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Basic Knowledge about Electromagnetism

  • Magnetism


The movement of a compass needle towards the North Pole and the attraction of a fridge magnet to the refrigerator are two examples of magnetism in our everyday lives. In this chapter we will examine a magnetic field in detail and begin to explore the sources of magnetism. We will also begin to see a connection between magnetism and electricity which, when developed further, is at the heart of one of the most successful theories in physics.

When we began to study electric forces we introduced the concept of the electric field, $\vec{E}$ . In particular, we had the fundamental relation between the force on a charge q in an electric field: $\vec{F}$ = q$\vec{E}$. We will introduce a magnetic field in a similar manner. Before doing this, however, we define what is known as the cross product $\vec{V_1}$ x $\vec{V_2}$ between two vectors $\vec{V_1}$ and $\vec{V_2}$ . Consider two such vectors in the plane of this page with an angle $\theta$ between them, as in Fig. 1.1.
  
Figure 1.1: Two vectors
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$\vec{V_1}$ x $\vec{V_2}$ , which itself is a vector, is defined as follows:

  • the magnitude of $\vec{V_1}$ x $\vec{V_2}$ is given by

     
    | $\displaystyle\vec{V_1}$ x $\displaystyle\vec{V_2}$ | = |$\displaystyle\vec{V_1}$|$\displaystyle\vec{V_2}$| sin $\displaystyle\theta$ $\displaystyle\equiv$ V1V2sin $\displaystyle\theta$, (1)

    where |$\vec{V}$| $\equiv$ V denotes the magnitude of a vector $\vec{V}$ .
  • $\vec{V_1}$ x $\vec{V_2}$ is a vector perpendicular to both $\vec{V_1}$ and $\vec{V_2}$ whose direction is given by the right-hand rule:
    Point the thumb of your right hand in the direction of $\vec{V_1}$ and your fingers in the direction of $\vec{V_2}$ . The direction of $\vec{V_1}$ x $\vec{V_2}$ then is directed out of the palm of your hand.
As the cross product of two vectors is three dimensional in nature, it is convenient to introduce the following notation for use on our two dimensional paper. We represent a vector perpendicular to the plane of this page which is directed out of the page as a circle with a black dot in the center - this can be thought of as the tip of an arrow coming towards you. On the other hand, a vector perpendicular to the plane of this page which is directed into the page is represented as a circle with a cross through the center - this can be thought of as the tail of an arrow heading away from you. Examples of this are in Fig. 1.2.
  
Figure 1.2: Cross product of two vectors
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We note two general features of the cross product:

  • $\vec{V_1}$ x $\vec{V_2}$ = - $\left[\vec {V_2}\times \vec {V_1}\right]$
  • $\vec{V_1}$ x $\vec{V_2}$ = 0 if $\vec{V_1}$ and $\vec{V_2}$ are parallel or antiparallel, as in these cases the angle $\theta$ between the two vectors is either 0 o or 180 o , for which the magnitude of $\vec{V_1}$ x $\vec{V_2}$ given by Eq.(1.1) vanishes.
We now consider a charge q moving at velocity $\vec{v}$ . If this charge is in a magnetic field $\vec{B}$ , then it experiences a force given by

 
$\displaystyle\vec{F}$ = q$\displaystyle\vec{v}$ x $\displaystyle\vec{B}$. (2)

From this we see that the units of $\vec{B}$ are N $\cdot$ s/(m $\cdot$ C), which are given the special name Tesla (T). We note that, analogous to the electric field, for a given velocity $\vec{v}$ and a given magnetic field $\vec{B}$ the force on a positive charge is in the opposite direction to the force on a negative charge.

