Download PHY 104 Quiz on Magnetic Field and Forces

PHY 104 Quiz on Magnetic Field and Forces
1) A solenoid is connected to a battery
N
S
B
as shown in the figure to the right
I
I
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 south 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
solenoid adjacent to it. The bar magnet will move to the left towards the solenoid.
2) Use the relation B 
0 I
N
and the definition of a Tesla (1T = 1
), to demonstrate,
2R
A m
N
by unit analysis, that the units for the permeability of free space 0 are  2  .
A 
First derive an expression for 0:
I
B 0
2R
B  2R
 0 
I
Next do the unit analysis
 N 
 m

T m  A  m 
N
Units of 0 

  2
A
A
A 
2
Thus the units of 0 are N/A
3) An electron shoots into a confined
uniform magnetic field directed into the
page as shown in the figure. Its velocity
is 3.72 x 105 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 what is the
strength of the magnetic field?
First set the magnetic force equal to

vi
X
X
X
B
X
X
X
X
X
d X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
b
a
c
the centripetal force as necessitated by circular motion principles.
mv2
 qvB
r
Derive an equation for B and solve.
m
9.111031 kg  3.72 105
mv
s  4.07 105 T
B

rq
0.052 m 1.6 1019 C
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 13C abundance
relative to the 12C abundance to date the
charcoal. The electric field and magnetic field
strengths are given in the diagram.
E1 and B1
B2
x
x
x
x
x
x





x
x
x
x
x
x

























E1 = 10,000 N/C
B1 = 1.67 mT
B2 = 6.00 T
A) What is the mass of the 13C atom?


Mass of 13C  13  1.67 1027 kg  2.17 1026 kg
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.
qE  qvB
N
E
C  5.99 10-6 m
v  
-3
B 1.67 10 T
s
10,000
C) What will be the radius of curvature of the trajectory of a singly ionized 13C ion?
First set the magnetic force equal to the centripetal force as necessitated by
circular motion principles.
mv2
 qvB
r
Derive an equation for r and solve.
m
2.17 1026 kg  5.99 106
mv
s  0.135 m  13.5 cm
r

19
Bq
6 T 1.6 10 C
D) Will the radius of curvature of a 12C atom be greater or less than the 13C atom?
By what percentage or fraction?
Since the radius of curvature of the trajectory is directly proportional to the mass
12
of the ion, the 12C ion will have a smaller radius of curvature by a factor of
or
13
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.
X
5.00 A
25 cm
X
X
F1 X
X
X
X
X
X
X
X
X
X
F3
X
X
B field
50 cm
X
X
X
F2
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
Identify the directions 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 F1 and F2 will cancel each other leaving only F3 as the sole
contributor to the net force. The magnitude of F3 is given by
F3  IlB  5 A  0.25 m  0.44 T  0.55 N
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.
I
A) What is the magnetic field strength 2.0 cm from the wire? Since B  0 , we
2R
see that B is inversely proportional 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, since
I
B  0 , we see that B is inversely proportional to R. So to decrease B by a
2R
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 B 
0 I
for I.
2R
I
B  2R
0

8 106 T  2  0.02 m
 1.6 A
-7 N
4 10
A2
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?
BLoop 
0 I
2R

N
2 A
A2
 2.5110-5 T
2  0.05 m
4 10-7
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?
N   1,000 

-4
BSolenoid  0 nI   4 10-7 2   
  2 A  5.03 10 T
A   5m 

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.)
BSolenoid 5.03 10 -4 T

 20 . The reason the
BLoop
2.5110 -5 T
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 single 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.
The ratio of the magnetic fields is