Chapter 21

__Magnetic Field:__

Magnetism is the result of electric charge motion.** As
soon as an electric charge moves, it creates a magnetic effect perpendicular to
its direction of motion.** This can be easily verified by the a

**The Right Hand Rule (RHR):**

If you hold your right hand such that the four bent fingers are in the direction
of the * positive current I*, then your thumb will point to
the

^{
}

Figure 1

The reason for naming the poles of the nail as * N *and

__Magnetic Field Lines:__

*
Magnetic field lines* are
generally accepted to

**Magnetic field lines around a bar-magnet. By convention, Field lines
emerge from the North pole of the magnet and enter its South pole.**

Figure 2

** It is not possible to separate the North and South poles of a magnet. They
coexist.** If a bar magnet is cut at its middle (its neutral line), each
piece becomes an independent magnet possessing a South pole and a North pole

__The Theory Behind Magnetism:__

As we know, atoms are made of negative electrons, positive
protons, and neutral neutrons**.** Protons and neutrons have almost the same
mass but are much heavier than electrons**.** Protons and neutrons form the
nucleus**.** Electrons orbit the nucleus**.** * Electrons are* considered
the

Visualize a single electron orbiting the nucleus of its
atom**.** For simplicity, visualize a sphere in which this electron spins at
a rate of say 10^{15} rotations
per second**.** This means one
thousand trillion turns every second**.**
Therefore, at every one-thousand-trillion**th** of
a second it possesses a particular plane of rotation in space**.** In other
words, the orientation of its plane of rotation changes roughly 10^{15 }times
every second**.** That is why we say it creates an electron cloud**.**
Only three of such orientations are sketched below**.** For each plane of
rotation, the magnetic field vector is shown to have its maximum effect at the
center of the circle of rotation and perpendicular to that circle**.**

Figure 3

An object of mass **1kg,** for
example, contains a very large number of atoms, and each atom depending on the
element contains several electrons and each electron at any given instant of
time has its own orientation of rotation and its own direction of magnetic field
vector**.** We are talking about hundreds of trillion trillions different
magnetic field vectors in a piece of material**.** There is no guarantee
that all such electrons have their magnetic field vectors oriented in a single
direction so that their magnetic effects add up**.** An **orbital **is
a space around a nucleus where the possibility of finding electron is high**.** An
orbital can be spherical, dumbbell-shaped, or of a few other geometric shapes**.**
We assumed a spherical orbital for simplicity**.** Each orbital can be
filled with 2 electrons**.** The two electrons in each orbital have opposite
rotation directions**.** This causes their magnetic field vectors
to have
opposite directions as well**.** The result is a zero net magnetic effect**.** This
way, each atom that contains an even number of electrons will have all of its
orbitals filled with pairs of electrons**.** Such atoms are magnetically
neutral**.** However, atoms that contain odd numbers of electrons will have
an orbital that is left with a single electron**.**
Such atoms are not magnetically neutral by themselves**.** They become
magnetically neutral, when they form molecules with the same or other atoms**.**
There are a few elements such as iron, cobalt, and nickel that have a particular
atomic structure**.** This
particular structure allows some outer shells orbitals to have single (unpaired)
electrons**.** Under normal circumstances, there is no guarantee that all
orbitals of the atoms in a piece of iron, for example, to have their magnetic
field vectors lined up parallel to each other**.** But if a piece of pure
iron is placed in an external magnetic field, the planes of rotation of those
single electrons line up in a manner that their magnetic fields line up with the
direction of the external field, and after the external field is removed, they
tend to keep the new orientation and therefore the piece of pure iron becomes a
magnet itself**.** The **overall
conclusion** however is the fact
that *magnetism is the
result of electric charge motion and that the magnetic field vector is
perpendicular to the plane of rotation of oriented charged particles *(*mainly
electrons*).

