Question

the conductor (CL) can slide without friction on the vertical conductors which are made of insulated material. The mass of the conductor is m = 200gr the length of l = 1m the intensity of the gravitational field g = 10 m / s ^ 2, while the intensity of the homogeneous magnetic field B = 0.1T Calculate the current intensity and the direction in KA pipeline so that: a) the pipeline is rising at a constant speed, b) the pipeline rises with a constant acceleration a = 2 m / s ^ 2 c) the pipeline descends at a constant speed; d) the pipeline descends with constant acceleration a = 12 m / s. 2.

Answer 1

Take into account that the magnetic force on a current line is given by:

$F_B=\text{ilBsin}\theta=\text{ilB}$

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:

$\text{ilB-mg}=ma$

where,

m: mass of the conductor = 200g = 0.2kg

a: acceleration of the conductor

When you solve the previous equation for i, you get:

$i=\frac{ma+mg}{lB}=\frac{m(a+g)}{lB}$

a) If the pipeline is rising at a constant speed, then a = 0m/s^2 and the current is:

$i=\frac{0.2kg\cdot\frac{10m}{s^2}}{1m\cdot0.1T}=20A$

b) For a = 2 m/s^2, the current on the conductor is:

$i=\frac{(0.2kg)(\frac{10m}{s^2}+\frac{2m}{s^2})}{(1m)(0.1T)}=24A$

and the current flows to the right.

c) If the pipeline descends with a constant speed, then, a = 0m/s^2. The current is:

$i=\frac{0.2kg\cdot\frac{10m}{s^2}}{1m\cdot0.1T}=20A$

d) If the pipeline descends with a constant acceleration a = 12m/s^2, then:

$i=\frac{(0.2kg)(\frac{10m}{s^2}-\frac{12m}{s^2})}{1m\cdot0.1T}=-4A$

and the current flows to the left.