1. Write dimensions of all given quantities in SI base dimensions

We use $M, L, T, I$ for mass, length, time, current.

• Magnetic field: $[B] = \frac{\text{force}}{\text{charge} \times \text{velocity}}$ Also, from Lorentz force $F=qvB$, $[B] = \frac{[F]}{[q][v]} = \frac{MLT^{-2}}{(IT)(LT^{-1})} = MT^{-2}I^{-1}$

• Electric charge: $[e] = IT$

• Electron mass: $[m_e] = M$

• Planck's constant: $[h] = [\text{energy}]\,[\text{time}] = (ML^2T^{-2})T = ML^2T^{-1}$

• Coulomb's constant $k$: From Coulomb's law, $F = k\frac{q^2}{r^2} \Rightarrow [k] = \frac{[F][r^2]}{[q]^2} = \frac{(MLT^{-2})L^2}{(IT)^2} = ML^3T^{-4}I^{-2}$

2. Set up the dimensional equation

Given $[B] = [e]^\alpha [m_e]^\beta [h]^\gamma [k]^\delta$

Substitute dimensions: $MT^{-2}I^{-1} = (IT)^\alpha (M)^\beta (ML^2T^{-1})^\gamma (ML^3T^{-4}I^{-2})^\delta$

Expanding powers: $MT^{-2}I^{-1} = M^{\beta+\gamma+\delta}L^{2\gamma+3\delta}T^{\alpha-\gamma-4\delta}I^{\alpha-2\delta}$

3. Compare exponents of $M, L, T, I$

For $M$: $\beta+\gamma+\delta = 1$

For $L$: $2\gamma+3\delta = 0$

For $T$: $\alpha-\gamma-4\delta = -2$

For $I$: $\alpha-2\delta = -1$

4. Solve the equations

From $\alpha-2\delta=-1 \Rightarrow \alpha = -1+2\delta$

From $2\gamma+3\delta=0 \Rightarrow \gamma = -\frac{3}{2}\delta$

Use in time equation: $\alpha-\gamma-4\delta=-2$ $(-1+2\delta)-\left(-\frac{3}{2}\delta\right)-4\delta=-2$ $-1+2\delta+\frac{3}{2}\delta-4\delta=-2$ $-1-\frac{1}{2}\delta=-2$ $\delta=2$

Then $\gamma=-\frac{3}{2}(2)=-3$ $\alpha=-1+2(2)=3$

Now from mass equation: $\beta+\gamma+\delta=1$ $\beta-3+2=1$ $\beta=2$

So, $\alpha=3,\quad \beta=2,\quad \gamma=-3,\quad \delta=2$

5. Compute the required sum

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

Final Answer

$\boxed{4}$

6. Comparison with stored correct answer

Stored correct answer = $4$.

Our derived answer also is $4$, so they agree.