Download KEY - Magnets Combo

Magnetic Combo: It’s Attractive
(force on particle, force on wire, Ampere’s Law, mass spectrometer)
1. A compass that has been calibrated in the Earth’s magnetic field is laced near the end of a
permanent bar magnet, and points away from that end of the magnet. It can be concluded that the
end of the magnet (a) acts as a north magnetic pole, (b) acts as a south magnetic pole, (c) you can’t
conclude anything about the magnetic properties of the permanent magnet. (Pick one) Why?
(A) Acts as a north magnetic pole because a compass points towards magnetic south and
away from magnetic north.
2. If you look directly at the south pole of a bar magnet, it’s magnetic field points (a) to the right, (b) to
the left, (c) away from you, (d) toward you. (Pick one)
(C) Away from you. Lines point towards a South pole, not away from.
3. If a charged particle moves in a straight line and there are no other forces on it except possibly from
a magnetic field, can you say with certainty that no magnetic field is present? Explain.
No, the particle could be moving parallel to the B-field lines.
4. A charge of 0.050 C moved vertically in the field of 0.080 T that is oriented 45o from the vertical.
What speed must the charge have such that the force acting on it is 10. N?
F = QvB sin "
10N = (0.050C)v(0.080T ) sin 90
v = 3536m / s # 3500m / s
5. An electron travels at a speed if 2.0x104 m/s through a uniform magnetic field whose magnitude is
1.2-10-3 T. What is the magnitude of the magnetic force on the electron it its velocity and the
magnetic field
a. Are perpendicular?
!
F = QvB sin "
F = (1.6 #10 $19 C)(2.0 #10 4 m / s)(1.2 #10 $3 T ) sin 90
F = 3.84 #10 $18 N
b. Make an angle of 45o?
F = QvB sin "
F = (1.6 #10 $19 C)(2.0 #10 4 m / s)(1.2 #10 $3 T ) sin 45
!
F = 2.72 #10 $18 N
c. Are parallel?
F =0
d. Are exactly opposite?
!
F =0
6. In a linear particle accelerator, a beam of proton travels horizontally northward. To deflect the
protons eastward with a uniform magnetic field, the field should point in which direction?
!
!
v
B-field is out of the page
F
7. A particle with a charge of -5.0x10-4 C and a mass of 2.0x10-9 kg moves at 1.0x103 m/s in the +xdirection. It enters a uniform magnetic field of 0.20 T that points in the +y-direction.
a. Which way will the particle deflect just as it enters the field?
B
v
Force is into the page (using
electrons)
b. What is the magnitude of the force on the particle just as it enters the field?
F = QvB sin "
F = (5.0 #10 $4 C)(1.0 #10 3 m / s)(0.2T ) sin 90
F = 0.10N
c. What is the radius of the circular arc that the particle will travel while in the field?
QBR = mv
(5.0 "10 #4 C)(0.20T )R = (2.0 "10 #9 kg)(1.0 "10 3 m / s)
R = 0.020m
!
8. One electron has been removed from a methane molecule before it enters the mass spectrometer.
After passing through the velocity selector, the ion has a speed of 1.00x103 m/s. It then enters the
main magnetic field region, in which the field strength is 6.70x10-3T. From there it follows a circular
path and lands 5.00 cm from the field entrance. Determine the mass of this molecule. (Neglect the
mass of the electron that is removed.)
!
QBR = mv
$ 0.05m '
3
(1.6 "10 #19 C)(6.7 "10 #3 T )&
) = m(1.0 "10 m / s)
% 2 (
m = 2.68 "10 #26 kg
9. A 2.0 m length of straight wire carried a current of 20 A in a uniform magnetic field of 50 mT whose
direction is at an angle of 37o from the direction of the current. Find the force on the wire (and
direction).
!
F = BIL sin "
F = (50 #10 $3 T )(20A)(2.0m) sin 37
F = 1.20N
!
!
!
10. A straight wire 25 cm long is oriented vertically in a uniform horizontal magnetic field of 0.30 T
pointing in the –x-direction. What current (including direction) would cause the wire to the subjected
to a force of 0.050 N in the +y-direction?
F
F = BIL sin "
B
0.050N = (0.30T )I(0.25m) sin 90
I = 0.67A _ in _ page
11. The maximum household current in a wire is about 15 A. Assume that this current exists in a long,
straight wire in a west-to-east direction. What are the magnitude and direction of the magnetic field
the current produces 1.0 cm directly below the wire?
!
µI
B= o
2"L
(4" #10 $7 Tm / A)(15A)
B=
2" (0.01m)
B = 0.00030T = 3.0 #10 $4 T _ in _ page
12. A current carrying wire is suspended 15 cm above Ms. Stevens’ desk. The wire is 1.00 m long and
has a mass of 0.0500kg. If the wire is in a magnetic field of 2.00 T, what is the current in the wire?
Fgravity = Fwire
mg = BIL sin "
(0.05kg)(9.8m / s 2 ) = (2.00T )I(1.0m)
I = 0.245A