Question

A conductor lies along the $z$-axis at $-1.5\le Z\le 1.5m$ and carries a fixed current of $10.0A$ in $-a_z$ direction as shown in figure for a field $B=3\times 10^{-4}{e}^{-0.2x}{a}_{y}T$, the total power required to move the conductor at constant speed to $x=2.0m,y=0m$ in $5\times 10^{-3}s$ is (Assume parallel motion along the $x$-axis)

• A. 1.57 W
• B. 2.97 W
• C. 4.45 W
• D. 9.87 W

Answer 1

Answer: (B)

Force exerted on current carrying conductor $(F)=BIl$
Given, $B=3\times 10^4{e}^{0.2x}$ ay $I=10.0A$
Average power $=\frac{\text{ Work done }}{\text{ Time taken }}$
$P=\frac{1}{t}\underset{0}{\overset{2}{\int }}F\cdot dx=\frac{1}{t}{\int }_0^{2}B(x)IL\cdot dx$
$P=\frac{1}{5\times 10^{-3}}\underset{0}{\overset{2}{\int }}3\times 10^{-4}{e}^{-0.2x}\times 10\times 3dx$
$=9\left[1-e^{-0.4}\right]$
$=9\left[1-\frac{1}{e^{0.4}}\right]$
$=2.967W=2.97W$