Question

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

• A. $29.7\text{W}$
• B. $14.85\text{W}$
• C. $2.97\text{W}$
• D. $1.57\text{W}$

Answer 1

The correct option is C $2.97\text{W}$


Given: $i=10\text{A}$, $\overrightarrow{B}=3.0\times {10}^{-4}{e}^{-0.2x}{\hat{a}}_y\text{T}$ , $x=2.0\text{m},y=0\text{m}$ in $t=5\times {10}^{-3}\text{s}$.

Work done in moving the conductor is,

$W={\int }_0^2{F}_{ext}.dx={\int }_0^{2}B\left(x\right)IL.dx$

Required power,

$P=\frac{\text{Work done}}{\text{time}}$

$P=\frac{1}{t}{\int }_0^2{F}_{ext}.dx=\frac{1}{t}{\int }_0^{2}B\left(x\right)ILdx=\frac{1}{5\times {10}^{-3}}{\int }_0^{2}3\times {10}^{-4}{e}^{-0.2x}\times 10\times 3dx$

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

$=9[1-\frac{1}{e^{0.4}}]=2.96\text{W}$

Hence, option $\left(B\right)$ is correct.