Let us consider a charge q with velocity $\vec{v}$ entering a region of space with a constant magnetic field $\vec{B}$ . We assume that initially $\vec{v}$ and $\vec{B}$ are at right angles. The charge will experience the force of Eq.(1.2) which, by definition, is perpendicular to the velocity $\vec{v}$ . Because of this, the force does no work on the charge (recall W = Fdcos $\theta$ = 0 if $\theta$ = 90 o ), and because W = $\Delta$K = $\Delta$(${\frac{1}{2}}$mv 2) the speed of the charge will not change. It turns out that the charge will move in a circular motion, with a (centripetal) acceleration $\vec{a}_{c}^{}$ directed toward the center of the circle, as illustrated in Fig. 1.3.
  
Figure 1.3: Motion of a charged particle in a magnetic field
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For such a motion we have the following relation:

F = qvB = m$\displaystyle{\frac{v^2}{r}}$ $\displaystyle\Rightarrow$ r = $\displaystyle{\frac{mv}{qB}}$. (3)

This behaviour of a charged particle in a magnetic field is the principle behind machines such as mass spectrometers, which can be used to measure the masses of charged particles by measuring their radii of curvature in a magnetic field.

Let us consider a long straight wire carrying a current $\vec{I}$ in a magnetic field $\vec{B}$ . Each charge q in the wire will experience a force, and it is possible to find the total force on the wire by the following arguments. Suppose there are n charges per unit volume in the wire of cross-sectional area A and length l , as in Fig. 1.4.
  
Figure 1.4: Current carrying wire in a magnetic field
\begin{figure} \begin{center} \leavevmode \epsfxsize=3 in \epsfbox{/export/home/fyde/randy/figs/fig19-4.eps}\end{center}\end{figure}

If each charge has a charge q and is moving with a drift velocity $\vec{v}_{d}^{}$, then there will be a total force on the wire given by

 
$\displaystyle\vec{F}$ = (nAlq)$\displaystyle\vec{v}_{d}^{}$ x $\displaystyle\vec{B}$ = l$\displaystyle\vec{I}$ x $\displaystyle\vec{B}$, (4)

where we recall that $\vec{I}$ = nq$\vec{v_d}$A and also that by convention $\vec{I}$ is in the direction of the flow of positive charges.

The fact that a force exists on a wire carrying a current in a magnetic field has an interesting application when we consider a current carrying loop. Consider a rectangular loop carrying a current $\vec{I}$ in a magnetic field $\vec{B}$ as in Fig. 1.5.
  
Figure 1.5: Torque on a current loop in a magnetic field
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We consider the force $\vec{F}$ = l$\vec{I}$ x $\vec{B}$ of Eq.(1.4) on each of the four sides of the loop. On the top and bottom sections this force vanishes, because $\vec{I}$ and $\vec{B}$ are parallel or antiparallel in these cases. The force $\vec{F}_{L}^{}$ on the left section will have a magnitude aIB in the direction indicated, while the force $\vec{F}_{R}^{}$ on the right section will have the same magnitude aIB but in the opposite direction. These two forces, since they are oppositely directed, do not give rise to a net linear acceleration, but they do tend to rotate the loop around the vertical axis. There will thus be a net torque $\tau$ on the loop, which is given by

$\displaystyle\tau$ = FL$\displaystyle{\frac{b}{2}}$ + FR$\displaystyle{\frac{b}{2}}$ = aIBb = BIA, (5)

where A = ab is the area of the loop.

Although we have considered a rectangular loop, the relation $\tau$ = BIA holds for an arbitrarily shaped loop of area A . The fact that current carrying loops experience a net torque in a magnetic field is the principle behind the electric motor, where the electrical energy involved in establishing a current is converted into the mechanical energy of rotating a shaft.

When we introduced the electric field $\vec{E}$ it was apparent that electric charges were the source of such a field. Experiments in the 19th century showed that the source of a magnetic field $\vec{B}$ was a moving charge, or current. A detailed mathematical relation between a charge moving at velocity $\vec{v}$ and the associated magnetic field $\vec{B}$ is known as Amp¨¨re's law or, in another form, the Biot-Savart law. We will consider three special cases of the results of this law.