__Like poles and Unlike poles:__

**Like poles of two magnets repel and their unlike poles
attract.** This is similar to electric charges**.** Recall that
like charges repel and unlike charges attract**.** One difference is that **separate
positive and negative charges are possible, but separate magnetic North poles
and South poles are not. ****See
Fig. 4.**

Figure 4

__Uniform Magnetic Fields:__

The magnetic field lines around a ** bar
magnet **is

Figure 5

__The Force of a Magnetic Field on a Moving Charge:__

When a moving charge enters a magnetic field such that magnetic field lines are
crossed, the charge finds itself under a force perpendicular
to its direction of motion that
gives it a circular motion**.** When a charge enters a magnetic field such that
its motion direction stays parallel to the field lines and no field line is
crossed, then the charge will not be affected by the field and keeps going
straight**.**

Imagine you are in a classroom facing the board and **visualize
a downward uniform
magnetic field B similar
to a rainfall.**

Also, visualize a **+*** q
charge* entering

Figure 6

*In general, with **B downward**,
and** **q moving to
the right at **velocity **V**,
the magnitude of F that
initially pushes
**q toward the board* is given
by:

If |

** q **can be
(

The SI unit
for magnetic field strength is "B"Tesla shown
by T.
a that
exerts a of
force magnetic field on a 1N charge 1C the
charge crosses the field at ifto its lines1m/s
normal . |

**Example 1: ** A
14μC charge enters a 0**.**030**T** magnetic
field at a speed of 1**.**8x10^{5}m**/**s normal to its
field lines**.** Find the magnitude of
the force on it**.**

**Solution:** ** F = qvB ; F =** 14μC(1

**Example 2:** A 14μC
charge moving to the left at 1**.**8x10^{5}m**/**s
enters a downward 0**.**030**T **magnetic
field**.** Its trajectory makes a 65^{o} angle
with respect to the field lines**.** Find the magnitude and direction of
the initial force on it as soon as it crosses a field line**.** Refer to **Fig.
6a.**

**Solution: **Referring to **Fig.
6a****,** it is clear that the charge
will initially be pushed away from us (the observers)**.** The magnitude of
this initial push is

** F = qvBsin θ ; F =** 14μC(1

**Example 3: ** An
electron enters a 0**.**013T magnetic field normal to its field lines and
experiences a 3**.**8x10^{-15}N force**.** Determine (a) its
speed and (b) its acceleration**. **The
mass and charge of electron are**:** ** M_{e} =** 9

**Solution: F =qvB ; v
=** F

*
F = Ma*

__Motion of a charged Particle in a Magnetic Field:__

So far, we have learned that when a charged particle crosses magnetic field
lines, it is forced to change direction**.** This change of direction does
not stop as long as there are field lines to be crossed in its pathway of motion**.** Magnetic
field lines keep changing the direction of motion of the charged particle and
if the field is constant in magnitude and direction, it
gives a circular motion to the charged particle**.** The reason
is that * F* is
perpendicular to

Figure 7

** This formula **is useful in
finding the radius of rotation of a charged particle caught in a magnetic field

**Example 4: ** A
proton * q
= e^{+}* =
1

**Solution:** *R = MV*/(*qB*)* ***;**
solving for *V***:**

** V =** RqB

**Example 5: ** In
a certain device, alpha-particles enter a 0**.**88-T magnetic field
perpendicular to its field lines**.** Find the radius of rotation they
attain if each carries an average kinetic energy of 520 ** keV.** An
alpha-particle is a helium nucleus

**Solution:** Since the * K.E.* of
each alpha-particle is given, knowing its mass (2M

**K.E. = (1/2)**Mv^{2} **;**
520(1**.**6x10^{-16}J) **=** **(1/2)**[2(1**.**672x10^{-27}kg)
**+**2(1**.**674x10^{-27}kg)] *V** ^{2}*.

