Problem#1
Positive point charges q = +8.00 μC and q’ = +3.00 μC are moving relative to an observer at point P, as shown in Fig. 1. The distance d is 0.120 m, v = 4.50 x 106 m/s and v’ = 9.00 x 106 m/s. (a) When the two charges are at the locations shown in the figure, what are the magnitude and direction of the net magnetic field they produce at point P? (b) What are the magnitude and direction of the electric and magnetic forces that each charge exerts on the other, and what is the ratio of the magnitude of the electric force to the magnitude of the magnetic force? (c) If the direction of v is reversed, so both charges are moving in the same direction, what are the magnitude and direction of the magnetic forces that the two charges exert on each other?
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| Fig.1 |
Answer:
Magnetic field of a point charge with constant velocity given by
B = (μ0/4π)(qv x r)/r3
(a) When the two charges are at the locations shown in the figure, the magnitude and direction of the net magnetic field they produce at point P is
Bnet = B + B’
With, B = (μ0/4π)(qv sin 900)/d2 (into the paper)
and Bel = (μ0/4π)(q’v’ sin 900)/d2 (into the paper), then
Bnet = (μ0/4π)(qv + q’v’)/d2
Bnet = [(4π x 10─7 T.m/A)/4π][(8.00 x 10─6 C)(4.50 x 106 m/s) + (3.00 x 10─6 C)(9.00 x 106 m/s)]/(0.120 m)2
Bnet = 4.38 x 10─4 T, into the paper
(b) we can find the magnetic force between the charges:
FB = (μ0/4π)(qq’vv’)/d2
FB = [(4π x 10─7 T.m/A)/4π][(8.00 x 10─6 C)(3.00 x 10─6 C)(4.50 x 106 m/s)(9.00 x 106 m/s)]/(0.240 m)2
FB = 1.69 x 10─3 N
The force on the upper charge points up and the force on the lower charge points down. The Coulomb force between the charges is
FC = (μ0/4π)(qq’/r2)
FC = [(4π x 10─7 T.m/A)/4π][(8.00 x 10─6 C)(3.00 x 10─6 C)]/(0.240 m)2
FC = 3.75 N
The force on the upper charge points up and the force on the lower charge points down. The ratio of the Coulomb force to the magnetic force is
FC/FB = 3.75 N/(1.69 x 10─3 N) = 2.22 x 103
the Coulomb force is much larger.
(c) the magnetic forces are reversed in direction when the direction of only one velocity is reversed but the magnitude of the forces is unchanged.
Problem#2
Figure 1 shows two point charges, q and q’, moving relative to an observer at point P. Suppose that the lower charge is actually negative, with q’ = ─q. (a) Find the magnetic field (magnitude and direction) produced by the two charges at point P if (i) v’ = v/2; (ii) v’ = v; (iii) v’ = 2v. (b) Find the direction of the magnetic force that q exerts on and find the direction of the magnetic force that exerts on q’. (c) If v = v’ = 3.00 x 105 m/s, what is the ratio of the magnitude of the magnetic force acting on each charge to that of the Coulomb force acting on each charge?
Answer:
Magnetic field of a point charge with constant velocity given by
B = (μ0/4π)(qv x r)/r3
(a) if q’ = ─q,
Bq = (μ0/4π)(qv/d2) into the page and
Bq’ = (μ0/4π)(qv’/d2) out the page
the magnetic field (magnitude and direction) produced by the two charges at point P if
(i) v’ = v/2;
B = Bq + Bq’
B = (μ0/4π)(qv/d2) ─ (μ0/4π)(qv/2d2)
B = μ0qv/8πd2, into the page
(ii) v’ = v;
Gives B = 0
(iii) v’ = 2v
B = Bq + Bq’
B = (μ0/4π)(qv/d2) ─ (μ0/4π)(2qv/d2)
B = μ0qv/4πd2, out the page
(b) the force that q exerts is given by
F = q’v x Bq
So, F = μ0q2v’v/16πd2
Bq into the page, so the force on q’ is toward q. the force that q’ exerts on q is is toward q’. the force between the two charges is attractive.
(c) the ratio of the magnitude of the magnetic force acting on each charge to that of the Coulomb force acting on each charge if v = v’ is
FB/FC = μ0ε0vv’ = μ0ε0(3.00 x 105 m/s)2 = 1.00 x 10─6
Because, FC = (q2/4πε0)(1/4d2)
Problem#3
An electron and a proton are each moving at 845 km/s in perpendicular paths as shown in Fig. 2. At the instant when they are at the positions shown in the figure, find the magnitude and direction of (a) the total magnetic field they produce at the origin; (b) the magnetic field the electron produces at the location of the proton; (c) the total electric force and the total magnetic force that the electron exerts on the proton.
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| Fig.2 |
Answer:
Magnetic field of a point charge with constant velocity given by
B = (μ0/4π)(qv sin φ)/r2
(a) Both moving charges produce magnetic fields, and the net field is the vector sum of the two.
Therefore,
Bnet = Bpro + Bel
With, Balpha = (μ0/4π)(2ev sin 900)/rpro2 (into the paper)
and Bel = (μ0/4π)(ev sin 900)/rel2 (into the paper), then
Bnet = (μ0ev/4π)(1/rpro2 + 1/rel2)sin 900
B = [(4π x 10─7 T.m/A)/4π][(1.60 x 10─19 C)(8.45 x 105 m/s)][1/(5.00 x 10-9 m)2 + 1/(4.00 x 10-9 m)2]
B = 1.39 x 10─3 T, into of the paper
(b) r2 = (5.00 x 10-9 m)2 + (4.00 x 10-9 m)2 = 4.10 x 10-17 m
and φ = 1800 ─ arctan(5/4) = 128.70, we get
B = [(4π x 10─7 T.m/A)/4π][(1.60 x 10─19 C)(8.45 x 105 m/s) sin 128.70]/(4.10 x 10-17 m)
B = 2.58 x 10─4 T, into of the paper
(c) the magnetic force on a moving charge is
FB = qvB sin φ
= (1.60 x 10─19 C)(8.45 x 105 m/s)(2.58 x 10─4 T) sin 900
FB = 3.48 x 10─11 N, in the +x direction
And the total electric force is
Felec = ke2/r2
= (9.0 x 109 Nm2/C2)(1.60 x 10─19 C)2/(4.10 x 10-17 m)
Felec = 5.62 x 10 ─12 N
Direction Felec is
θ = arc tan (5/4) = 51.30 below the +x─axis measured clockwise.
Problem#4
A negative charge q = ─3.60 x 10─6 C is located at the origin and has velocity v = (7.50 x 104 m/s)i + (─4.90 x 104 m/s)j . At this instant what are the magnitude and direction of the magnetic field produced by this charge at the point x = 2.00 m, y = ─0.300 m, z = 0?
Answer:
Velocity, v = (7.50 x 104 m/s)i + (─4.90 x 104 m/s)j
Magnetic field of a point charge with constant velocity given by
B = (μ0/4π)(qv x r)/r3
so that, the magnetic-field vector B it produces at the following points:
for x = 2.00 m, y = ─0.300 m, z = 0; or r = (2.00 m)i + (─0.300 m)j, then r = 2.022 m is
B = [(4π x 10─7 T.m/A)(─3.60 x 10─6 C)/4π]{[(7.50 x 104 m/s)i + (─4.90 x 104 m/s)j] x [(2.00 m)i + (─0.300 m)j]/(2.022 m)3
B = (─4.35 x 10─14)[(─2.25 x 104)k + (─9.80 x 104 m/s)(─k)] T
B = (─3.28 x 10─9 T)k = (─3.28 nT)k
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