We first consider a very long straight wire carrying a current I . In Fig. 1.6 we illustrate this for a wire with current coming out of the page.
  
Figure 1.6: Magnetic field of a long straight wire
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At a distance r from the wire, as shown, the magnitude of the magnetic field is given by

 
B = $\displaystyle{\frac{\mu_0I}{2\pi r}}$, (6)

where the constant $\mu_{0}^{}$ is called the permeability of free space, and is given (exactly) by

 
$\displaystyle\mu_{0}^{}$ = 4$\displaystyle\pi$ x 10- 7 T $\displaystyle\cdot$ m / A . (7)

We will see shortly the reason why this is an exact number. The direction of the magnetic field of this wire is indicated in the above figure, and can be remembered by the following rule:
Point the thumb of your right hand in the direction of the current along the wire - the curl of your fingers then indicates the direction of the magnetic field.
As a second example of the magnetic field of a moving charge, we consider a circular loop of radius r carrying a current I , as in in Fig. 1.7.
  
Figure 1.7: Magnetic field of a current loop
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At the center of the loop, the magnitude of the magnetic field is given by

 
B = $\displaystyle{\frac{\mu_0I}{2 r}}$, (8)

and the direction of the magnetic field indicated can be remembered by the following rule:
Curl the fingers of your right hand in the direction of the current around the loop - your thumb then indicates the direction of the magnetic field.
Note that this is the magnetic field just at the center of the loop, and away from the center the magnetic field changes in both magnitude and direction.

The association of a magnetic field with a current loop enables us to understand qualitatively the formation of permanent magnets. At the atomic level materials are composed of essentially stationary nuclei around which electrons orbit. The orbiting electrons can be considered as current loops, and thus each atom has its own magnetic field. In non-magnetic materials the magnetic fields of all the atoms are randomly oriented, resulting in no net magnetic field, but in permanent magnets interactions between the atoms favour the individual atomic magnetic fields to be aligned, producing a net macroscopic magnetic field.

An extension of the previous case of a single current loop is to consider a large number N of such loops tightly packed together over a distance L - such a device is called a solenoid. This is illustrated in Fig. 1.8.
  
Figure 1.8: A solenoid
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This device has many important applications for two reasons: one is that fairly strong magnetic fields can be produced by using a large number of turns, and secondly, the magnetic field is fairly constant in magnitude and direction throughout the solenoid, except near the ends. The magnitude of the magnetic field of a solenoid is given by

B = $\displaystyle\mu_{0}^{}$$\displaystyle{\frac{N}{L}}$I $\displaystyle\equiv$ $\displaystyle\mu_{0}^{}$nI, (9)

where n = N/L is the number of turns per unit length. The direction of the magnetic field is given by the same rule for determining the direction of a magnetic field of a single current loop.

An interesting effect occurs if we consider two long straight parallel wires separated by a distance d carrying currents I1 and I2 . Let us examine the case where the two currents are in the same direction, as in Fig. 1.9.
  
Figure 1.9: Force between two long straight parallel wires
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Wire #2 will experience a magnetic field of Eq.(1.6) due to wire #1 given by

B1 = $\displaystyle{\frac{\mu_0I_1}{2\pi r}}$ (10)

in the direction indicated, and hence will experience a force per unit length given by Eq.(1.4):

 
$\displaystyle{\frac{F_2}{l}}$ = $\displaystyle{\frac{\mu_0I_1I_2}{2\pi d}}$. (11)

The direction of $\vec{F}_{2}^{}$ indicated shows that wire #2 will be attracted towards wire #1. In a similar manner, one can show that wire #1 will experience a force due to the magnetic field $\vec{B}_{2}^{}$ of wire #2, and that this force $\vec{F}_{1}^{}$ will have a magnitude equal to that of F2 given in Eq.(1.11) but opposite in direction. Thus, wire #1 will be attracted towards wire #2.