** V** = 5

**R =**[2(1**.**672x10^{-27}kg)**+**2(1**.**674x10^{-27}kg)**]**(5**.**0x10^{6}m/s)**/****[**(**2**x1**.**6x10^{-19}C)(0**.**88T)**]** **=
0.12m.**

Note that each alpha-particle has **2** protons
and carries **2**x1**.**6x10^{-19}C of electric charge**.**

__Test Yourself 1:__

1) In magnetizing a nail that is wrapped around with a coil of wire, the
direction of the electric current in the loops of the wire is (a) parallel to
the nail (b) perpendicular to the nail if the loops are closely packed (c)
almost perpendicular to the nail if the loops are not closely packed (d) b & c**.** click
here**.**

2) The direction of the magnetic field in a magnetized nail is (a) along the
nail (b) perpendicular to the nail (c) neither a nor b**.**

3) If the four bent fingers of the right hand point in the direction of current
in the loops of a magnetized coil, then the thumb points to (a) the South pole
of the magnet (b) the North pole of the magnet (c) the direction normal to the
magnet**.** click
here**.**

4) The magnetized nail experiment shows that (a) magnetic field occurs anywhere
that there is an iron core (b) anywhere a charged particle moves, magnetic
effect develops in all directions (c) anywhere a charged particle moves,
it develops a magnetic effect that is normal to the direction of the its motion**.**

5) An electron orbiting the nucleus of an atom (a) does not develop a magnetic
field because its radius of rotation is extremely small (b) generates a
magnetic effect that is of course normal to its plane of rotation at any
instant (c) cannot generate any magnetic effect because of its extremely small
charge**.** click
here**.**

6) In a hydrogen molecule (**H _{2}**)
the net magnetic effect caused by the rotation of its two electrons is zero
because (a) at any instant, the two electrons spin in opposite directions
creating opposite magnetic effects (b) the instant its two electrons pass by
each other, they repel and change planes of rotation that are opposite to each
other causing opposite magnetic effects

7) The reason atoms, **in
general**, are magnetically neutral is that (a) electrons of atoms exist in
pairs spinning in opposite directions in every orbital thereby neutralizing each
other's magnetic effect**.** (a) True (b) False click
here**.**

8) The reason iron and a few other elements can retain magnetism is
that (a) these elements have orbitals that contain unpaired electrons
(b) under an external magnetic field, the orbitals in these element with single
electrons in them can orient themselves to the direction of the external field
and stay that way (c) both a & b**.**

9) For a bar-magnet, the magnetic field lines (a) emerge from its South pole and
enter its North pole (b) emerge from its North pole and enter its South pole
(c) emerge from its poles and enter its middle, the neutral zone**.** click
here**.**

10) If a bar magnet is cut at its middle (the neutral zone), (a) one piece
becomes a pure North pole and the other piece a pure South pole (b) both pieces
will have their own South and North poles because magnetic poles coexist (c)
neither a nor b**.**

11) The magnetic field strength around a bar magnet is (a) uniform (b)
non-uniform and therefore varies with distance from its poles (c) uniform at
points far from the poles**.** click
here**.**

12) The magnetic field in between the poles of a horseshoe magnet is (a)
uniform (b) non-uniform (c) zero**.**

13) The magnetic field in between the poles of a horseshoe magnet (a) varies
with distance from its either pole (b) is directed from N to S (c) has a
constant magnitude and direction and is therefore uniform**.** (d) b &
c**.** click
here**.**

**Problem:** Visualize
you are sitting in a class facing the board**.** Suppose that the ceiling is
the North pole of a huge horseshoe magnet and the floor is its South pole; therefore, you
are sitting inside a uniform
downward magnetic field**.** Also visualize a fast moving positive
charge emerging from the left wall and is heading toward the right wall**;** in
other words, the
velocity vector of the positive charge acts to the right**.** Answer
questions** 14 **through

**14)** The
charge will **initially** be
pushed (a) toward you (b) downward (c) toward the board**.**

**15)** The
charge will take a path that (a) is straight toward the board (b) is circular at a
certain radius of rotation (c) curves upward**.** click
here**.**

