Problem#1
A particle with a charge of -1.24 x 10-8 C is moving with instantaneous velocity v = (4.9 x 104 m/s)i + (─3.85 x 104 m/s)j. What is the force exerted on this particle by a magnetic field (a) B = (1.40 T)i and (b) B = (1.40 T)k.
Answer:
known:
velocity v = (4.9 x 104 m/s)i + (─3.85 x 104 m/s)j
charge of q = ─1.24 x 10-8 C
(a) the force exerted on this particle by a magnetic field B = (1.40 T)i is
F = qv x B = (─1.24 x 10-8 C)[(4.9 x 104 m/s)i + (─3.85 x 104 m/s)j] X [(1.40 T)i]
F = (─1.24 x 10-8 C) (1.40 T)[(4.9 x 104 m/s)i x i + (─3.85 x 104 m/s)j x i]
Because: i x i = 0, j x i = ─k¸ then
F = (─1.24 x 10-8 C) (1.40 T)[0 + (─3.85 x 104 m/s)(─k)]
F = (─6.68 x 104 N)k
The right─hand rule gives that v x B is directed out the paper (+z direction). The charge is negative so F is opposite to v x B.
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| Fig.1 |
(b) the force exerted on this particle by a magnetic field B = (1.40 T)k is
F = qv x B = (─1.24 x 10-8 C)[(4.9 x 104 m/s)i + (─3.85 x 104 m/s)j] X [(1.40 T)k]
F = (─1.24 x 10-8 C) (1.40 T)[(4.9 x 104 m/s)i x k + (─3.85 x 104 m/s)j x k]
Because: i x k = ─j, j x k = i¸ then
F = (─1.24 x 10-8 C) (1.40 T)[0 + (─3.85 x 104 m/s)(─k)]
F = (─7.27 x 104 N)(─j) + (6.68 x 104 N)i
F = (7.27 x 104 N)j + (6.68 x 104 N)i
The direction of F is opposite to v x B since q is negative. The direction of F comuted from the right─hand rule agrees ualitatively with the direction calculated with vectors.
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| Fig.2 |
Problem#2
A particle of mass 0.195 g carries a charge of ─2.5 x 10─8 C. The particle is given an initial horizontal velocity that is due north and has magnitude What are the magnitude 4.00 x 104 m/s and direction of the minimum magnetic field that will keep the particle moving in the earth’s gravitational field in the same horizontal, northward direction?
Answer:
Known:
Mass, m = 0.195 g = 0.195 x 10─3 kg
Charge, q = ─2.5 x 10─8 C
Magnitude velocity, v = 4.00 x 104 m/s
The gravity force is downward so the force from the magnetic field must be upward. The charge’s velocity and the force are shown figure 3. Since the charge is negative, the magnetic force is opposite to the right─hand rule direction. The minimum magnetic field is when the field is perpendicular to v. The force is also perpendicular to B, so B is either eastward or westward.
If B is eastward, the right─hand rule direction is into the page and FB is out of th e page, as required. Therefore, B is eastward
mg = lqlvB sin θ, with θ = 900
mg = lqlvB
B = mg/lqlv = (0.195 g = 0.195 x 10─3 kg)(9.8 m/s2)/[(─2.5 x 10─8 C)(v = 4.00 x 104 m/s )]
B = 1.91 T
Answer:
Known:
Mass, m = 0.195 g = 0.195 x 10─3 kg
Charge, q = ─2.5 x 10─8 C
Magnitude velocity, v = 4.00 x 104 m/s
The gravity force is downward so the force from the magnetic field must be upward. The charge’s velocity and the force are shown figure 3. Since the charge is negative, the magnetic force is opposite to the right─hand rule direction. The minimum magnetic field is when the field is perpendicular to v. The force is also perpendicular to B, so B is either eastward or westward.
If B is eastward, the right─hand rule direction is into the page and FB is out of th e page, as required. Therefore, B is eastward
mg = lqlvB sin θ, with θ = 900
mg = lqlvB
B = mg/lqlv = (0.195 g = 0.195 x 10─3 kg)(9.8 m/s2)/[(─2.5 x 10─8 C)(v = 4.00 x 104 m/s )]
B = 1.91 T
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| Fig.3 |
Problem#3
In a 1.25-T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50 μC and initially moving northward 4.75 km/s at is deflected toward the east. (a) What is the sign of the charge of this particle? Make a sketch to illustrate how you found your answer. (b) Find the magnetic force on the particle.
