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(B) The dimensional formula for magnetic field $B$ is $[M^1 T^{-2} A^{-1}]$.The dimensions of the constants are:$[e] = [A^1 T^1]$$[m_e] = [M^1]$$[h] =

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$4$

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(B) The dimensional formula for magnetic field $B$ is $[M^1 T^{-2} A^{-1}]$.The dimensions of the constants are:$[e] = [A^1 T^1]$$[m_e] = [M^1]$$[h] = [M^1 L^2 T^{-1}]$$[k] = [M^1 L^3 T^{-4} A^{-2}]$Equating the dimensions:$[M^1 T^{-2} A^{-1}] = [A^1 T^1]^\alpha [M^1]^\beta [M^1 L^2 T^{-1}]^\gamma [M^1 L^3 T^{-4} A^{-2}]^\delta$$[M^1 T^{-2} A^{-1}] = M^{\beta + \gamma + \delta} L^{2\gamma + 3\delta} T^{\alpha - \gamma - 4\delta} A^{\alpha - 2\delta}$Comparing the powers of $M, L, T, A$ on both sides:$1) \beta + \gamma + \delta = 1$$2) 2\gamma + 3\delta = 0$$3) \alpha - \gamma - 4\delta = -2$$4) \alpha - 2\delta = -1$From $(4)$,$\alpha = 2\delta - 1$. Substituting into $(3)$:$(2\delta - 1) - \gamma - 4\delta = -2 \implies -\gamma - 2\delta = -1 \implies \gamma + 2\delta = 1$.Solving with $(2)$ $(2\gamma + 3\delta = 0)$:$2(1 - 2\delta) + 3\delta = 0 \implies 2 - 4\delta + 3\delta = 0 \implies \delta = 2$.Then $\gamma = 1 - 2(2) = -3$.Then $\alpha = 2(2) - 1 = 3$.Then $\beta = 1 - (-3) - 2 = 2$.Thus,$\alpha + \beta + \gamma + \delta = 3 + 2 - 3 + 2 = 4$.

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