**16)** If
the radius of curvature is small such that the charge does not leave the space
in
between the poles of the magnet it will have a circular motion that looking from
the top will be (a) counterclockwise (b) clockwise (c) oscillatory**.**

**17)** If
instead, a negative charge enters from the left side, looking from the top, it
spins (a) counterclockwise (b) clockwise**.**

**18)** If
a positive charge enters from the right side heading left, looking from
the top again, it spins (a) clockwise (b) counterclockwise**.** click
here**.**

19) If the polarity of a magnetic field is reversed, the spin direction of a
moving charged particle caught in it will (a) remain the same (b) reverse as
well**.**

20) The force * F *of
a magnetic field

21) The force * F* of
a magnetic field

22) In F
= qvBsinθ,
if * q* is
in Coulombs,

23) The magnitude of the
force that
a 0**.**0025**T** magnetic
field exerts on a proton that enters it normal to its field lines at a speed of
3**.**7x10^{6}m**/**s
is (a) 1**.**5x10^{15}N (b) 0 (c) 1**.**5x10^{-15}N**.**

24) An electron moves at 7**.**4x10^{7}m**/**s
parallel to a uniform magnetic field**.** The force on the electron
by the field
is (a) 3**.**2x10^{-19}N (b) 0 (c) 4**.**8x10^{-19}N**. **
click
here**.**

25) The force that keeps a particle in circular
motion is (a) circular force (b) centripetal force (c) tangential force**.**

26) When a charged particle is caught in a magnetic
field and keeps spinning at a certain radius of rotation, the necessary
centripetal force is (a) the force of the magnetic field on it (b) the
electric force of the charged particle itself (c) both a & b**.** click
here**.**

27) Equating the force of magnetic field **F**_{m }and
the centripetal force * F_{c}* is
like (a) Mv

28) In the previous question, solving for * R* yields

29) The radius of rotation that a 4**.**0μC
charge carried by a 3**.**4μg mass moving at 360m**/**s normal to a 0**.**78T
magnetic field attains is (a) 39cm (b) 3**.**9m (c) 7**.**8m. click
here**.**

30) One electron-volt of energy (* 1eV*) is the
energy of (

31) Knowing that **1eV
= 1.6x10**^{-19}* J*, if a moving proton has an energy of 25000eV, its
energy in Joules is (a) 4

32) A 25-keV proton enters a 0**.**014T magnetic
field normal to its field lines**.** * M_{p}* =
1

__Velocity Selector:__

*
It is possible* to run a
charged particle through a magnetic field perpendicular to the field lines

q
v
B = q
E ; v
B = E , or *V **
= E*/*B**.*

If there is a large number of charged particles traveling at different speeds
but in the same direction, and we want to separate the ones with a certain speed
from the rest, this device proves useful**.**

**Example 6: ** In
a left-to-right flow of * alpha-rays *(Helium nuclei = 2p+2n) coming out of a
radioactive substance a 0

**Solution:** * K.E. =
*0

Since **each eV** is
equal to **1.6x10**^{-19}** Joules ;** therefore**,**

* K.E. =* 0

To find **v** from
* K.E.*, use

Using the *velocity selector
formula*:* *

*V* =
E/B ;** E = vB**

__Cyclotron:__

An application of the foregoing discussion is in the **"cyclotron."**
A cyclotron is a device that accelerates charged particles for nuclear
experiments**.** It works on the basis of the motion of charged particles in
electric and magnetic fields**.** When a particle of mass * M* and
charge

*R*
= Mv/(qB).