Answer:
Known:
magnetic field, B = 1.25 T
charge of magnitude, q = 8.50 μC = 8.50 x 10─6 C
Magnitude velocity, v = 4.75 km/s = 4.75 x 103 m/s
(a) when you apply the right─hand to v and B, your thumb points east. F is in this direction, so the charge is positive
(b) Use F = lqlvB sin θ, with θ = 900
F = lqlvB = (8.50 x 10─6 C)(4.75 x 103 m/s)(1.25 T)
F = 0.0505 N
Problem#4
A particle with mass 1.81 x 10─3 kg and a charge of 1.22 x 10─8 C has, at a given instant, a velocity v = (3.00 x 106 m/s)j. What are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field B = (1.63 T)i + (0.980 T)j?
Answer:
Known:
uniform magnetic field, B = (1.63 T)i + (0.980 T)j
velocity, v = (3.00 x 106 m/s)j
mass, m = 1.81 x 10─3 kg
charge, q = 1.22 x 10─8 C
Apply Newton’s second law, with the force being the magnetic force
F = ma = qv x B
Gives
a = qv x B/m
a = (1.22 x 10─8 C)((3.00 x 106 m/s)j)[(1.63 T)i + (0.980 T)j]/(1.81 x 10─3 kg)
= (1.22 x 10─8 C)(3.00 x 106 m/s) (1.63 T)(j x i)/(1.81 x 10─3 kg)
a = ─(0.330 m/s2)k
the acceleration is in the ─z─direction and is perpendicular to both v and B.
In a 1.25-T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50 μC and initially moving northward 4.75 km/s at is deflected toward the east. (a) What is the sign of the charge of this particle? Make a sketch to illustrate how you found your answer. (b) Find the magnetic force on the particle.
Answer:
Known:
magnetic field, B = 1.25 T
charge of magnitude, q = 8.50 μC = 8.50 x 10─6 C
Magnitude velocity, v = 4.75 km/s = 4.75 x 103 m/s
(a) when you apply the right─hand to v and B, your thumb points east. F is in this direction, so the charge is positive
(b) Use F = lqlvB sin θ, with θ = 900
F = lqlvB = (8.50 x 10─6 C)(4.75 x 103 m/s)(1.25 T)
F = 0.0505 N
Problem#4
A particle with mass 1.81 x 10─3 kg and a charge of 1.22 x 10─8 C has, at a given instant, a velocity v = (3.00 x 106 m/s)j. What are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field B = (1.63 T)i + (0.980 T)j?
Answer:
Known:
uniform magnetic field, B = (1.63 T)i + (0.980 T)j
velocity, v = (3.00 x 106 m/s)j
mass, m = 1.81 x 10─3 kg
charge, q = 1.22 x 10─8 C
Apply Newton’s second law, with the force being the magnetic force
F = ma = qv x B
Gives
a = qv x B/m
a = (1.22 x 10─8 C)((3.00 x 106 m/s)j)[(1.63 T)i + (0.980 T)j]/(1.81 x 10─3 kg)
= (1.22 x 10─8 C)(3.00 x 106 m/s) (1.63 T)(j x i)/(1.81 x 10─3 kg)
a = ─(0.330 m/s2)k
the acceleration is in the ─z─direction and is perpendicular to both v and B.
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| Fig.4 |
Problem#5
A particle with charge is 7.80 μC moving with velocity v = ─(3.80 x 103 m/s)j. The magnetic force on the particle is measured to be F = +(7.60 x 10─3 N)i ─ (5.20 x 10─3 N)k. (a) Calculate all the components of the magnetic field you can from this information. (b) Are there components of the magnetic field that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product B.F . What is the angle between B and F.
Answer:
Known:
Charge, q = 7.80 μC = 7.80 x 10─6 C
Velocity, v = ─(3.80 x 103 m/s)j
Magnetic force, F = +(7.60 x 10─3 N)i ─ (5.20 x 10─3 N)k
Apply F = qv x B
(a) Fx = q(vyBz ─ vzBy) = qvyBz
Bz = Fx/(qvy)
Bz = (7.60 x 10─3 N)/[(7.80 x 10─6 C)(─3.80 x 103 m/s)]
Bz = ─0.256 T
Fy = 0 = q(vzBx ─ vxBz), with is consistent with F as given in the problem. There is no force component along the direction of the velocity.
