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Notes for July 3, 2008 discussion prepared by Lisa Albrecht
Reviewed the right hand rule:
To find the direction of the magnetic field force employ right hand rule number 1:
1. Point fingers of right hand in the direction of the velocity v.
2. Curl fingers in the direction of the magnetic field B moving through the smallest angle.
3. Thumb is pointing in the direction of the magnetic force F exerted on a positive charge.
(The force will be pointed in the opposite direction for an electron.)
When a wire is carrying a current I, to find the direction of the magnetic field B use right
hand rule number two:
Point the thumb of your right hand along a wire in the direction of the positive
current.The natural way to curl fingers curt in the direction of the magnetic field B.
B= ((µ0I)/2πR)
where µ0 (the permeability of free space) = 4 π x 10^-7 Tm/A.
Conceptual problems (Saved #3 and #4 for Monday)
2007 Midterm 2 #4
4. If a charged particle is moving in a uniform magnetic field, its path can be:
a. a straight line.
b. a circle.
c. a helix.
d. any of the above.
e. none of the above.
If the magnetic field is parallel to the path of velocity, the particle moves in a straight line
(A). A magnetic field also causes the particle to follow a curved path (B) because the F
causes centripetal acceleration. And, (C) the particle travels in a helical path if velocity is at
an angle to the uniform magnetic field. Since A, B, and C are all correct, D is the correct
answer.
2007 Midterm 2 #6
6. A current in a solenoid coil creates a magnetic field inside that coil. The field strength is
directly proportional to:
a. the coil area.
b. the current.
c. the coil diameter.
d. Both A and B are valid choices.
e. None of the above choices are valid.
The equation to use is B= µ0 n I where n=number of turns N/unit length and I is the
current. A solenoid coil is a tightly wound coil that creates a magnetic field that is constant
everywhere inside the coil. If I increases, then B increases. (Also, a note, B outside of the
solenoid is equal to zero.)
Quantitative Problems
Ben Kuntz Physics Text Pg 517 prob. 25
A proton moving freely in a circular path perpendicular to a constant magnetic field takes 1.00
nano-seconds to complete one revolution. Determine the magnitude of the magnetic field.
F=mac=qvB (Because force of magnetic field causes centripetal acceleration.)
v= 2πr/(1E-9sec)
ac=v^2 / r
q= 1.6E-19 C
Solved for B. Found B= 65.58 T.
Brie Henry From textbook p. 518, #36
A long, straight wire lies on a horizontal table and carries a current of 1.2E-9A. In a vacuum, a
proton moves parallel to the wire (opposite the direction of the current) with a constant velocity
of 2.30E4 m/s at a constant distance “d” above the wire. Determine the value of “d.” You may
ignore the magnetic field due to the Earth.
FB=Fg Fg=mg
FB= qvB
m=1.67E-27 kg
g=9.81m/s^2
So, B=(mg)/(qv).
Because of equation 19.11 from page 505, B=((µ0I)/2πd.
So, (mg/qv) = (µ0I)/2πd.
d= (µ0Iqv)/(mg2π). We calculated d= 5.39e-5 m.
Michael McLoughlin from the text, page 519, #44
An electron is moving at a speed of 1.0x10^4 m/s in a circular path
of radius of 2.0 cm inside a solenoid. The magnetic field of the
solenoid is perpendicular to the plane of the electron's path. Find (a)
the strength of the magnetic field inside the solenoid and (b) the
current in the solenoid if it has 25 turns per centimeter.
Part A:
B= mv/qr (If you rearrange equation 19.10 from page 503)
m=9.11E-31kg
v= 1.0E4 m/s
r= .02 m
q=1.67E-19C
Calculated B=2.85E-6 T
Part B:
Use B= µ0nI rearranged to find I=B/µ0n.
n= 25 turns/ .01m = 2500
µ0 = 4 π x 10^-7 Tm/A
B=2.85E-6 T
Calculated I= 9.06E-4 A