Question

A conductor lies along the $z$-axis at$–1.5\le z\le 1.5\text{m}$ and carries a fixed current of $10.0\text{A}$ in $–\hat{k}$ direction (see figure). For a field $\overrightarrow{B}=3.0\times {10}^{-4}{e}^{0.2x}\hat{j}\text{T}$, find the power required to move the conductor at a 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. (Assume $e^{0.4}=1.5$)

• A. $14.85\text{W}$
• B. $29.7\text{W}$
• C. $1.57\text{W}$
• D. $4.5\text{W}$

Answer 1

The correct option is D $4.5\text{W}$

According to question, $\overrightarrow{B}=3.0\times {10}^{-4}{e}^{0.2x}\hat{j}\text{T}$, $l=3\text{m}$, $I=10.0\text{A}$
We know that magnetic force, $\overrightarrow{F}=I(\overrightarrow{l}\times \overrightarrow{B})$
$\overrightarrow{F}=I(-l\hat{k}\times B\hat{j})=FIB\hat{i}$
So, the work done by force due to magnetic field,
$W={\int }_0^{2}F.dx={\int }_0^{2}BIldx$
$\Rightarrow W={\int }_0^{2}10\times 3\times 3\times {10}^{-4}{e}^{0.2x}dx$
$=4.5\times {10}^{-2}(e^{0.4}-1)\text{J}$
Therefore, power is
$P=\frac{W}{T}=\frac{4.5\times {10}^{-2}\times 0.5}{5\times {10}^{-3}}=4.5\text{W}$