Consider two parallel conducting frictionless tracks kept in a gravity-free space as shown in Fig. 3.200. A movable conducting `PQ`, initially kept at `OA`, is given a velocity `10 ms^(-1)` toward right. The space contains a magnetic field which depends upon the distance moved by conductor `PQ` from the `OA` line and given by

`vec(B) = cx(-hat(k)) [c = constant = 1 SI unit]`
The mass of the conductor `PQ` is `1 kg` and length of `PQ` is 1m. Answer the following question based on the passage.
The work done by magnetic force acting on the conductor `PQ` during its motion in the time interval `t = 0 to t = t` second when the speed of conductor is `5 ms^(-1)` is

Options

A. (a) zero
B. (b) `50 J`
C. ( c) `10 J`
D. (d) `30 J`

Answer

A

Explanation

(a) Let anticlockwise direction is positive

`phi = BA cos 180^(@) = cxlx cos 180^(@) = -clx^(2)`
`e = -(dphi)/(dt) = cl2x(dx)/(dt), I = (e)/(R ) = (2clxv)/(1) = 2clxv`
This is positive so current is anticlockwise.
`F = m(-a) `rarr` IBl = -mv(dv)/(dx) `rarr` sc^(2)l^(2)x^(2)v = -mv(dv)/(dx)`
`rarr` `2c^(2)l^(2)int_(0)^(x)x^(2)dx = -int_(10)^(5)mdv `rarr` 2c^(2)l^(2)(x^(3))/(3) = m(5)`
`rarr` `x^(3) = (15m)/(2c^(2)l^(2)) `rarr` x((15)/(2))^(1)/(3)`
Heat produced = loss in `KE` of rod `= (1)/(2) xx 1[10^(2) - 5^(2)] = (75)/(2) = 37.5 J`
Magnetic force is not doing any work.