It is a good exercise to show that if the wires were carrying currents in the opposite directions that the resulting forces will have the same magnitude as in Eq.(1.11) but are such as to cause a repulsion between the wires.

This force between two current carrying wires gives rise to the fundamental definition of the Amp¨¨re:

If two long parallel wires 1 m apart each carry a current of 1 A, then the force per unit length on each wire is 2 x 10- 7 N/m.
This definition of the Amp¨¨re then gives rise to the basic definition of the unit of charge, the Coulomb:
A wire carrying a current of 1 A transports past a given point 1 C of charge per second.

This definition also explains the reason why the constant $\mu_{0}^{}$ of Eq.(1.7) was given exactly as 4$\pi$ x 10- 7 T $\cdot$ m/A.

% latex2html id marker 1258 $\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

What is the net force (magnitude and direction) on the electron moving in the magnetic field in Fig. 1.10 if B = 2 T, v = 4 x 104 m/s, and $\theta$ = 30 o ?

  
Figure 1.10: Point charge in a magnetic field
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Solution:
We first find the magnitude of the force as

F = qvBsin $\displaystyle\theta$ = 1.6 x 10- 19 $\displaystyle\cdot$ 4 x 104 $\displaystyle\cdot$ 2 $\displaystyle\cdot$ sin 30 o = 6.4 x 10- 15 N .

To determine the direction, we note that $\vec{v}$ x $\vec{B}$ is directed out of the page, and so the force on a negative charge is opposite to this. Thus, the force on the electron is 6.4 x 10- 15 N directed into the page.



% latex2html id marker 1273 $\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Suppose a very long straight wire of linear mass density 20 g/m is immersed in a constant magnetic field B = 3 T, as in Fig. 1.11. What current I would be required so that the wire will be suspended?

  
Figure 1.11: Long straight wire in a magnetic field
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Solution:
We note that the force per unit length due to gravity,

$\displaystyle{\frac{F_g}{l}}$ = $\displaystyle{\frac{m}{l}}$g,

is directed downwards, while the force per unit length due to the magnetic field,

$\displaystyle{\frac{F_B}{l}}$ = IB,

is acting upwards. In order that they balance, we demand

$\displaystyle{\frac{m}{l}}$g = IB $\displaystyle\Rightarrow$ I = $\displaystyle{\frac{m}{l}}$$\displaystyle{\frac{g}{B}}$ = $\displaystyle{\frac{0.02\times 9.8}{3}}$ = 0.065 A .

Thus, the wire must carry a current of 0.065 A in order to be suspended.



% latex2html id marker 1288 $\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

Consider the mass spectrometer in Fig. 1.12. The electric field between the plates of the velocity selector is E = 950 V/m, and the magnetic field B in both the velocity selector and in the deflection chamber has a magnitude of 0.9 T. Find the radius r for a singly charged ion of mass m = 2.18 x 10- 26 kg in the deflection chamber.
  
Figure 1.12: Mass spectrometer
\begin{figure} \begin{center} \leavevmode \epsfxsize=2.0 in \epsfbox{/export/home/fyde/randy/figs/figpr19-3.eps}\end{center}\end{figure}

Solution:
We first use the fact that in order for the ion to pass through the velocity selector undeflected the force due to the electric and magnetic fields must balance. Note that the force on a positive charge is downwards due to the electric field and upwards due to the magnetic field; to balance, we then have

qE = qvB $\displaystyle\Rightarrow$ v = $\displaystyle{\frac{E}{B}}$.

When the ion enters the deflection chamber it experiences a magnetic force, causing it to go in a circular orbit. The magnetic force then gives rise to the centripetal acceleration as

F = qvB = m$\displaystyle{\frac{v^2}{r}}$ $\displaystyle\Rightarrow$ qB = $\displaystyle{\frac{mv}{r}}$.

