Motion of Charged Particles in a Magnetic Field Problems and Solutions 2
Problem#1
In an experiment with cosmic rays, a vertical beam of particles that have charge of magnitude 3e and mass 12 times the proton mass enters a uniform horizontal magnetic field of 0.250 T and is bent in a semicircle of diameter 95.0 cm, as shown in Fig. 1. (a) Find the speed of the particles and the sign of their charge. (b) Is it reasonable to ignore the gravity force on the particles? (c) How does the speed of the particles as they enter the field compare to their speed as they exit the field? |
| Fig.1 |
Answer:
Known:
charge of magnitude, q = 3e = 4.8 x 10-19 C
total proton mass, mp = 12 x 1.67 x 10-27 = 2.004 x 10-26 kg
magnetic field, B = 0.250 T
diameter, d = 95.0 cm = 0.95 m
(a) the speed of the particles and the sign of their charge
The centripetal (Lorentz) force exerted by the magnetic field has magnitude
F = qvB
F is also equal to m times the centripetal acceleration v2/R
Therefore, F = mv2/R = qvB which boils down to:
mv = qBR
v = qBR/m
v = (4.8 x 10-19 C)(0.250 T)(0.95 m/2)/(2.004 x 10-26 kg) = 2.84 x 106 m/s
(b) The gravitational force mg = 1.97 x 10−25 N, while the magnetic force qvB = 6.82 x 10−13 N. Therefore, the gravitational force can be safely ignored.
(c) the speed of the particles as they enter the field compare to their speed as they exit the field, because the magnetic force F = qv × B is always perpendicular to the particle’s velocity v, the magnetic force will never change the speed of a particle. Only the direction of motion is affected. So the particle leaves the field with the same speed it entered with.
Problem#2
A physicist wishes to produce electromagnetic waves of frequency 3.0 THz using a magnetron. (a) What magnetic field would be required? Compare this field with the strongest constant magnetic fields yet produced on earth, about 45 T. (b) Would there be any advantage to using protons instead of electrons in the magnetron? Why or why not?
Answer:
Known:
Frequency, f = 3.0 THz
magnetic field, B = 45 T
The angular speed that corresponds to the frequency f is
ω = 2πf = 2π x 3.0 x 1012 Hz = 1.88 x 1013 rad/s
magnetic field would be required is
B = mω/│q│
B = (9.11 x 10-31 kg)(1.88 x 1013 rad/s)/(1.6 x 10-19 C)
B = 107.32 T
this magnetic field is 2.38 times the magnetic field produced by the earth
(b) there is no excess using protons
Problem#3
A beam of protons traveling at 1.20 km/s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction (Fig. 2). The beam travels a distance of 1.18 cm while in the field. What is the magnitude of the magnetic field?
charge of magnitude, q = 3e = 4.8 x 10-19 C
total proton mass, mp = 12 x 1.67 x 10-27 = 2.004 x 10-26 kg
magnetic field, B = 0.250 T
diameter, d = 95.0 cm = 0.95 m
(a) the speed of the particles and the sign of their charge
The centripetal (Lorentz) force exerted by the magnetic field has magnitude
F = qvB
F is also equal to m times the centripetal acceleration v2/R
Therefore, F = mv2/R = qvB which boils down to:
mv = qBR
v = qBR/m
v = (4.8 x 10-19 C)(0.250 T)(0.95 m/2)/(2.004 x 10-26 kg) = 2.84 x 106 m/s
(b) The gravitational force mg = 1.97 x 10−25 N, while the magnetic force qvB = 6.82 x 10−13 N. Therefore, the gravitational force can be safely ignored.
(c) the speed of the particles as they enter the field compare to their speed as they exit the field, because the magnetic force F = qv × B is always perpendicular to the particle’s velocity v, the magnetic force will never change the speed of a particle. Only the direction of motion is affected. So the particle leaves the field with the same speed it entered with.
Problem#2
A physicist wishes to produce electromagnetic waves of frequency 3.0 THz using a magnetron. (a) What magnetic field would be required? Compare this field with the strongest constant magnetic fields yet produced on earth, about 45 T. (b) Would there be any advantage to using protons instead of electrons in the magnetron? Why or why not?
Answer:
Known:
Frequency, f = 3.0 THz
magnetic field, B = 45 T
The angular speed that corresponds to the frequency f is
ω = 2πf = 2π x 3.0 x 1012 Hz = 1.88 x 1013 rad/s
magnetic field would be required is
B = mω/│q│
B = (9.11 x 10-31 kg)(1.88 x 1013 rad/s)/(1.6 x 10-19 C)
B = 107.32 T
this magnetic field is 2.38 times the magnetic field produced by the earth
(b) there is no excess using protons
Problem#3
A beam of protons traveling at 1.20 km/s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction (Fig. 2). The beam travels a distance of 1.18 cm while in the field. What is the magnitude of the magnetic field?
 |
| Fig.2 |
Answer:
Known:
Speed, v = 1.20 km/s
Distance, d = 1.18 cm = 0.0118 m
Curved path length = 1/4 circumference
= 0.25 x 2πR
= 0.50π x 0.0118 m = 0.0075 m
Centripetal force FC = mv²/R is provided by the magnetic force, Fm = Bqv
mv²/R = Bqv (where, m = mass of proton, q = proton charge)
B = mv/qR
B = (1.67 X 10-27kg x 1.20 x 103 m/s)/(1.60 x 10-19 C x 0.0075m)
B = 1.67 x 10-3 T
Problem#4
A proton (q = 1.60 x 10-19 C, m = 1.67 x 10-27 kg) moves in a uniform magnetic field B = (0.500T)i. At t = 0 the proton has velocity components vx = 1.50 x 105 m/s, vy = 0 and vz = 2.00 x 105 m/s. (a) What are the magnitude and direction of the magnetic force acting on the proton? In addition to the magnetic field there is a uniform electric field in the +x-direction, E = (+2.00 x 104 V/m)i. (b) Will the proton have a component of acceleration in the direction of the electric field? (c) Describe the path of the proton. Does the electric field affect the radius of the helix? Explain. (d) At t = T/2 where is the period of the circular motion of the proton, what is the x-component of the displacement of the proton from its positions at t = 0?
