PHY 104 Quiz on Magnetic Field and Forces

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18 Οκτ 2013 (πριν από 3 χρόνια και 7 μήνες)

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PHY 104 Quiz on Magnetic Field and Forces



1)

A solenoid is connected to a battery
as shown in the figure to the right
and a bar magnet is placed nearby.
In what direction will the bar
magnet move is response to the
magnetic field created by the
solenoid?


Explain your answer by describing, in words, the magnetic polarity of the
solenoid.

Using the Right
-
hand Rule, the magnetic field of the solenoid will be
directed towards the left as shown. This implies that the right end of the solenoid acts
like a sou
th pole of a magnet while the left end of the solenoid acts like a north pole
(Rule: magnetic field lines exit the north pole and travel into the south pole).Thus the
bar magnet to the right will feel an attraction to the apparent opposite pole of the
sole
noid adjacent to it. The bar magnet will move to the left towards the solenoid.


2)

Use the relation

and the definition of a Tesla (1
T

= 1
)
, to demonstrate,
by unit analysis, that the units for the permeability
of free space

0

are
.

First derive an expression for

0
:



Next do the unit analysis



Thus the units of

0

are N/A
2


3)

An electron shoots into a
confined
uniform m
agnetic field directed into the
page as shown in the figure. Its velocity
is 3
.72
x

10
5

m/s

directed horizontally as
shown.


A)

Which of the illustrated paths (
a
,
b
,
c
, or
d
) will the electron follow?

B)

If the radius of curvature of the
electron is 5.2 cm wha
t is the
strength of the magnetic field?


First set the magnetic force equal to
+

_

N

S

I

I

B



a

c

b

d

X

X

X

X

X


X

X

X

X

X


X

X

X

X

X


X

X

X

X

X


X

X

X

X

X


X

X

X

X

X


the centripetal force as necessitated by circular motion principles.



Derive an equation for
B

and solve.




4)

Carbon atoms from
a prehistoric sample of
charcoal found in an ancient cave
are

ionized
by intense heating
and injected
,

with various
velocities,
into a velocity selector/mass
spectrometer illustrated to the right
, for the
purpose of measuring the
13
C abundance
relative to
the
12
C abundance
to date the
charcoal
. The

electric field and magnetic field
strengths are given in the diagram.


A)

What is the mass of the
13
C atom?




B)

What velocity ions will emerge from the velocity selector and enter the turning
field?

Derive the velocity selector relation by equating the downward electric
force on the charges
F = qE

with the upward magnetic force on the charges
F =
qvB
.




C)

What will be the radius of curvature of
the trajectory of
a singly
ionized
13
C ion?

First set the magnetic force equal to the centripetal force as necessitated by
circular motion principles.



Derive an equation for
r

and solve.



x x x x x x


x x x x x x

E
1

and B
1

B
2

E
1

=
10,000 N/C

B
1

=

1.67 mT

B
2

=

6.00 T








































































D)

Will the radius of curvature
of a
12
C atom b
e greater or less than the
13
C atom?
By what percentage

or fraction
?


Since the radius of curvature of the trajectory is directly proportional to the mass
of the ion, the
12
C ion will have a smaller radius of curvature by a factor of

or
92% smaller.


5)

A long wire carrying a 5.00 A current penetrates into a region of a 0.44 T magnetic
field as shown below. Using the dimensions in the figure determine the magnitude
and direction of the
net
force on the wire.










Identify the di
rections of the forces consistent with the right
-
hand rule and use
symmetry to eliminate any equal and opposite forces.

You can see that the two
opposing forces
F
1

and
F
2

will cancel each other leaving only
F
3

as the sole
contributor to the net force. Th
e magnitude of
F
3

is given by


Thus the net force on the wire is 0.55 N directed to the right.


6)

A current in a long straight wire produces a magnetic field of 8.0

T at 4.0 cm from
the wire’s center.

A)

What is the magnetic field streng
th 2.0 cm from the wire?

Since
, we
see that B is
inversely
proportiona
l

to
R. So by halving R you will double B.
Thus, B = 16

T when R = 2 cm.

B)

How far from the wire’s center will the magnetic field be 1.0

T?

Again, s
ince
, we see that B is
inversely
proportional

to R. So to
decrease

B by a
factor of 8 you must
increase

R by the same factor. Thus, R= 8∙4 cm = 32 cm for
B to equal 1.0

T.


C)

What is the current in the wire?

Solve

for

I.


5.00 A

25 cm

50

cm

B field

F
1

F
3

F
2

X X X X X X X


X X X X X X X


X X X X X X X


X X X X X X X


X X X X X X X






7)

A single circular loop 10.0 cm in diameter carries a 2.00 A current.

A)

What is the magnetic field at the center of this loop?




B)

Now connect 1,000 of these single loops in series within a length of 500

cm to
create a 500 cm long solenoid. What is the magnetic field strength in the center
of the solenoid?




C)

Why is the magnetic field strength in the solenoid not 1,000 times the magnetic
field strength of a single loop? (Answer in

a few sentences.)



The ratio of the magnetic fields is
. The reason the
solenoid’s B field is not simply 1,000 times the B field of the single loop is that the
loops of the solenoid are distributed over a 5 m length. Thus a sing
le loop of the
solenoid does not fully contribute to the B field at the center of another loop in the
solenoid. The more distance between the two loops the less the contribution. Thus
the 1,000 loops of the solenoid do not all equally of fully contribute

to the magnetic
field at a particular point in the center of the solenoid.