A conductor lies along the ` z`-axis at ` -1.5 le z lt 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 xx 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 `5xx 10^(-3)s` . Assume parallel motion along the ` x-axis`.

Options

A. `14.85 W`
B. `29.7 W`
C. `1.57 W`
D. `2.97 W`

Answer

D

Explanation

Here, `l = 1.5 + 1.5 = 3.0 m , I = 10.0 A`,
Average power `= (w o r k)/(t I m e)`
`W = int_(0)^(2) F dx = int_(0)^(2) BI l dx`
`= int_(0)^(2) 3.0 xx 10^(-4) e^(-0.2 x) xx 10 xx 3 dx`
`9 xx 10^(-3) xx int_(0)^(2) e^(-0.2x) dx`
`= 9 xx 10^(-3) [(e^(-0.2x))/(-0.2)]_(0)^(2) = 9 xx 10^(-3) [(e^(-0.2 xx 2) - 1)/(-0.2)]`
`= (9 xx 10^(-3))/(0.2) J xx [1 - r^(-0.4)] = (9 xx 10^(-3))/(0.2) xx 0.33`
`W = (2.97 xx 10^(-3))/(0.2) j`
`P = (W)/(t) = (2.97 xx 10^(-3))/((02) xx 5 xx 10^(-3)) = 2.97 W`