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Question

(A) The force on a current-carrying conductor in a magnetic field is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$.Here,the length vector is $\vec{L}

Vedclass · Accepted Answer

$2.97$

Vedclass · Answer

(A) The force on a current-carrying conductor in a magnetic field is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$.Here,the length vector is $\vec{L} = 3 \hat{a}_z \ m$ and the current is $I = 10 \ A$ in the $-\hat{a}_z$ direction,so $\vec{L} = -3 \hat{a}_z \ m$.The magnetic force is $\vec{F} = 10 	imes (-3 \hat{a}_z 	imes 3.0 	imes 10^{-4} e^{-0.2x} \hat{a}_y) = -30 	imes 3.0 	imes 10^{-4} e^{-0.2x} (\hat{a}_z 	imes \hat{a}_y) = -90 	imes 10^{-4} e^{-0.2x} (-\hat{a}_x) = 9.0 	imes 10^{-3} e^{-0.2x} \hat{a}_x \ N$.To move the conductor at a constant speed,an external force $\vec{F}_{ext} = -\vec{F} = -9.0 	imes 10^{-3} e^{-0.2x} \hat{a}_x \ N$ must be applied.The work done $W$ in moving the conductor from $x = 0$ to $x = 2.0 \ m$ is:$W = \int_{0}^{2} |F_{ext}| dx = \int_{0}^{2} 9.0 	imes 10^{-3} e^{-0.2x} dx$$W = 9.0 	imes 10^{-3} \left[ \frac{e^{-0.2x}}{-0.2} ight]_{0}^{2} = \frac{9.0 	imes 10^{-3}}{0.2} (1 - e^{-0.4}) = 45 	imes 10^{-3} (1 - 0.6703) = 45 	imes 10^{-3} 	imes 0.3297 \approx 14.836 	imes 10^{-3} \ J$.The power required is $P = \frac{W}{t} = \frac{14.836 	imes 10^{-3}}{5 	imes 10^{-3}} = 2.967 \ W \approx 2.97 \ W$.

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