T−1Q−1=Mβ+γ+δ⋅L2γ+3δ⋅T−γ−2δ⋅Qα−2δM^1 L^0 T^{-1} Q^{-1} = M^{\beta + \gamma + \delta} \cdot L^{2\gamma + 3\delta} \cdot T^{-\gamma - 2\delta} \cdot Q^{\alpha - 2\delta}

Equating the exponents of MM, LL, TT, and QQ on both sides:

  • For MM: β+γ+δ=1— (1)\beta + \gamma + \delta = 1 \quad \text{--- (1)}

  • For LL: 2γ+3δ=0  ⟹  γ=−32δ— (2)2\gamma + 3\delta = 0 \implies \gamma = -\frac{3}{2}\delta \quad \text{--- (2)}

  • For TT: −γ−2δ=−1  ⟹  γ+2δ=1— (3)-\gamma - 2\delta = -1 \implies \gamma + 2\delta = 1 \quad \text{--- (3)}

  • For QQ: α−2δ=−1  ⟹  α=2δ−1— (4)\alpha - 2\delta = -1 \implies \alpha = 2\delta - 1 \quad \text{--- (4)}

Substituting equation (2) into equation (3): −32δ+2δ=1-\frac{3}{2}\delta + 2\delta = 1 12δ=1  ⟹  δ=2\frac{1}{2}\delta = 1 \implies \delta = 2

Using δ=2\delta = 2 in equations (2), (4), and (1):

  • From (2): γ=−32(2)=−3\gamma = -\frac{3}{2}(2) = -3
  • From (4): α=2(2)−1=3\alpha = 2(2) - 1 = 3
  • From (1): β+(−3)+2=1  ⟹  β=2\beta + (-3) + 2 = 1 \implies \beta = 2

Now, calculating the value of α+β+γ+δ\alpha + \beta + \gamma + \delta: α+β+γ+δ=3+2+(−3)+2=4\alpha + \beta + \gamma + \delta = 3 + 2 + (-3) + 2 = 4

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