Take into account that the magnetic force on a current line is given by:
[tex]F_B=\text{ilBsin}\theta=\text{ilB}[/tex]where,
i: current = ?
l: length of conductor CL = 1m
B: magnitude of magnetic field = 0.1T
θ: angle between magnetic and current vector = 90 degrees
Furthermore, by using the Newton second law, you have:
[tex]\text{ilB-mg}=ma[/tex]where,
m: mass of the conductor = 200g = 0.2kg
a: acceleration of the conductor
When you solve the previous equation for i, you get:
[tex]i=\frac{ma+mg}{lB}=\frac{m(a+g)}{lB}[/tex]a) If the pipeline is rising at a constant speed, then a = 0m/s^2 and the current is:
[tex]i=\frac{0.2kg\cdot\frac{10m}{s^2}}{1m\cdot0.1T}=20A[/tex]b) For a = 2 m/s^2, the current on the conductor is:
[tex]i=\frac{(0.2kg)(\frac{10m}{s^2}+\frac{2m}{s^2})}{(1m)(0.1T)}=24A[/tex]and the current flows to the right.
c) If the pipeline descends with a constant speed, then, a = 0m/s^2. The current is:
[tex]i=\frac{0.2kg\cdot\frac{10m}{s^2}}{1m\cdot0.1T}=20A[/tex]d) If the pipeline descends with a constant acceleration a = 12m/s^2, then:
[tex]i=\frac{(0.2kg)(\frac{10m}{s^2}-\frac{12m}{s^2})}{1m\cdot0.1T}=-4A[/tex]and the current flows to the left.