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 $k = 1/(4 \pi$ eps0), where eps0 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 ____.

Step 1: Determine the dimensions of the given physical constants.

Express the dimensions of electric charge $e$, electron mass $m_e$, Planck's constant $h$, and Coulomb's constant $k$ in the $M, L, T, Q$ system, where $Q$ represents charge.

$$[e] = Q$$ $$[m_e] = M$$ $$[h] = M L^2 T^{-1}$$ $$[k] = M L^3 T^{-2} Q^{-2}$$

Step 2: Determine the dimension of the magnetic field $B$.

Use the magnetic force formula $F = qvB$ to find the dimension of $B$.

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

Step 3: Set up the dimensional equation.

Substitute the dimensions of the constants into the given relation $[B] = [e]^{\alpha} [m_e]^{\beta} [h]^{\gamma} [k]^{\delta}$.

$$M T^{-1} Q^{-1} = (Q)^{\alpha} (M)^{\beta} (M L^2 T^{-1})^{\gamma} (M L^3 T^{-2} Q^{-2})^{\delta}$$ $$M^1 L^0 T^{-1} Q^{-1} = M^{\beta + \gamma + \delta} L^{2\gamma + 3\delta} T^{-\gamma - 2\delta} Q^{\alpha - 2\delta}$$

Step 4: Equate the exponents of $M, L, T,$ and $Q$ on both sides.

This yields a system of four linear equations.

$$M: \beta + \gamma + \delta = 1$$ $$L: 2\gamma + 3\delta = 0$$ $$T: -\gamma - 2\delta = -1$$ $$Q: \alpha - 2\delta = -1$$

Step 5: Solve the system of equations.

Express $\gamma$ in terms of $\delta$ using the equation for $L$.

$$\gamma = -\frac{3}{2}\delta$$

Substitute this expression for $\gamma$ into the equation for $T$.

$$\frac{3}{2}\delta - 2\delta = -1$$ $$\delta = 2$$

Substitute $\delta = 2$ back to find the remaining exponents.

$$\gamma = -\frac{3}{2}(2) = -3$$ $$\beta = 1 - \gamma - \delta = 1 - (-3) - 2 = 2$$ $$\alpha = 2\delta - 1 = 2(2) - 1 = 3$$

Step 6: Calculate the sum of the exponents.

Sum the values of $\alpha$, $\beta$, $\gamma$, and $\delta$.

$$\alpha + \beta + \gamma + \delta = 3 + 2 - 3 + 2$$ $$\alpha + \beta + \gamma + \delta = 4$$

Answer: 4