Fz = q(vxBy ─ vyBx) = ─qvyBx
Bx = ─Fz/(qvy)
Bx = ─(─5.20 x 10─3 N)/[(7.80 x 10─6 C)(─3.80 x 103 m/s)]
Bx = ─0.175 T
(b) By is not determined. No force due to this component of B along v, measurement of the force tells us nothing about By.
(c) B.F = (Bxi + Byj + Bzk).(Fxi + Fyj +
B.F = (─0.175 T)(7.60 x 10─3 N) + (─0.256 T)(─5.20 x 10─3 N) = 0
B and F are perpendicular (angle 900). The force is perpendicular to both v and B, so v.F is also zero.
Problem#6
A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50 km/s in the +x─direction experiences a force of 2.25 x 10─16 N in the +y─direction, and an electron moving at 4.75 km/s in the ─z─direction experiences a force of 8.50 x 10─16 N in the +y-direction. (a) What are the magnitude and direction of the magnetic field? (b) What are the magnitude and direction of the magnetic force on an electron moving in the +y-direction at 3.20 km/s.
Answer:
Known:
F is perpendicular to both v and B. Since the force on the proton is in the +y─direction, By = 0 and B = Bxi + Bzk.
For the proton,
vp = (1.50 km/s)i = vpi and Fp = (2.25 x 10─16 N)j = Fpj.
For the electron,
ve = ─(4.75 km/s)k = vek and Fe = (8.50 x 10─16 N)j = Fej.
the magnetic force is
F = qv x B
(a) For the proton,
Fp = qvp x Bp gives,
Fpj = evpi x (Bxi + Bzk) = ─evpBzj.
Solving for BZ gives
Bz = ─Fp/(evp) = (2.25 x 10─16 N)/[(1.60 x 10─19 C)(1500 m/s)] = ─0.9375 T
For the electron,
Fe = qve x Be
Which gives
Fej = ─e(─vek) x (Bxi + Bzk) = ─eveBxj
Solving for Bx gives
Bx = Fe/(eve) = (8.50 x 10─16 N)/[(1.60 x 10─19 C)(4750 m/s)] = 1.118 T
Therefore,
B = 1.118 Ti ─ 0.9375 Tk
The magnitude of the field is
B = [(Bx)2 + (Bz)2]1/2
B = [(1.118 T)2 + (─ 0.9375 T)2]1/2 = 1.46 T
Calling θ the angle that the magnetic field makes with the +x─axis, we have
Tan θ = Bz/Bx = (─ 0.9375 T)/1.118 T = ─0.8386
So, θ = tan─1(─0.8386) = ─40.00.
Therefore, the magnetic field is in the xz─plane directied at 40.00 from the +x─axis toward the ─z─axis, having a magnitude of 1.46 T.
(b) B = Bxi + BZk and v = (3.2 km/s)(─j)
F = qv x B
F = (─e)(3.2 km/s)(─j) x (Bxi + Bzk) = e(3.2 x 103 m/s)[Bx(─k) + Bzi)
F = e(3.2 x 103 m/s)[(─1.118 Tk ─ 0.9375 Ti)
F = ─4.80 x 10─16 Ni ─ 5.726 x 10─16 Nk
The magnitude of the force is
F = [(Fx)2 + (Fz)2]1/2
B = [(─4.80 x 10─16 N)2 + (─ 5.726 x 10─16 N)2]1/2 = 7.47 x 10─16 N
Calling θ the angle that the magnetic force makes with the +x─axis, we have
Tan θ = Fz/Fx = (─ 5.726 x 10─16 N)/(─4.80 x 10─16 N)
So, θ = 50.00
Therefore, the force is in the xz─plane directied at 50.00 from the +x─axis toward the ─z─axis.
The force on the electrons in parts (a) and (b) are comparable in magnitude because the electron speeds are comparable in both cases.
