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

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

Work done in moving the conductor is, $W=\int_{0}^{2} F d x$ $=\int_{0}^{2} 3.0 \times 10^{-4} e^{-0.2 x} \times 10 \times 3 d x$ $=9 \times 10^{-3} \int_{0}^{2} e^{-0.2 x} d x$ $=\frac{9 \times 10^{-3}}{0.2}\left[-e^{-0.2 \times 2}+1\right] B=3.0 \times 10^{-4} e^{-0.2 x}$ (By exponential function) $=\frac{9 \times 10^{-3}}{0.2} \times\left[1-e^{-0.4}\right]$ $=9 \times 10^{-3} \times(0.33)=2.97 \times 10^{-3}\, \mathrm{J}$ Power required to move the conductor is, $P=\frac{W}{t}$ $P=\frac{2.97 \times 10^{-3}}{(0.2) \times 5 \times 10^{-3}}=2.97 \,\mathrm{W}$

Question

JEE MAIN 2014, Diffcult

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Solution

Work done in moving the conductor is,

$W=\int_{0}^{2} F d x$

$=\int_{0}^{2} 3.0 \times 10^{-4} e^{-0.2 x} \times 10 \times 3 d x$

$=9 \times 10^{-3} \int_{0}^{2} e^{-0.2 x} d x$

$=\frac{9 \times 10^{-3}}{0.2}\left[-e^{-0.2 \times 2}+1\right] B=3.0 \times 10^{-4} e^{-0.2 x}$

(By exponential function)

$=\frac{9 \times 10^{-3}}{0.2} \times\left[1-e^{-0.4}\right]$

$=9 \times 10^{-3} \times(0.33)=2.97 \times 10^{-3}\, \mathrm{J}$

Power required to move the conductor is, $P=\frac{W}{t}$

$P=\frac{2.97 \times 10^{-3}}{(0.2) \times 5 \times 10^{-3}}=2.97 \,\mathrm{W}$

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