A conductor lies along the \(z-\)axis \(a\) \(-1.5\)\( \le Z < 1.5\,m\) 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........\( W\)

\(W=\int_{0}^{2} F d x\)

\(=\int_{0}^{2} 3.0 \times 10^{-4} e^{-0.2 x} \times 10 \times 3 d x\)

\(=9 \times 10^{-3} \int_{0}^{2} e^{-0.2 x} d x\)

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

(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, \(P=\frac{W}{t}\)

\(P=\frac{2.97 \times 10^{-3}}{(0.2) \times 5 \times 10^{-3}}=2.97 \,\mathrm{W}\)