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  4. A conductor lies along the z -axis at -1.5 ≤ z<1.5 m and carries a fixed current of 10.0 A in - hataz direction (see figure). For a field vecB=3.0 × 10-4 e-0.2 x hatay 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 <img class=img-fluid question-image alt=image src=https://cdn.tardigrade.in/img/question/physics/48b51dc3057bed9f940de0389b0cabcc-.png />

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Q. 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 $\vec{B}=3.0 \times 10^{-4} e^{-0.2 x} \hat{a}_{y} T$, 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
image

JEE MainJEE Main 2014Moving Charges and Magnetism

Solution:

Average Power $=\frac{\text { work }}{\text { time }} $
$W =\int_{0}^{2} F dx $
$=\int_{0}^{2} 3.0 \times 10^{-4} e^{-0.2 x} \times 10 \times 3 \,d x$
$=9 \times 10^{-3} \int_{0}^{2} e^{-0.2 x} \,dx$
$=\frac{9 \times 10^{-3}}{0.2}\left[-e^{-0.2 \times 2}+1\right]$
$B=3.0 \times 10^{-4} e^{-0.2 x}$
image
$=\frac{9 \times 10^{-3}}{0.2} \times\left[1-e^{-0.4}\right] $
$=9 \times 10^{-3} \times(0.33) $
$=2.97 \times 10^{-3} \,J$
$P=\frac{2.97 \times 10^{-3}}{(0.2) \times 5 \times 10^{-3}}$
$=2.97 \,W$