The space in which such particles spin is cylindrical**.** To accelerate the
spinning particles to higher and higher speeds, the cylinder is divided into
two semi-cylinders called the** "Dees"** as shown below

are forced to spin under the influence of Charged particles
perpendicular to the deesmagnetic field B. The "dees" are connected to
an alternating voltage. The AC voltage keeps
alternating the dees polarity at a high frequency forcing the
charged particles to be repelled from one dee and attracted to the
other repeatedly. This process accelerates the
particles. As each particle speeds up, its radius of
rotation increases spirally and finally hits the intended target at
the edge. |

Figure 8

According to the Einstein theory of relativity, as the a particle's
speed increases, its energy and therefore mass increases**.** This mass
increase is * insignificant for the regular speeds* we deal with
daily

__Period of Rotation:__

The period * T*, the time it
takes for a charged particle to
travel one full circle or

** T = 2πR/V**
(

* V* may be
found from the formula for radius of rotation

*T* = 2πM/(qB)

**Example 7:** In a cyclotron, protons
are to be accelerated**.** The strength of the magnetic field is 0**.**024**T.**
(a) Find the period of rotation of the protons, (b) their frequency, (c) their
final speed if the final radius before hitting the target is 2**.**0m, and
(d) their * K.E. *in
Joule and

**Solution: ** (a) *T*
= 2πM/(qB) ;

(a) * T *= 2π
x 1

(b) ** f
= 1/T ; f =** 3

(c) *V* = Rω ; *V*** =** R(2πf) **;** *V*** ****=** (2**.**0m)(2π)(3**.**7x10^{5}s^{-1})
= 4**.**6x10^{6} m/s**.**

This speed (although very high for us) is small
enough compared to the light
speed ( 3**.**00x10^{8} m**/**s
) that we may neglect the
relativistic effect**.**

(d) **K.E****.****=**
(**1/2**)Mv^{2} **=** **(1/2)**(1**.**672x10^{-27}kg)(4**.**6x10^{6}m/s)^{2} **=**
1**.**8x10^{-14}J**.**

**K.E.****=** 1**.**8x10^{-14}**J**
(1**eV/**1**.**6x10^{-19}**J**)
**=** 110,000**eV**** =** 0**.**11
**MeV****.**

__An Easy Relativistic Calculation:__

According to Einstein's
theory of relativity, when
mass * M* travels close to speed of light it becomes more massive
(energetic) and more
difficult to accelerate further

**γ is pronounced "Gamma."**

**Example 8: ** In
a cyclotron electrons are accelerated to a speed of 2**.**95x10^{8}m**/**s**.**
(a) By what factor does the electron mass increase**?** Knowing that the rest
mass of electron * M_{o}* =
9

**Solution: ** (a) Let's find γ step
by step**.** Let's first find **v/c**, then (**v/c**)** ^{2}** that
is the same as

(v**/**c)
= 2**.**95**/**3**.**00
= 0**.**98333... (Note that 10^{8} powers
cancel)**.**

v^{2}**/**c^{2}
= (v**/**c)^{2}
= (0**.**98333...)^{2} = 0**.**96694...

1 - v^{2}**/**c^{2}
= 1 - 0**.**96694...
= 0**.**0330555...

[1 - v^{2}**/**c^{2}]^{1/2} ** =** 0**.**182

γ **=** 1**/**[1 - v^{2}**/**c^{2}]^{1/2} **
=** 1**/**0**.**182 = **5.50** ( # of times
mass increases)**.**

(b) M = M_{o}γ **;** M
= 9**.**108x10^{-31}kg ( 5**.**50 ) = 5**.**01x10^{-30} kg**.**

**Example 9: ** Find
the value of γ for
the protons in Example
7**.**

**Solution: ** To
be done by students**.**

__Sources of Magnetic Field:__

Aside from permanent magnets, magnetic fields are mostly generated by coils**.** A coil is a wire that most often is wrapped around a cylinder**.**
Most coils are cylindrical**.** Long coils produce a fairly uniform magnetic field
inside of them specially toward their middle and along their axis of symmetry**.**
To understand the magnetic field inside a coil, we need to first know the magnetic
field around a long straight wire as well as that of a single circular loop**.**

__Magnetic Field Around a Straight and Long Wire:__

For a very * long and straight wire that carries current I*,
we expect the magnetic field

Figure 9

As the distance from the wire, * r*, increases,

**Example 10:** If
in the above figure, I
= 9**.**00A, determine the magnitude of B at
r = 10**.**0cm, 20**.**0cm, and 30**.**0cm.

