Solving Problems on Magnetic Fields Vector
According to electromagnetic theory, any conductor which has current flow
through it will produce magnetic field. For a typical straight wire conductor, the
magnetic field that is produced is shown below :
Figure 3-7 (p. 102)
Oersted’s experiment with a compass placed in several positions in close proximity
Figure 1 the cross section
to a current-carrying wire. The inset shows  used to represent
for current coming out of the paper: this represents the head of an arrow. A 
symbol would represent the feathered end of an arrow and would correspond to a
current heading into the paper.
Fundamentals of Electromagnetics With Engineering Applications by Stuart M. Wentworth
The direction
of magnetic field
lines© is
determined
by right
hand
thumbs up rules.
Copyright
2005
by John Wiley & Sons.
All rights
reserved.
Direction pointed by the thumb is the direction of current, while the direction of
magnetic fields is shown by the other four fingers.
Figure 2

Most of the equation for magnetic field intensity vector H has vector unit â 

which indicates that H is in  direction. Why is it so? Remember again that
there are three types of co-ordinate system i.e. Cartesian, cylindrical and
spherical (please refer to your vector calculus notes). A cylindrical co-ordinate
system is shown in Figure 3. Any point P in a cylindrical system is expressed in
term of ρ, Ф and z.
ρ is distance between point P and z axis where it is actually the radius of the
cylinder. z is the location of point P at z-axis, whereas Ф is the angle between
positive x-axis and the line which join point P to z-axis. Direction for Ф is anticlockwise from positive x-axis to point P.
The direction of unit vector for point P is indicated by az, aФ and aρ .
Figure 3
Figure 4
Now, suppose we have an infinite straight wire on z-axis which produce magnetic
field as shown in Figure 4 above. The magnetic field forms a circular loop around
the wire with direction according to the right hand thumbs up rules. In the case of
figure 4 above, current moving along positive z-axis produces magnetic field in
anti-clockwise direction.
If we put the wire above in a cylindrical co-ordinate system, comparing Figure 4
with Figure 3, we could see that the direction of magnetic field intensity is along
aФ . So that is why the equation for vector of magnetic field intensity shows that
the vector is in aФ direction.
The equation for MAGNITUDE of magnetic field intensity for infinite wire is given
by :
H
I
2
 (1)
Based on this equation, we can say that the magnitude of magnetic field
intensity, H, is uniform at any point located at position ρ from the conductor. To
illustrate this let us consider Figure 4 again. Imagine that we view Figure 4 from
top of z-axis, which looks like Figure 5.
Figure 5
Magnetic field line, H, is seen ‘orbiting’ the wire in anti-clockwise direction.
Magnitude of magnetic field intensity is different for different value of ρ.
According to equation (1), H will decrease if the location of ρ goes further from
the wire.
Consider Figure 5 again. Suppose we choose 4 points i.e. a, b, c and d which are
located on the same ‘track’ of H. Since the magnetic field line is a full circle, the
value of ρ is the same for each of the point. Thus, magnitude of magnetic field
intensity are the same at point a, b, c and d. Please note that we are talking
about MAGNITUDE! If it is a vector, the direction should be taken into
consideration.
For infinite wire, the VECTOR of magnetic field intensity is :

I
H
aˆ
2
 (2)

Applying equation (2) to find H at point a, b, c and d in Figure 5 will give the
same result since the magnitude is the same and the direction is â . So how can

we differentiate the location of point a, b, c and d? Given H at each of the point,
how could we know that they are actually different point in the same magnetic
filed line?
Going back to Figure 3, we notice that point P on cylindrical co-ordinate can be
divided into three unit vector component i.e. az, aФ and aρ . The same concept
can be applied to magnetic field line in Figure 5 only that this time we only have
aФ component (see Figure 6).
Figure 6
These vector units in Ф direction can be divided into Cartesian co-ordinate
components i.e. ax, ay and az . For the case of Figure 6, since the aФ is in x-y
plane, there is only ax and ay component need to be considered (see Figure 7).
Figure 7
After figuring out the x and y component of the unit vector aФ graphically, we then
calculate the value. Figure 8 and Figure 9 illustrates how to solve the unit vector
component.
Unit vector aФ is always tangent of the point on the magnetic field line, thus it is
perpendicular to the line connecting the point and the wire (examine Figure 7 and
Figure 8 again carefully to understand this!).
The expression for unit vector at point a is :
aФ = - ax cos (90O-Ф) + ay sin (90O-Ф)
= - ax sin Ф + ay cos Ф
Please note that aФ is a unit vector with magnitude of 1 (one). See Figure 9 to
understand how we obtain the expression for aФ in cartesian form. Figure 9
shows example for finding the component of aФ at point a.
The plus(+) and minus(-) sign for component ax and ay is determined by the
direction of that component. If the direction is towards positive x or y, then the
sign will be plus and vice versa.
Similarly for other points,
Point b : aФ = - ax cos Ф - ay sin Ф
Point c : aФ = ax sin Ф - ay cos Ф
Point d : aФ = ax sin Ф + ay cos Ф
Figure 8
Figure 9
For problems where we have to find the net value of two or more magnetic field
added together, we apply the steps described above. Find the Cartesian coordinate component of the unit vector aФ for each magnetic field line and add all
the vectors involved. Question 2 in the quiz is a typical example for this type of
problem.
As an exercise, please do again Question 2 and Question 3 from the quiz based
on your understanding on this note. The working might be different from the one
given in the quiz’s solution but you will still obtain the same answer.
Achtung!
If you still unable to understand any of this (You should understand it! It has been
covered in Engineering Maths and elementary physics back in your school days),
please feel free to see your lecturer at any time (if you are fortunate enough to
catch him in his room!) Remember! If you don’t see your lecturer, then they’ll
assume you understand perfectly and would succeed in the exam.
Ahmad Nasrul – Human System Lab