A conductor lies along the z-axis at - 1.5 ≤ z < 1.5 m and carries a fixed current of 10.0 A in - a _ z direction (see figure). For a field B = 3.0 ×…

$$\text{Power P} = \frac{ \text{W} }{ \text{Δ} \text{t} }$$Power P=ΔtW​

Force on conductor

$$\vec{F}=I(\vec{l}\times \vec{b})$$F=I(l×b)

$$\vec{F} =-10( 3{\hat{a}}_{z} )\times ( 3\times { 10 }^{ -4 } {e}^{ -0.2x } {\hat{a}}_{y} )$$F=−10(3a^z​)×(3×10−4e−0.2xa^y​)

$\Rightarrow \vec{F}=90\times {10}^{-4}({e}^{-0.2x})$⇒F=90×10−4(e−0.2x) along with x-axis

work done on the conductor in moving along x-axis.

$$W=\int _{x=0}^{2}\vec{F}.\vec{dx}$$W=∫x=02​F.dx

$\displaystyle \text{W} = \int _{0}^{2} \text{Bi} \ell \text{dx} = \int _{0}^{2} 3 \times {\text{10}}^{ - 4 } {\text{e}}^{ - \text{0.2x} } \times 3 \times \text{10dx}$W=∫02​Biℓdx=∫02​3×10−4e−0.2x×3×10dx$=90\times {10}^{-4}{[\frac{{e}^{-0.2x}}{-0.2}]}_{0}^{2}$=90×10−4[−0.2e−0.2x​]02​

$$= \text{450} \times {\text{10}}^{ - 4 } [ 1 - {\text{e}}^{ - \text{0.4} } ] = \text{180} \times {\text{10}}^{ - 4 } ( \text{Using} {\text{e}}^{ - \text{x} } = 1 - \text{x} )$$=450×10−4[1−e−0.4]=180×10−4(Usinge−x=1−x)

$$\text{P} = \frac{ \text{W} }{ \text{Δ} \text{t} } = \frac{ \text{180} \times {\text{10}}^{ - 4 } }{ 5 \times {\text{10}}^{ - 3 } } = \text{3.6 W.}$$P=ΔtW​=5×10−3180×10−4​=3.6 W.

Hence, closest option is 2.97 W.

(if we used exact value of exponential term will get exact answer).