A conductor lies along the z -axis at -1.5 ≤ z < 1.5 m and carries a fixed current of 10.0 A in -az direction (see figure). For a field B = 3.0 × 10⁻⁴ e-0.2x ay T, find the power required to move the conductor at constant speed to x = 2.0 m, y = 0 m in 5 × 10⁻³ s. Assume parallel motion along the x -axis. (a) 1.57 W (b) 2.97 W (c) 14.85 W (d) 29.7 W

Answer: (b). Force on the conductor: F = I l B = 10×3×3×10⁻⁴ e-0.2x = 9×10⁻³ e-0.2x N, along x. Work against it: W = ∫₀² 9×10⁻³ e-0.2x dx = 9×10⁻³×1-e-0.40.2 W = 9×10⁻³×1.648 = 1.48×10⁻² J P = W/t = 1.48×10⁻²5×10⁻³ = 2.97 W So the answer is (b) 2.97 W.

A conductor lies along the z -axis at -1.5 ≤ z < 1.5 m and carries a… | JEE Main 2014

Q57JEE Main 2014Difficult

Force on Current-Carrying Conductors

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

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Answer: (b)

Force on the conductor: $F = I l B = 10\times3\times3\times10^{-4} e^{-0.2x} = 9\times10^{-3} e^{-0.2x}\text{ N}$ , along $x$ .
Work against it: $W = \int_0^{2} 9\times10^{-3} e^{-0.2x}\,dx = 9\times10^{-3}\times\frac{1-e^{-0.4}}{0.2}$
$W = 9\times10^{-3}\times1.648 = 1.48\times10^{-2}\text{ J}$
$P = \frac{W}{t} = \frac{1.48\times10^{-2}}{5\times10^{-3}} = 2.97\text{ W}$
So the answer is (b) 2.97 W.

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