**Solution:** Using the formula ** B
= μ_{o}I/(2πr)**, we get

** B_{1} =** 18x10

__Magnetic Field of a Current-Carrying Circular Loop:__

The magnetic field produced by a current-carrying circular loop is necessarily
perpendicular to the plane of the loop**.** The reason is that * B* must
be perpendicular to the current

Figure 10

**Example 11:** In the above figure, I
= 6**.**80Amps**.** Determine the magnitude of * B* for
r = 10

**Solution: ** Using
the formula ** B
= μ_{o}I/(2r)**, we get

** B_{1} =** 4

__Magnetic Field Inside a Solenoid:__

* A solenoid is a long coil of wire* for which the length-to-radius
ratio is not less than 10

* B = μ_{o}nI * where

Figure 11

In SI,
*
n is*

**Example 11:** A
solenoid is 8**.**0cm long and contains 2400 loops**.** A 1**.**2A current
flows through it**.** Find (a) the strength of the magnetic field inside it
and along its centerline
toward its middle**.** (b) If an iron core is inserted in the solenoid, what
will the field strength be**?**

**Solution:** (a) *B
***= μ _{o}nI**

(b) Iron increases **μ _{o}** by
a factor of

__One Application of Solenoid:__

Anytime you start you car, a solenoid similar to the one in the above example
gets magnetized and pulls an iron rod in**.** The strong magnetic field of
the solenoid exerts a strong force on the iron rod (core) and gives it a great
acceleration and high speed within a short distance**.** The rod is
partially in the solenoid to begin with and gets fully pulled in after the
solenoid is connected to the battery when the ignition key is turned to crank the
engine**.** The current that
feeds the solenoid may not be even 1A, but the connection it causes between
battery and starter pulls several amps from the battery**.** The forceful
moving rod inside the solenoid collides with a copper connector that connects
the starter to the battery**.** In an average car, this connection allows a current of 30A
to 80A to flow through the starter motor and crank the vehicle**.** The
variation of the amperage depends also on how cold the engine is**.** The
colder the engine, the less viscous the oil, and the more power is needed to
turn the crankshaft**.**

**Example 12:** The
magnetic field inside a 16**.**3cm long solenoid is 0**.**027T when a
current of 368mA flows through it**.** How many loops does it have**?**

**Solution: ****B
****= μ _{o}nI**

** n
=** 0

This is the number of loops per meter**.** If
the solenoid were 1**.**00m long, it would have 58400 loops**.** It is
only 0**.**163m long, and therefore it has less number of loops**.** If * N* is
the number of loops, we may write

*N
= nl* ; ** N
=** (958400 loops

__Definition of Ampere:__

We defined * 1A* to
be the flow of

If two parallel wires that are
1m apart
carry equal same direction currents, and the attraction force between them is 2x10^{ -7}N
per meter of the wires in vacuum, the
current through each wire is defined as 1A.

__Test Yourself 2:__

1) A velocity selector takes advantage of (a) two perpendicular electric fields
(b) a set of perpendicular electric and magnetic fields (c) two perpendicular
magnetic fields**.** click
here**.**

2) The forces * F_{m}* and

3) * F_{m}* and

4) In a velocity selector, setting F_{m} = F_{e} and
solving for v results
in (a) V = B**/**E
(b) V = E**/**B
(c) V =EB**.**

5) The formula V = E**/**B
(a) depends on the amount of charge (b) does not depend on the amount of the
charge (c) does not depend on the sign of the charge (d) b & c**.**

6) What strength uniform electric field must be placed normal to a 0**.**0033* T* uniform
magnetic field such that only charged particles at a speed of 2

7) A cyclotron is a device that is used to (a) accelerate charged particles to
high speeds and energies (b) accelerate charged particles to speeds close to
that of light (c) perform experiments with the nuclei of atoms (d) a, b, & c**.**

