Question

A conductor lies along the $z$ -axis at $-1.5\le z<1.5m$ and carries a fixed current of $10.0A$ in $-{\hat{a}}_z$ direction (see figure). For a field $\vec{B}=3.0\times 10^{-4}{e}^{-0.2x}{\hat{a}}_{y}T$, find the power required to move the conductor at constant speed to $x=2.0m,y=0m$ in $5\times 10^{-3}s$. Assume parallel motion along the $x$ -axis

• A. 1.57 W
• B. 2.97 W
• C. 14.85 W
• D. 29.7 W

Answer 1

Answer: (B)

Average Power $=\frac{\text{ work }}{\text{ time }}$
$W={\int }_0^{2}Fdx$
$={\int }_0^{2}3.0\times 10^{-4}{e}^{-0.2x}\times 10\times 3dx$
$=9\times 10^{-3}{\int }_0^2{e}^{-0.2x}dx$
$=\frac{9\times 10^{-3}}{0.2}\left[-e^{-0.2\times 2}+1\right]$
$B=3.0\times 10^{-4}{e}^{-0.2x}$

$=\frac{9\times 10^{-3}}{0.2}\times \left[1-e^{-0.4}\right]$
$=9\times 10^{-3}\times (0.33)$
$=2.97\times 10^{-3}J$
$P=\frac{2.97\times 10^{-3}}{(0.2)\times 5\times 10^{-3}}$
$=2.97W$