Excercises Electric Machinery Fundamentals

Chapter 1

Problem 1-9 to 1-11



In [16]:

    
%pylab inline
%precision 4









    



Populating the interactive namespace from numpy and matplotlib






    Out[16]:





'%.4f'

Problem 1-9

Description

A wire is shown in Figure P1-6 below which is carrying 2.0 A in the presence of a magnetic field.

Calculate the magnitude and direction of the force induced on the wire.



In [17]:

    
B = 0.5  #[T]
i = 2.0  #[A]
l = 1.0  #[m]

SOLUTION

The force on this wire can be calculated from the equation $$ \vec{F} = i(\vec{l}\times\vec{B}) = ilB$$



In [18]:

    
F = i*l*B 
F









    Out[18]:





1.0000

Newton, into the page.

Problem 1-10

Description

The wire is shown in Figure P1-7 is moving in the presence of a magnetic field.

With the information given in the figure, determine the magnitude and direction of the induced voltage in the wire.

SOLUTION



In [19]:

    
v = 10.0    # [m/s]
B =  0.2    # [T] 
l =  0.25   # [m]
theta = 45  # [deg]

The induced voltage on this wire can be calculated from the equation shown below. The voltage on the wire is positive downward because the vector quantity $\vec{v}\times \vec{B}$ points downward. $$ e_\text{ind} = (\vec{v}\times \vec{B}) \cdot \vec{l} = vBl\cos(\theta)$$



In [20]:

    
e_ind = v*B*l* cos(deg2rad(theta))
e_ind









    Out[20]:





0.3536

Volt, positive down.

Problem 1-11

Description

Repeat Problem 1-10 for the wire in Figure P1-8 below:



In [21]:

    
v = 1.0    # [m/s]
B =  0.5   # [T] 
l =  0.5   # [m]
theta = 90 # [deg]

SOLUTION

The induced voltage on this wire can be calculated from the equation shown below. The total voltage is zero, because the vector quantity $\vec{v}\times \vec{B}$ points into the page, while the wire runs in the plane of the page.

$$ e_\text{ind} = (\vec{v}\times \vec{B}) \cdot \vec{l} = vBl\cos(\theta)$$



In [22]:

    
e_ind = v*B*l* cos(deg2rad(90))
e_ind









    Out[22]:





0.0000

Volts.