8) Speed of light is (a) 3**.**00x10^{8}m**/**s (b) 3**.**00x10^{-8}m**/**s
(c) 3**.**00x10^{5 }km**/**s
(d) a & c**.**

9) A speed of 3**.**00x10^{-8}m/s is (a) faster than speed of light
(b) slower than the motion of an ant (c) even germs may not move that slow
(d) extremely slow, almost motionless (e) b, c, and d. click
here**.**

10) In cyclotrons, a charged particle released near the center (a) finds itself
in a perpendicular magnetic field and starts spinning (b) spins at a certain
period of rotation given by T = 2πM**/**(qB)
(c) is under an accelerating electric field that alternates based on the
period of rotation of the charged particle (d) a, b, & c**.**

11) As the particles in a cyclotron accelerate to high speeds comparable to that
of light (a) a mass increase must be taken into account (b) the mass increase
affects the period of rotation (c) a & b**.**

12) The magnetic field around a current-carrying long wire (a) is perpendicular
to the wire**, **and at equal distances from the wire has the same magnitude
(b) is parallel to the wire (c) both a and b**.**
click
here**.**

13) The magnetic field around a current-carrying long wire (a) may be pictured as concentric circles at which the field vectors act radially outward (b) may be pictured as concentric circles at which the field vectors act tangent to the circles (c) has a constant magnitude that does not vary with distance from the wire (d) b & c.

14) The formula for the field strength around a current carrying long wire is
(a) B = μ_{o}I**/**(2πR) (b) B
= μ_{o}I**/**(2R)
(c) B = I**/**(2R)**.** click
here**.**

15) μ_{o}=
4π x10^{-7}Tm/A
is called (a) the permittivity of free space for the passage of electric field
effect (b) the permeability of free space for the passage of the magnetic
field effect (c) neither a nor b**.**

16) The farther from a straight and long
current-carrying wire, the (a) stronger the magnetic effect (b) the more
constant the magnetic effect (c) the weaker the magnetic effect**.** click
here

**Problem:** Draw
two concentric circles in a plane perpendicular to a wire that passes through
the center of the circles**.** Suppose that the wire carries a constant
electric current ** I
**upward

17) The magnitude of the field strength at the
smaller radius is (a) a vector of length 1 inch (b) a vector of length 1**/**4
inch (c) a vector of length 1**/**16 inch**.** click
here**.**

18) Based on the upward current in the wire, and
looking from the top, the direction of vectors you draw must be (a) clockwise
(b) counterclockwise**.**

19) What should be the length of the tangent vector
you may draw at another circle which radius is 5 times that of the smaller
circle**?** (a) 1**/**25 inch (b) 1**/**125 inch (c) 1**/**5
inch**.**

20) The magnetic field that a current carrying loop
of wire (circular) generates is (a) perpendicular to the plane of the loop (b)
has its maximum effect at the center of the loop and normal to it (c) has an
upward direction if the current flows in the circular loop horizontally in the
counterclockwise direction (d) a, b, & c**.** click
here**.**

21) A solenoid is a coil which length is (a) at
most 5 times its radius (b) at least about 10 times its radius**.**

22) The magnetic field inside a solenoid and in the
vicinity of its middle (a) is fairly uniform (b) is non-uniform (c) has a
magnitude of B
= μ_{o}nI where n is
its number of loops (d) has a magnitude of B
= μ_{o}n
I where n is
its number of loops per unit length (e) a & d**.**

23) A solenoid is 14**.**0cm long and has 280**0 ** loops**.**
A current of 5**.**00A flows through it**.** The magnetic field strength
inside and near its middle is (a) 0**.**0176T (b) 0**.**126T (c) 0**.**00126T**.**
click
here**.**

24) The magnitude of * B* inside
and at the middle of a 8

25) The magnetic field strength, B,
at the center of a coil that
has N loops and carries a current I is (a) B
= Nμ_{o}I **
/**(2R)
(b) B = Nμ_{o}I** ****/**(2πR)**.**
click
here**.**