Magnetic Effects & Magnetism

JEE Main level, single correct, medium

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Question 233 of 276

A conductor lies along the z-axis at −1.5≤z<1.5m-\text{1.5}\leq \text{z}\lt \text{1.5}m and carries a fixed current of 10.0 A in −a^z{-\hat{\text{a}}}_{\text{z}} direction (see figure). For a field B⃗=3.0×10−4e−0.2xa^y\vec{\text{B}}=\text{3.0}\times 1{0}^{-4}{\text{e}}^{-\text{0.2x}}{\hat{\text{a}}}_{\text{y}} T, find the power required to move the conductor at constant speed to x = 2.0 m, y = 0 m in 5×10−3s5\times 1{0}^{-3}\text{s}. Assume parallel motion along the x-axis.

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Answer: (B) 2.97 W

Power P=WΔt\text{Power P}=\dfrac{\text{W}}{\text{Δ}\text{t}}

Force on conductor
F⃗=I(l⃗×b⃗)\vec{F}=I(\vec{l}\times \vec{b})

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

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

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

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

W=∫02Biℓdx=∫023×10−4e−0.2x×3×10dx\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}=90×10−4[e−0.2x−0.2]02=90\times {10}^{-4}{\left[\dfrac{{e}^{-0.2x}}{-0.2}\right]}_{0}^{2}

=450×10−4[1−e−0.4]=180×10−4(Usinge−x=1−x)=\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})

P=WΔt=180×10−45×10−3=3.6 W.\text{P}=\dfrac{\text{W}}{\text{Δ}\text{t}}=\dfrac{\text{180}\times {\text{10}}^{-4}}{5\times {\text{10}}^{-3}}=\text{3.6 W.}

Hence, closest option is 2.97 W.
(if we used exact value of exponential term will get exact answer).

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