We thus find

r = $\displaystyle{\frac{m}{qB}}$v = $\displaystyle{\frac{mE}{qB^2}}$ = $\displaystyle{\frac{2.18\times 10^{-26}\cdot 950}{1.6\times 10^{-19}\cdot (0.9)^2}}$ = 1.6 x 10- 4 m .

Thus, the radius of the orbit is 0.16 mm.


% latex2html id marker 1305 $\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

The two long straight wires in Fig. 1.13 each carry a current of I = 5 A in opposite directions and are separated by a distance d = 30 cm. Find the magnetic field a distance l = 20 cm to the right of the wire on the right.
  
Figure 1.13: Magnetic field of two long straight wires
\begin{figure} \begin{center} \leavevmode \epsfxsize=2.0 in \epsfbox{/export/home/fyde/randy/figs/figpr19-4.eps}\end{center}\end{figure}

Solution:
We shall use the expression for the magnetic field of a long straight wire a distance r away as

B = $\displaystyle{\frac{\mu_0I}{2\pi r}}$,

with direction given by the appropriate right-hand-rule. Due to the wire on the right, the magnetic field is

BR = $\displaystyle{\frac{\mu_0I}{2\pi l}}$ = $\displaystyle{\frac{4\pi\times 10^{-7}\cdot 5}{2\pi\cdot 0.1}}$ = 1.0 x 10- 5 T ,

which is directed out of the page. Due to the wire on the left, the magnetic field is

BL = $\displaystyle{\frac{\mu_0I}{2\pi (l+d)}}$ = $\displaystyle{\frac{4\pi\times 10^{-7}\cdot 5}{2\pi\cdot 0.5}}$ = 2.0 x 10- 6 T ,

which is directed into the page. Thus, the net magnetic field is 1.0 x 10- 5 - 2.0 x 10- 6 = 8.0 x 10- 6 T directed out of the page.



% latex2html id marker 1320 $\fbox{\bf PROBLEM \thechapter.\arabic{probcount}}$

For the arrangement shown in Fig. 1.14, the long straight wire carries a current of I1 = 5 A. This wire is a distance d = 0.1 m away from a rectangular loop of dimensions a = 0.3 m and b = 0.4 m which carries a current I2 = 10 A. Find the net force exerted on the rectangular loop by the long straight wire.

  
Figure 1.14: Force on a current loop due to a long straight wire
\begin{figure} \begin{center} \leavevmode \epsfxsize=2.0 in \epsfbox{/export/home/fyde/randy/figs/figpr19-5.eps}\end{center}\end{figure}

Solution:
The force $\vec{F}$ = l$\vec{I}$ x $\vec{B}$ on the rectangular loop arises due to the magnetic field B = $\mu_{0}^{}$I/2$\pi$r of the long straight wire. We first of all note that the force cancels between the top and bottom portions of the loop. Now, on the left portion of the loop,

BL = $\displaystyle{\frac{\mu_0I_1}{2\pi d}}$,

directed into the page, which gives rise to a force

FL = bI2BL = $\displaystyle{\frac{bI_2\mu_0I_1}{2\pi d}}$ = $\displaystyle{\frac{0.4\cdot10\cdot4\pi\times10^{-7}\cdot 5}{2\pi\cdot 0.1}}$ = 4 x 10- 5 N ,

which is directed to the left. On the right portion of the loop,

BR = $\displaystyle{\frac{\mu_0I_1}{2\pi (a+d)}}$,

directed into the page, which gives rise to a force

FR = bI2BR = $\displaystyle{\frac{bI_2\mu_0I_1}{2\pi (a+d)}}$ = $\displaystyle{\frac{0.4\cdot10\cdot4\pi\times10^{-7}\cdot 5}{2\pi\cdot (0.3+0.1)}}$= 1 x 10- 5 N ,

which is directed to the right. Thus, the net force is 4 x 10- 5 - 1 x 10- 5 = 3 x 10- 5 N directed to the left.