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

Answer

c

Explanation

Force on conductor can be calculated as
` vecF = vec (IL) xx vec B`
Here force acting on conductor can be written when it is at a distance x from origin :
`vecF = vec(IL) xx vecB = (10) (-3 hat k) xx (3 xx 10^(-4) e^(-0.2x)) hat j `
` = 9 xx 10^(-3) e^(-0.2x) hat i`
Let further displacement be `vec (dr) = dx hat i`
`dW = vec F * vec (dr) = 9 xx 10^(-3) e^(-0.2x) dx `
Total work done :
`W = 9 xx 10^(-3) underset(0)overset(2) int e^(-0.2x) dx `
`W = 9 xx 10^(-3) [(e^(-0.2x))/(-0.2)]_(0)^(2) = 0.045 (e^(-0.4) -1) `
` = - 0.045(0.670 -1) = 0.045(0.3297) = 0.01484`
Time interval given to us `Delta t = 5 xx 10^(-3) s`
Power , `p = W/(Delta t) = (0.01484)/(5 xx 10^(-3)) = 2.97` W
Hence, option (c ) is correct.