A conductor lies along the z-axis at $$-1.5 \leq z < 1.5$$ m and 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.

$$d\vec{F}_m = I (d\vec{l} \times \vec{B})$$
$$d\vec{F}_m = (-10 dz \hat{a}_z) \times (3.0 \times 10^{-4} e^{-0.2x} \hat{a}_y)$$

$$d\vec{F}_m = 3.0 \times 10^{-3} e^{-0.2x} dz \hat{a}_x$$

$$\vec{F}_m = \left( \int_{-1.5}^{1.5} 3.0 \times 10^{-3} e^{-0.2x} dz \right) \hat{a}_x$$

$$\vec{F}_m = 9.0 \times 10^{-3} e^{-0.2x} \hat{a}_x\text{ N}$$

$$W = \int_{0}^{2} F_{\text{ext}} dx = \int_{0}^{2} 9.0 \times 10^{-3} e^{-0.2x} dx$$

$$W = 9.0 \times 10^{-3} \left[ \frac{e^{-0.2x}}{-0.2} \right]_{0}^{2}$$

$$W = \frac{9.0 \times 10^{-3}}{-0.2} \left( e^{-0.4} - e^{0} \right)$$

$$W = -45 \times 10^{-3} \times (-0.3297) \approx 14.836 \times 10^{-3}\text{ J}$$

$$P = \frac{W}{\Delta t} = \frac{14.836 \times 10^{-3}\text{ J}}{5 \times 10^{-3}\text{ s}}$$

$$P \approx 2.97\text{ W}$$