\(W=\int_{0}^{2} F d x\)
\(=\int_{0}^{2} 3.0 \times 10^{-4} e^{-0.2 x} \times 10 \times 3 d x\)
\(=9 \times 10^{-3} \int_{0}^{2} e^{-0.2 x} d x\)
\(=\frac{9 \times 10^{-3}}{0.2}\left[-e^{-0.2 \times 2}+1\right] B=3.0 \times 10^{-4} e^{-0.2 x}\)
(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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