Question

A conductor lies along the $z$-axis at $-1.5 \leq Z \leq 1.5\, m$ and carries a fixed current of $10.0\, A$ in $-a_{z}$ direction as shown in figure for a field $B=3 \times 10^{-4} e^{-0.2 x} a_{y} T$, the total power required to move the conductor at constant speed to $x=2.0 m , y=0 m$ 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 $(F)=B I l$
Given, $B=3 \times 10^{4} e^{0.2 x}$ ay $I=10.0\, A$
Average power $=\frac{\text { Work done }}{\text { Time taken }}$
$P =\frac{1}{t} \int\limits_{0}^{2} F \cdot d x=\frac{1}{t} \int_{0}^{2} B(x) I L \cdot d x$
$P =\frac{1}{5 \times 10^{-3}} \int\limits_{0}^{2} 3 \times 10^{-4} e^{-0.2 x} \times 10 \times 3 d x$
$=9\left[1-e^{-0.4}\right]$
$=9\left[1-\frac{1}{e^{0.4}}\right]$
$=2.967\, W =2.97\, W$