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(A) The force on a current-carrying conductor in a magnetic field is $\vec{F} = I(\vec{L} 	imes \vec{B})$.Here,$\vec{L} = 3 \hat{a}_z 	ext{ m}$ (len

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$2.97$

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(A) The force on a current-carrying conductor in a magnetic field is $\vec{F} = I(\vec{L} 	imes \vec{B})$.Here,$\vec{L} = 3 \hat{a}_z 	ext{ m}$ (length of the conductor from $-1.5$ to $1.5$) and $I = 10 	ext{ A}$ in the $-\hat{a}_z$ direction,so $\vec{I} = -10 \hat{a}_z 	ext{ A}$.The magnetic field is $\vec{B} = 3.0 	imes 10^{-4} e^{-0.2x} \hat{a}_y 	ext{ T}$.The magnetic force is $\vec{F} = I \vec{L} 	imes \vec{B} = (-10 \hat{a}_z) 	imes (3.0 	imes 10^{-4} e^{-0.2x} \hat{a}_y) = -30 	imes 10^{-4} e^{-0.2x} (\hat{a}_z 	imes \hat{a}_y) = 3.0 	imes 10^{-3} e^{-0.2x} \hat{a}_x 	ext{ N}$.To move the conductor at a constant speed,an external force $\vec{F}_{ext} = -\vec{F} = -3.0 	imes 10^{-3} e^{-0.2x} \hat{a}_x 	ext{ N}$ must be applied.The work done is $W = \int_0^2 F_{ext} dx = \int_0^2 3.0 	imes 10^{-3} e^{-0.2x} dx$.$W = 3.0 	imes 10^{-3} \left[ \frac{e^{-0.2x}}{-0.2} ight]_0^2 = \frac{3.0 	imes 10^{-3}}{-0.2} (e^{-0.4} - 1) = 0.015 (1 - e^{-0.4}) \approx 0.015 (1 - 0.6703) = 0.0049455 	ext{ J}$.The power is $P = \frac{W}{t} = \frac{0.0049455}{5 	imes 10^{-3}} = 0.989 	ext{ W}$.Re-evaluating the integral based on the provided options: If the force magnitude is taken as $F = I L B = 10 	imes 3 	imes 3 	imes 10^{-4} e^{-0.2x} = 9 	imes 10^{-3} e^{-0.2x}$,then $W = \int_0^2 9 	imes 10^{-3} e^{-0.2x} dx = 9 	imes 10^{-3} [\frac{e^{-0.2x}}{-0.2}]_0^2 = 0.045(1 - e^{-0.4}) \approx 0.01485 	ext{ J}$.$P = \frac{0.01485}{5 	imes 10^{-3}} = 2.97 	ext{ W}$.

List-$I$	List-$II$
$a$. Fleming's left-hand rule	$e$. Direction of induced current
$b$. Fleming's right-hand rule	$f$. South pole
$c$. Clockwise current	$g$. North pole
$d$. Anticlockwise current	$h$. Direction of force

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Match the following:
List-$I$ List-$II$
$a$. Fleming's left-hand rule $e$. Direction of induced current
$b$. Fleming's right-hand rule $f$. South pole
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