Discussion Forum

1)

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

Answer:

Option B

Explanation:

 When force exerted on a current carrying conductor

 F_ext=BIL

average power= work done /time taken

 $P=\frac{1}{t}\int_{0}^{2} F_{ext}.dx$

    $=\frac{1}{t}\int_{0}^{2} B(x)IL.dx$

=$\frac{1}{5\times 10^{-3}}\int_{0}^{2} 3\times 10^{-4}e^{-0.2x}\times10\times3 dx$

= $9[1-e^{-0.4}]$

=   $9\left[1-\frac{1}{e^{0.4}}\right]=2.967=2.97W$