Problem#1
(a) What is the speed of a beam of electrons when the simultaneous influence of an electric field of 1.56 x 104 V/m and a magnetic field of 4.62 x 10-3 T, with both fields normal to the beam and to each other, produces no deflection of the electrons? (b) In a diagram, show the relative orientation of the vectors v, E and B, (c) When the electric field is removed, what is the radius of the electron orbit? What is the period of the orbit?Answer:
Known:
electric field, E = 1.56 x 104 V/m
magnetic field, B = 4.62 x 10-3 T
(a) the speed of a beam of electrons for no deflection
v = E/B = (1.56 x 104 V/m)/(4.62 x 10-3 T) = 3.38 x 106 m/s
(b) show the relative orientation of the vectors v, E and B, shown below
 |
| Fig.1 |
(c) When the electric field is removed, the radius of the electron orbit is
lqlBv = mv2/R
R = mv/lqlB = (9.11 x 10-31 kg)(3.38 x 106 m/s)/[(1.6 x 10-19 C)(4.62 x 10-3 T)]
R = 4.17 x 10-3 m = 4.17 mm
the period of the orbit is given by
T = 2πR/v
T = 2π(4.17 x 10-3 m)/(3.38 x 106 m/s) = 7.74 x 10-9 s = 7.79 ns
Problem#2
In designing a velocity selector that uses uniform perpendicular electric and magnetic fields, you want to select positive ions of charge +5e that are traveling perpendicular to the fields at 8.75 km/s. The magnetic field available to you has a magnitude of 0.550 T. (a) What magnitude of electric field do you need? (b) Show how the two fields should be oriented relative to each other and to the velocity of the ions. (c) Will your velocity selector also allow the following ions (having the same velocity as the +5e ions) to pass through undeflected: (i) negative ions of charge –5e (ii) positive ions of charge different from +5e?
Answer:
Known:
Speed, v = 8.75 km/s = 8.75 x 103 m/s
magnetic field, B = 0.550 T
(a) magnitude of electric field do you need is
E = vB = 8.75 x 103 m/s x 0.550 T = 4.8 kN/C
(b) here, the magnetic field and electric field should be perpendicular to each other in such a way that the net force on the ions should be zero.
(c) yes.
Problem#3
A particle with initial velocity v0 = (5.85 x 103 m/s)j enters a region of uniform electric and magnetic fields. The magnetic field in the region is B = –(1.35 T)k. Calculate the magnitude and direction of the electric field in the region if the particle is to pass through undeflected, for a particle of charge (a) +0.640 nC and (b) –0.320 nC. You can ignore the weight of the particle.
Answer:
Known:
initial velocity, v0 = (5.85 x 103 m/s)j
magnetic field, B = –(1.35 T)k
(a) the magnitude and direction of the electric field in the region if the particle is to pass through undeflected, for a particle of charge q = +0.640 nC,
E = v x B = (5.85 x 103 m/s)j x [–(1.35 T)k] = 7898 N/C i
So, if q = +0.640 nC, the direction of the electric field in the direction of the x-axis is positive, since it must point in the opposite direction to the magnetic force.
(b) the magnitude and direction of the electric field in the region if the particle is to pass through undeflected, for a particle of charge q = –0.320 nC,
E = v x B = (5.85 x 103 m/s)j x [–(1.35 T)k] = 7898 N/C i
So, if q = +0.640 nC, the direction of the electric field in the direction of the x-axis is positive, since the electric force must point in the opposite direction as the magnetic force. Since the particle has negative charge, the electric force is opposite to the direction of the electric field and the magnetic force is opposite to the direction it has in parti (a).
