Question

A conductor lies along the $z-$axis at $-1.5\le z<1.5m$ and carries a fixed current of $10.0$ A in ${-\hat{a}}_z$ direction (see figure). For a field $\overrightarrow{B}=3.0\times {10}^{-4}{e}^{-0.2x}{\hat{a}}_{y}T$ 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. $14.85W$
• B. $29.7W$
• C. $1.57W$
• D. $2.97W$

Answer 1

The correct option is D $2.97W$

Power $=\frac{w.r.t}{time}$

also force $F.\overrightarrow{il}\times \overrightarrow{B}=ilB$

$=10\times 3\times 3\times {10}^{-4}{e}^{-0.2x}$ Newtons

$F=9e^{-0.2x}\times {10}^{-3}N$

then Work $=F.dx={\int }_0^{2}9e^{-2.2x}\times {10}^{-3}dx$

$\Rightarrow \frac{9}{-0.2}{e}^{-0.2x}\times {10}^{-3}{]}_0^2$

$\Rightarrow -(4.5{.10}^{-2}{e}^{-0.4}-4.5\times {10}^{-2})$

Then, Power $=\frac{\Delta E}{\Delta t}=-\frac{45(e^{-0.4}-1){10}^{-3}}{5\times {10}^{-3}}$

where $\Delta =5\times {10}^{-3}$ seconds

Power $=-9(e^{-0.9}-1)$

$=2.967\sim 2.97watt$