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=(ma+mg)/(lB)=(m(a+g))/(lB)`

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

`i=(0.2kg\cdot(10m)/(s^2))/(1m\cdot0.1T)=20A`

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

`i=((0.2kg)((10m)/(s^2)+(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=(0.2kg\cdot(10m)/(s^2))/(1m\cdot0.1T)=20A`

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

`i=((0.2kg)((10m)/(s^2)-(12m)/(s^2)))/(1m\cdot0.1T)=-4A`

and the current flows to the left.