In a particular system of units, a physical quantity can be expressed in terms of the electric charge $$e$$, electron mass $$m_e$$, Planck's constant $$h$$, and Coulomb's constant $$k = \dfrac{1}{4\pi \epsilon_0}$$, where $$\epsilon_0$$ is the permittivity of vacuum. In terms of these physical constants, the dimension of the magnetic field is [B] = [e]$$^\alpha$$ [m$$_e$$]$$^\beta$$ [h]$$^\gamma$$ [k]$$^\delta$$. The value of $$\alpha + \beta + \gamma + \delta$$ is _______.

The required fundamental dimensions are chosen as mass $$M$$, length $$L$$, time $$T$$ and electric charge $$Q$$ (instead of current).
Magnetic field $$B$$ is defined from the Lorentz force $$\vec F = q\,\vec v \times \vec B$$, giving

$$[B] = \frac{[F]}{[q][v]} = \frac{M\,L\,T^{-2}}{Q\,L\,T^{-1}} = M^{1}L^{0}T^{-1}Q^{-1}.$$

The four given constants have the following dimensions:

• Electric charge $$e:\;[e] = Q$$
• Electron mass $$m_e:\;[m_e] = M$$
• Planck’s constant $$h:\;[h] = M\,L^{2}T^{-1}$$ (energy × time)
• Coulomb constant $$k:\;[k] = M\,L^{3}T^{-2}Q^{-2}$$ (from $$F = k\,q_1q_2/r^{2}$$)

Assume $$[B] = [e]^{\alpha}[m_e]^{\beta}[h]^{\gamma}[k]^{\delta}.$$ Comparing the exponents of each fundamental dimension:

Mass $$M: \;\; \beta + \gamma + \delta = 1 \quad -(1)$$
Length $$L: \;\; 2\gamma + 3\delta = 0 \quad -(2)$$
Time $$T: \;\; -\gamma - 2\delta = -1 \quad -(3)$$
Charge $$Q: \;\; \alpha - 2\delta = -1 \quad -(4)$$

From $$(2):\; 2\gamma + 3\delta = 0 \;\Rightarrow\; \gamma = -\tfrac{3}{2}\delta.$

Insert this in $$(3):\; -$$\gamma$$ - 2$$\delta$$ = -1$$
$$-\!$$\left$$(-\tfrac{3}{2}$$\delta$$$$\right$$) - 2$$\delta$$ = -1 \;$$\Rightarrow$$\; \tfrac{3}{2}$$\delta$$ - 2$$\delta$$ = -1$$
$$-\tfrac{1}{2}$$\delta$$ = -1 \;$$\Rightarrow$$\; $$\delta$$ = 2.$$

Then $$$$\gamma$$ = -\tfrac{3}{2}$$\times$$ 2 = -3.$$

Use $$(1):\; $$\beta + \gamma + \delta$$ = 1$$
$$$$\beta$$ + (-3) + 2 = 1 \;$$\Rightarrow$$\; $$\beta$$ = 2.$$

Finally from $$(4):\; $$\alpha$$ - 2$$\delta$$ = -1$$
$$$$\alpha$$ - 4 = -1 \;$$\Rightarrow$$\; $$\alpha$$ = 3.$$

Thus $$$$\alpha$$ = 3,\;$$\beta$$ = 2,\;$$\gamma$$ = -3,\;$$\delta$$ = 2.$$

The required sum is
$$$$\alpha + \beta + \gamma + \delta$$ = 3 + 2 - 3 + 2 = 4.$$

Answer: 4