Problem#4
A 150-V battery is connected across two parallel metal plates of area 28.5 cm2 and separation 8.20 mm. A beam of alpha particles (charge +2e, mass 6.64 x 10-27 kg) is accelerated from rest through a potential difference of 1.75 kV and enters the region between the plates perpendicular to the electric field, as shown in Fig. 2. What magnitude and direction of magnetic field are needed so that the alpha particles emerge undeflected from between the plates?
lqlBv = mv2/R
R = mv/lqlB = (9.11 x 10-31 kg)(3.38 x 106 m/s)/[(1.6 x 10-19 C)(4.62 x 10-3 T)]
R = 4.17 x 10-3 m = 4.17 mm
the period of the orbit is given by
T = 2πR/v
T = 2π(4.17 x 10-3 m)/(3.38 x 106 m/s) = 7.74 x 10-9 s = 7.79 ns
Problem#2
In designing a velocity selector that uses uniform perpendicular electric and magnetic fields, you want to select positive ions of charge +5e that are traveling perpendicular to the fields at 8.75 km/s. The magnetic field available to you has a magnitude of 0.550 T. (a) What magnitude of electric field do you need? (b) Show how the two fields should be oriented relative to each other and to the velocity of the ions. (c) Will your velocity selector also allow the following ions (having the same velocity as the +5e ions) to pass through undeflected: (i) negative ions of charge –5e (ii) positive ions of charge different from +5e?
Answer:
Known:
Speed, v = 8.75 km/s = 8.75 x 103 m/s
magnetic field, B = 0.550 T
(a) magnitude of electric field do you need is
E = vB = 8.75 x 103 m/s x 0.550 T = 4.8 kN/C
(b) here, the magnetic field and electric field should be perpendicular to each other in such a way that the net force on the ions should be zero.
(c) yes.
Problem#3
A particle with initial velocity v0 = (5.85 x 103 m/s)j enters a region of uniform electric and magnetic fields. The magnetic field in the region is B = –(1.35 T)k. Calculate the magnitude and direction of the electric field in the region if the particle is to pass through undeflected, for a particle of charge (a) +0.640 nC and (b) –0.320 nC. You can ignore the weight of the particle.
Answer:
Known:
initial velocity, v0 = (5.85 x 103 m/s)j
magnetic field, B = –(1.35 T)k
(a) the magnitude and direction of the electric field in the region if the particle is to pass through undeflected, for a particle of charge q = +0.640 nC,
E = v x B = (5.85 x 103 m/s)j x [–(1.35 T)k] = 7898 N/C i
So, if q = +0.640 nC, the direction of the electric field in the direction of the x-axis is positive, since it must point in the opposite direction to the magnetic force.
(b) the magnitude and direction of the electric field in the region if the particle is to pass through undeflected, for a particle of charge q = –0.320 nC,
E = v x B = (5.85 x 103 m/s)j x [–(1.35 T)k] = 7898 N/C i
So, if q = +0.640 nC, the direction of the electric field in the direction of the x-axis is positive, since the electric force must point in the opposite direction as the magnetic force. Since the particle has negative charge, the electric force is opposite to the direction of the electric field and the magnetic force is opposite to the direction it has in parti (a).
Problem#4
A 150-V battery is connected across two parallel metal plates of area 28.5 cm2 and separation 8.20 mm. A beam of alpha particles (charge +2e, mass 6.64 x 10-27 kg) is accelerated from rest through a potential difference of 1.75 kV and enters the region between the plates perpendicular to the electric field, as shown in Fig. 2. What magnitude and direction of magnetic field are needed so that the alpha particles emerge undeflected from between the plates?
 |
| Fig.2 |
Answer:
Known:
Area, A = 28.5 cm2 = 2.85 x 10-3 m2
Mass alpha particles, mα = 6.64 x 10-27 kg
potential difference, V = 1.75 kV = 1.75 x 103 V
First use energy conservation to find the speed of the alpha particles as they enter the plates:
qV = ½ mv2
2eV = ½ mv2
vα = [4eV/m]1/2 = [4 x 1.6 x 10-19 C x 1.75 x 103 V/6.64 x 10-27 kg]1/2 = 4.11 x 105 m/s
The electric field between the plates the plates, produced by the battery, is
E = Vb/d = 150 V/(0.00820 m) = 18,300 V/m = 18.3 kV/m
The magnetic force must cancel the electric force:
B = E/vα = (18,300 V)/(4.11 x 105 m/s) = 0.0445 T
The magnetic field is perpendicular to the electric field. If the charges are moving to the right and the electric field points upward, the magnetic field is out of the page.