Answer:
Known:
Charge, q = 7.80 μC = 7.80 x 10─6 C
Velocity, v = ─(3.80 x 103 m/s)j
Magnetic force, F = +(7.60 x 10─3 N)i ─ (5.20 x 10─3 N)k
Apply F = qv x B
(a) Fx = q(vyBz ─ vzBy) = qvyBz
Bz = Fx/(qvy)
Bz = (7.60 x 10─3 N)/[(7.80 x 10─6 C)(─3.80 x 103 m/s)]
Bz = ─0.256 T
Fy = 0 = q(vzBx ─ vxBz), with is consistent with F as given in the problem. There is no force component along the direction of the velocity.
Fz = q(vxBy ─ vyBx) = ─qvyBx
Bx = ─Fz/(qvy)
Bx = ─(─5.20 x 10─3 N)/[(7.80 x 10─6 C)(─3.80 x 103 m/s)]
Bx = ─0.175 T
(b) By is not determined. No force due to this component of B along v, measurement of the force tells us nothing about By.
(c) B.F = (Bxi + Byj + Bzk).(Fxi + Fyj +
B.F = (─0.175 T)(7.60 x 10─3 N) + (─0.256 T)(─5.20 x 10─3 N) = 0
B and F are perpendicular (angle 900). The force is perpendicular to both v and B, so v.F is also zero.
Problem#6
A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50 km/s in the +x─direction experiences a force of 2.25 x 10─16 N in the +y─direction, and an electron moving at 4.75 km/s in the ─z─direction experiences a force of 8.50 x 10─16 N in the +y-direction. (a) What are the magnitude and direction of the magnetic field? (b) What are the magnitude and direction of the magnetic force on an electron moving in the +y-direction at 3.20 km/s.
Answer:
Known:
F is perpendicular to both v and B. Since the force on the proton is in the +y─direction, By = 0 and B = Bxi + Bzk.
For the proton,
vp = (1.50 km/s)i = vpi and Fp = (2.25 x 10─16 N)j = Fpj.
For the electron,
ve = ─(4.75 km/s)k = vek and Fe = (8.50 x 10─16 N)j = Fej.
the magnetic force is
F = qv x B
(a) For the proton,
Fp = qvp x Bp gives,
Fpj = evpi x (Bxi + Bzk) = ─evpBzj.
Solving for BZ gives
Bz = ─Fp/(evp) = (2.25 x 10─16 N)/[(1.60 x 10─19 C)(1500 m/s)] = ─0.9375 T
For the electron,
Fe = qve x Be
Which gives
Fej = ─e(─vek) x (Bxi + Bzk) = ─eveBxj
Solving for Bx gives
Bx = Fe/(eve) = (8.50 x 10─16 N)/[(1.60 x 10─19 C)(4750 m/s)] = 1.118 T
Therefore,
B = 1.118 Ti ─ 0.9375 Tk
The magnitude of the field is
B = [(Bx)2 + (Bz)2]1/2
B = [(1.118 T)2 + (─ 0.9375 T)2]1/2 = 1.46 T
Calling θ the angle that the magnetic field makes with the +x─axis, we have
Tan θ = Bz/Bx = (─ 0.9375 T)/1.118 T = ─0.8386
So, θ = tan─1(─0.8386) = ─40.00.
Therefore, the magnetic field is in the xz─plane directied at 40.00 from the +x─axis toward the ─z─axis, having a magnitude of 1.46 T.
(b) B = Bxi + BZk and v = (3.2 km/s)(─j)
F = qv x B
F = (─e)(3.2 km/s)(─j) x (Bxi + Bzk) = e(3.2 x 103 m/s)[Bx(─k) + Bzi)
F = e(3.2 x 103 m/s)[(─1.118 Tk ─ 0.9375 Ti)
F = ─4.80 x 10─16 Ni ─ 5.726 x 10─16 Nk
The magnitude of the force is
F = [(Fx)2 + (Fz)2]1/2
B = [(─4.80 x 10─16 N)2 + (─ 5.726 x 10─16 N)2]1/2 = 7.47 x 10─16 N
Calling θ the angle that the magnetic force makes with the +x─axis, we have
Tan θ = Fz/Fx = (─ 5.726 x 10─16 N)/(─4.80 x 10─16 N)
So, θ = 50.00
Therefore, the force is in the xz─plane directied at 50.00 from the +x─axis toward the ─z─axis.
The force on the electrons in parts (a) and (b) are comparable in magnitude because the electron speeds are comparable in both cases.
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