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← JEE Main 2014 · 6 Apr · Shift 1
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JEE Main 2014 · 6 Apr · Shift 1, Physics question 19

A conductor lies along the z-axis at −1.5≤z<1.5-1.5\le z<1.5 m and carries a fixed current of 10.0 A in −a^z-\hat a_z direction (see figure). For a field B⃗=3.0×10−4 e−0.2x a^y\vec B=3.0\times10^{-4}\,e^{-0.2x}\,\hat a_y T, find the power required to move the conductor at constant speed to x=2.0x=2.0 m, y=0y=0 m in 5×10−35\times10^{-3} s. Assume parallel motion along the xx-axis.

Axes x (out, to the lower left), y (right) and z (up). A conductor along the z-axis from z = −1.5 to 1.5 carries current I downward (−z); the field B points along +y. Dashed lines show the conductor moved parallel to itself to x = 2.0.
Choose one option
Answer and worked solution

Official answer: B

The force on the conductor (length 3 m) is F=BIL=3×10−4e−0.2x×10×3=9×10−3e−0.2xF=BIL=3\times10^{-4}e^{-0.2x}\times10\times3=9\times10^{-3}e^{-0.2x} N, along xx.

Work from x=0x=0 to 22:
W=9×10−3∫02e−0.2xdx=9×10−30.2(1−e−0.4)=0.045×0.330=1.48×10−2 J.W=9\times10^{-3}\int_0^2e^{-0.2x}dx=\frac{9\times10^{-3}}{0.2}\left(1-e^{-0.4}\right)=0.045\times0.330=1.48\times10^{-2}\text{ J}.
Average power: P=1.48×10−25×10−3≈2.97P=\dfrac{1.48\times10^{-2}}{5\times10^{-3}}\approx2.97 W.

Topics:Force on a current-carrying conductorPower
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