Answer:
(a) with B = (0.500T)i and v = vxi + vyj + vzk = (1.50 x 105 m/s)i + (2.00 x 105 m/s)k
the magnitude and direction of the magnetic force acting on the proton is
F = qv x B = (1.60 x 10-19 C)[(1.50 x 105 m/s)i + (2.00 x 105 m/s)k] x (0.500Ti) = 1.6 x 10-14 Nj
Then, magnitude of the magnetic force is 1.6 x 10-14 N and direction of the magnetic force in the direction of the positive +y-axis.
(b) the proton have a component of acceleration in the direction of the electric field, yes, because the electric field exerts a force in the direction of the electric, since the charge of the proton is positive and there is a component of acceleration in this direction
(c) In the plane perpendicular B (the xy-plane) the motion is circular. But there is a velocity component in the direction of B, so the motions is a helix. The electric field in the +i directions exerts a force in the +i direction. This force produces an acceleration in the +i direction and their causes the pitch of the helix to vary the force does not effect the circular motion in the yz-plane. So the electric field does not effect the radius of the helix.
(d) period the motion is
T = 2π/ω = 2πm/qB
T = 2π x (1.67 x 10-27 kg)/[(1.60 x 10-19 C)(0.500T)] = 1.31 x 10-7 s
Then, t = T/2 = 6.56 x 10-8 s
constant acceleration is given by
ax = Fx/m = (1.60 x 10-19 C)(2.00 x 104 V/m)/(1.67 x 10-27 kg)
ax = +1.92 x 1012 m/s2
the x-component of the displacement of the proton from its positions at t = 0
Δx = v0xt + ½ axt2
Δx = (1.50 x 105 m/s)(6.56 x 10-8 s) + ½(1.92 x 1012 m/s2)(6.56 x 10-8 s)2
Δx = 1.40 cm
Speed, v = 1.20 km/s
Distance, d = 1.18 cm = 0.0118 m
Curved path length = 1/4 circumference
= 0.25 x 2πR
= 0.50π x 0.0118 m = 0.0075 m
Centripetal force FC = mv²/R is provided by the magnetic force, Fm = Bqv
mv²/R = Bqv (where, m = mass of proton, q = proton charge)
B = mv/qR
B = (1.67 X 10-27kg x 1.20 x 103 m/s)/(1.60 x 10-19 C x 0.0075m)
B = 1.67 x 10-3 T
Problem#4
A proton (q = 1.60 x 10-19 C, m = 1.67 x 10-27 kg) moves in a uniform magnetic field B = (0.500T)i. At t = 0 the proton has velocity components vx = 1.50 x 105 m/s, vy = 0 and vz = 2.00 x 105 m/s. (a) What are the magnitude and direction of the magnetic force acting on the proton? In addition to the magnetic field there is a uniform electric field in the +x-direction, E = (+2.00 x 104 V/m)i. (b) Will the proton have a component of acceleration in the direction of the electric field? (c) Describe the path of the proton. Does the electric field affect the radius of the helix? Explain. (d) At t = T/2 where is the period of the circular motion of the proton, what is the x-component of the displacement of the proton from its positions at t = 0?
Answer:
(a) with B = (0.500T)i and v = vxi + vyj + vzk = (1.50 x 105 m/s)i + (2.00 x 105 m/s)k
the magnitude and direction of the magnetic force acting on the proton is
F = qv x B = (1.60 x 10-19 C)[(1.50 x 105 m/s)i + (2.00 x 105 m/s)k] x (0.500Ti) = 1.6 x 10-14 Nj
Then, magnitude of the magnetic force is 1.6 x 10-14 N and direction of the magnetic force in the direction of the positive +y-axis.
(b) the proton have a component of acceleration in the direction of the electric field, yes, because the electric field exerts a force in the direction of the electric, since the charge of the proton is positive and there is a component of acceleration in this direction
(c) In the plane perpendicular B (the xy-plane) the motion is circular. But there is a velocity component in the direction of B, so the motions is a helix. The electric field in the +i directions exerts a force in the +i direction. This force produces an acceleration in the +i direction and their causes the pitch of the helix to vary the force does not effect the circular motion in the yz-plane. So the electric field does not effect the radius of the helix.
(d) period the motion is
T = 2π/ω = 2πm/qB
T = 2π x (1.67 x 10-27 kg)/[(1.60 x 10-19 C)(0.500T)] = 1.31 x 10-7 s
Then, t = T/2 = 6.56 x 10-8 s
constant acceleration is given by
ax = Fx/m = (1.60 x 10-19 C)(2.00 x 104 V/m)/(1.67 x 10-27 kg)
ax = +1.92 x 1012 m/s2
the x-component of the displacement of the proton from its positions at t = 0
Δx = v0xt + ½ axt2
Δx = (1.50 x 105 m/s)(6.56 x 10-8 s) + ½(1.92 x 1012 m/s2)(6.56 x 10-8 s)2
Δx = 1.40 cm
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