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(B) The current is given by $i = \alpha_{0} t + \beta t^{2}$.Substituting the given values,we have $i = 20t + 8t^{2}$.We know that current $i = \frac{

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$11250$

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(B) The current is given by $i = \alpha_{0} t + \beta t^{2}$.Substituting the given values,we have $i = 20t + 8t^{2}$.We know that current $i = \frac{dq}{dt}$,so the charge $q$ is given by the integral $q = \int i \, dt$.To find the total charge crossed in $15 \, s$,we integrate from $t = 0$ to $t = 15 \, s$:$q = \int_{0}^{15} (20t + 8t^{2}) \, dt$$q = \left[ \frac{20t^{2}}{2} + \frac{8t^{3}}{3} \right]_{0}^{15}$$q = \left[ 10t^{2} + \frac{8}{3}t^{3} \right]_{0}^{15}$$q = 10(15)^{2} + \frac{8}{3}(15)^{3}$$q = 10(225) + \frac{8}{3}(3375)$$q = 2250 + 8(1125)$$q = 2250 + 9000$$q = 11250 \, C$.

$A$ current through a wire depends on time as $i = \alpha_{0} t + \beta t^{2}$ where $\alpha_{0} = 20 \, A/s$ and $\beta = 8 \, As^{-2}$. Find the charge crossed through a section of the wire in $15 \, s$ (in $C$).

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