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ES2C6 Week 6 Examples - qwcdwq

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Electromechanical System Design (ES2C6)

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ES2C6 Week 6 Examples: Electromagnetics

and Motors

Lecture 6: Relating Current, Magnetism and Motion

Question 1: The diagram below shows a current carrying wire suspended in free space. The current

is 10A, and the permeability of free space is 𝜇 0 = 4𝜋 × 10−7 𝐻 𝑚⁄ . State: the direction of the

current (into or out of the page), the direction of the flux lines (clockwise or anticlockwise) and the

magnitude of the B field (flux density) at 10 mm and 1000 mm from the wire:

Solution 1: There is a cross in the wire, telling us that current flows into the page. Using the right

hand grip rule, we can see that the direction of the field is clockwise.

The equation for the magnitude of the flux density is:

0

2

I

B

r

=

We are told the current, the permeability and are given two distances:

( )

( )

7

4

10mm 3

7

6

1000mm

4 10 10

2 10 T

2 10 10

4 10 10

2 10 T

2 1

r

r

B

B

   

= −

=

 

= = 

 

= = 

You can see the magnetic flux density decreasing the further away from the wire we get.

Question 2: The diagram below shows a current carrying wire suspended in free space. The current

is 0𝐴, and the permeability of free space is 𝜇 0 = 4𝜋 × 10−7 𝐻 𝑚⁄ . State: the direction of the

current (into or out of the page), the direction of the flux lines (clockwise or anticlockwise) and the

magnitude of the B field (flux density) at 1 mm and 10 mm from the wire:

Numerical Answer: B r =1mm = 3  10 − 5 T and Br=10mm= 3  10 − 6 T

####### Solution 2: There is a cross in the wire, telling us that current flows into the page. Using the right-

####### hand grip rule, we can see that the direction of the field is clockwise.

####### The equation for the magnitude of the flux density is:

0

2
I
B
r
=

####### We are told the current, the permeability and are given two distances:

( )

( )

7 5 1mm 3 7 6 10mm 3

4 10 0.
3 10 T
2 1 10
4 10 0.
3 10 T
2 10 10

r r

B
B

− − = − − − = −

 
= = 
 
= = 

####### You can see the magnetic flux density decreasing the further away from the wire we get.

####### Question 3: A 5𝑚 long wire is placed at an angle of 𝜃 = 90° to a uniform magnetic field of flux

####### density 𝐵 = 2 𝑇. The wire is carrying a current of 3 𝐴. Calculate the magnitude of the force on the

####### wire. In which direction is the force if the current flows upwards in the diagram? What is this force at

####### 𝜃 = 10° and 𝜃 = 30°.

####### Now we have the angular velocity, we can find the current:

0 29 0.
248 mA
0.
i b T L
K
 +  +
= = =

####### The equation for the efficiency is given in the lectures as:

( )( )

2 2 L L PMDC L

T K V KRT
V b T Rb K
=
+ +

####### This is an important equation which can be useful for optimisation and understanding how the

####### parameters affect the efficiency of the motor. However, when you are asked to calculate efficiency

####### and you know the current (as we do here), you can use this simpler method:

29 0.
20%
30 0.

out L PMDC in

P T
P Vi
= = = =

####### Question 6: You measure the armature resistance of a permanent magnet DC motor to be 15 Ω. At a

####### torque load 𝑇𝐿 = 0 Nm, you measure the steady state angular velocity as 𝜔 = 50 rad/s, and the

####### current as 𝑖 = 0 A, with at an input voltage of 𝑉 = 35 V. State the following:

  • The motor constant, 𝐾. (Numerical answer: 0 V/(rad/s))
  • The damping, 𝑏. (Numerical answer: 0 Nm/(rad/s))

####### Solution 6: This is a similar question to the one given in the lecture, however there is the slight

####### complication of a torque load. First, we need to find the motor constant, because it is used in the

####### calculation for the damping:

####### The steady state equation for the electrical behaviour of the PMDC motor is:

V = Ri +K e M

####### This is rearranged to solve for 𝐾𝑒:

( )

35 15 0.
0 V/ rad/s
50

e M

V Ri
K
− − 
= = =

####### Now the damping can be found. The steady state equation for the mechanical behaviour of the

####### PMDC motor (with a torque load) is:

K i T = b  M +TL

####### Which can be solved for the damping:

0 0 0 5 10 3 Nm/(rad/s)
50

T L M

K i T
b
= − =  − =  −

####### End of Questions

####### Mark Dooner, 2021/

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ES2C6 Week 6 Examples - qwcdwq

Module: Electromechanical System Design (ES2C6)

28 Documents
Students shared 28 documents in this course
Was this document helpful?
1
ES2C6 Week 6 Examples: Electromagnetics
and Motors
Lecture 6.1: Relating Current, Magnetism and Motion
Question 1: The diagram below shows a current carrying wire suspended in free space. The current
is 10A, and the permeability of free space is 𝜇0= 4𝜋 × 10−7 𝐻 𝑚
. State: the direction of the
current (into or out of the page), the direction of the flux lines (clockwise or anticlockwise) and the
magnitude of the B field (flux density) at 10 mm and 1000 mm from the wire:
Solution 1: There is a cross in the wire, telling us that current flows into the page. Using the right
hand grip rule, we can see that the direction of the field is clockwise.
The equation for the magnitude of the flux density is:
0
2
I
B
r
=
We are told the current, the permeability and are given two distances:
( )
( )
7
4
10mm 3
7
6
1000mm
4 10 10 2 10 T
2 10 10
4 10 10 2 10 T
21
r
r
B
B
=
=

==

==
You can see the magnetic flux density decreasing the further away from the wire we get.
Question 2: The diagram below shows a current carrying wire suspended in free space. The current
is 0.15𝐴, and the permeability of free space is 𝜇0= 4𝜋 × 10−7 𝐻 𝑚
. State: the direction of the
current (into or out of the page), the direction of the flux lines (clockwise or anticlockwise) and the
magnitude of the B field (flux density) at 1 mm and 10 mm from the wire: