1.
A mass spectrometer accelerates a singly ionized ion through a potential difference of 2.00 kV to the proper speed. What
is the final speed of the particle? Then a 0.400 T magnetic field bends the ion into a circular path of radius 0.226 m. What
is the mass of the ion?
, mv2 = 2 qV = qvBr;
2 (2000) = v(0.4)(0.226); v = 44248 m/s
mv=qrB
m(44248) = 1.6E-19(0.226)(0.400)
m = 3.27 E -25 kg
2.
An electric power line carries a current of 1400 A in a location where the earth’s magnetic field is 0.50 E –4 T. The line
makes an angle of 75 with respect to the field. Determine the magnitude of the magnetic force on a 120-m length of line.
This question is badly worded. On the left is what I think it looks like with the current being the
horizontal line and the magnetic field being the vector at an angle.
F = BIL = (0.5 E -4)(1400)(sin75)(120) = 8.11 N
3.
The electrons in the beam of a television tube have an energy of 15 keV. The tube is oriented so that the electrons move
horizontally from east to west. The earth’s magnetic field points down INTO THE GROUND and has a magnitude of 4.5
E-5T. (a) In what direction will the beam deflect? (b) What is the acceleration of a given electron?
15000 eV *1.6 E -19 = 2.4 E -15 J = 1/2mv2; v = 7.26 E 7 m/s
mv2 =qvBr
v2/r = qvB/m
2
2
, a = v /r= 1.6E -19(7.26 E7)(4.5 E-5)/9.11 E -31 = 4.75 E 14 m/s (b) North
4.
Protons in a magnetic field of 0.80 T follow a circular trajectory with a 75-cm radius. (a) What is the speed of the protons?
(b) If electrons traveled at the same speed in this field, what would the radius of their trajectory be?
, mv = qrB, v = qrB/m = 1.6 E -19(0.75)(0.8)/1.67 E -27 = 5.75 E 7 m/s
, r = mv/qB = 9.11 E -31(5.75E7)/1.6 E -19/0.8 = 4.09 E -4 m or 4 tenths of a millimeter!
Two isotopes of carbon, carbon-12 and carbon-13, have masses of 19.92 E –27 kg and
21.59 E-27 kg, respectively. These two isotopes are singly ionized and each is given a
speed of 6.667 E 5 m/s. The ions then enter the bending region of a mass spectrometer
where the magnetic field is 0.8500 T. Determine the spatial separation between the two
isotopes after they have traveled through a half-circle.
, r = mv/qB
r1 = 0.09765
r2 = 0.1058
After they have gone a half circle they have traveled 2 radii (1 diameter) linearly
x for 1 = 0.1953 x for 2 = 0.2117
separation = 0.0164 m
5.
6.
A weighing scale supports a battery, to which a rigid wire loop is attached, as shown in the
figure at the right. The lower part of the loop is in a magnetic field of 0.20 T. If the
combined mass of the battery and wire loop is 0.25 kg, what current in the wire is
necessary for the scale to indicate zero weight? Which pole of the battery is
positive?
7.
Two wires parallel to one another are separated by 1.0 m as shown in the figure at
the right. Each carries a current of 3.0, but in opposite directions. Find the
magnitude and direction of the magnetic field at point P midway between the
wires.
B = 0I/2r = 2E-7I/r
Using Fonzie RHR both B fields are in the same direction,
south at that point. So we find one B field and double it 2E-7 (3)/0.5 X 2 = 2.4 E-6T
8.
Two parallel straight wires are separated by a distance of 0.75 m and carry currents I 1 = 25 and I2 = 35 A. Find the force
(magnitude & direction) that each wire exerts on a 2.5 m length of the other wire if the currents are (a) in the same
direction and (b) in the opposite direction.
F = 0I I ℓ/2r = 2 E -7 I I ℓ/r = 3.28 E -4 N (a) attractive (b) repulsive
9.
Four long parallel conductors all carry currents of 5.0 A. An end view of the conductors is
shown in the figure at the right. The current directions for wires A and D are into the page
(crosses) while the current directions for B and C are out of the page (dots). Calculate the
magnitude and direction of the magnetic field at point P, located at the center of the square of
edge length 0.40 m.
Ba = Bb= Bc= Bd = 2E-7(5)/(0.40*square root of 2/2) = 3.535E-6
Ba and Bc point directly southwest. Bb and Bd point directly southeast. AND since all the
vectors have the same magnitude, all the x parts of all the vectors cancel out and only the y
components are left. Total magnetic field at P = 4(3.535E-6)sin45 degrees = 1E –5 T SOUTH