Question

A linear machine is connected as shown in Fig. 1. The input source voltage Vin=10V, the internal resistance Rin=0.05Ω. The applied force F=10N. The magnetic field density B=0.5T. The rail width l=200mm. (50 pts)

Answer 1

It is important to note that the direct of flow of the electricity current in the Linear Machine is anti-clockwise direction. The resultant effect of this is such that is stand at variance with the change in flux as force is going the right direction. See attached Image (Illustration I).

What is the magnitude of the current?

Recall that Force (f) = il xB

Given that Force (F) = 10N; and

Magnetic Field Density B = 0.5T

Hence,

10 = i x 0.2 x 0.5  

[Note that Width is converted to Meters from millimeters that is 200 mmm = 0.2m]

10 = 0.1i

i = 10/0.1

i = 100A

What is the Induced Voltage Across the metal bar?

Note that Vg = Vde + IR

Vg = 10 + 100 x 0.05

Vg = 10 + 5

Vg = 15v

What is the final speed of the metal bar?

The final speed of the bar is given as V

Recall that Induced Voltage is VB x L

That is:

15 = V x 0.5 x 0.2;

15 = 0.1 V

make V the subject of the formula:

V = 15/0.1

V = 150m/s

What is the status of the Machine? a motor or generator?

Given that the current being supplied is  in DC hence the machine is a generator. [Vdc x i]

What is the electric power of the source?

The electric power of the source is:

Vdc x i

= 10 x 100

= 1,000 watts

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