Problem#5
A singly ionized (one electron removed) 40K atom passes through a velocity selector consisting of uniform perpendicular electric and magnetic fields. The selector is adjusted to allow ions having a speed of 4.50 km/s to pass through undeflected when the magnetic field is 0.0250 T. The ions next enter a second uniform
magnetic field (B’) oriented at right angles to their velocity. 40K contains 19 protons and 21 neutrons and has a mass of 6.64 x 10-26 kg. (a) What is the magnitude of the electric field in the velocity selector? (b) What must be the magnitude of B’ so that the ions will be bent into a semicircle of radius 12.5 cm?
Answer:
Known:
magnetic field, B = 0.0250 T
speed, v = 4.50 km/s = 4.50 x 103 m/s
Velocity selector allows having desired velocity and then magnetic field bends the ion in a circular .
Velocity selector v = E/B, velocity is perpendicular to B,
mv2/R = qvB
(a) the magnitude of the electric field in the velocity selector is
E = vB = (4.50 x 103 m/s)(0.0250 T) = 112 V/m
(b) magnitude of so that the ions will be bent into a semicircle of radius 12.5 cm is
B’ = mv/qR
B’ = (6.64 x 10-26 kg)(4.50 x 103 m/s)/[(1.6 x 10-19 C)(0.125 m)]
B’ = 0.0149 T
Area, A = 28.5 cm2 = 2.85 x 10-3 m2
Mass alpha particles, mα = 6.64 x 10-27 kg
potential difference, V = 1.75 kV = 1.75 x 103 V
First use energy conservation to find the speed of the alpha particles as they enter the plates:
qV = ½ mv2
2eV = ½ mv2
vα = [4eV/m]1/2 = [4 x 1.6 x 10-19 C x 1.75 x 103 V/6.64 x 10-27 kg]1/2 = 4.11 x 105 m/s
The electric field between the plates the plates, produced by the battery, is
E = Vb/d = 150 V/(0.00820 m) = 18,300 V/m = 18.3 kV/m
The magnetic force must cancel the electric force:
B = E/vα = (18,300 V)/(4.11 x 105 m/s) = 0.0445 T
The magnetic field is perpendicular to the electric field. If the charges are moving to the right and the electric field points upward, the magnetic field is out of the page.
Problem#5
A singly ionized (one electron removed) 40K atom passes through a velocity selector consisting of uniform perpendicular electric and magnetic fields. The selector is adjusted to allow ions having a speed of 4.50 km/s to pass through undeflected when the magnetic field is 0.0250 T. The ions next enter a second uniform
magnetic field (B’) oriented at right angles to their velocity. 40K contains 19 protons and 21 neutrons and has a mass of 6.64 x 10-26 kg. (a) What is the magnitude of the electric field in the velocity selector? (b) What must be the magnitude of B’ so that the ions will be bent into a semicircle of radius 12.5 cm?
Answer:
Known:
magnetic field, B = 0.0250 T
speed, v = 4.50 km/s = 4.50 x 103 m/s
Velocity selector allows having desired velocity and then magnetic field bends the ion in a circular .
Velocity selector v = E/B, velocity is perpendicular to B,
mv2/R = qvB
(a) the magnitude of the electric field in the velocity selector is
E = vB = (4.50 x 103 m/s)(0.0250 T) = 112 V/m
(b) magnitude of so that the ions will be bent into a semicircle of radius 12.5 cm is
B’ = mv/qR
B’ = (6.64 x 10-26 kg)(4.50 x 103 m/s)/[(1.6 x 10-19 C)(0.125 m)]
B’ = 0.0149 T
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