Ch 34 Physics for Scientists and Engineers

Ch 34 Physics for Scientists and Engineers -...

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ELECTROMAGNETIC FIELDS AND WAVES w.1. Model: The net magnetic flux over a closed surface is zero. Visualize: Please refer to Ex34.1. Solve: Because we can't enclose a "net pole" within a surface, Q,,, = fB.di = 0. Since the magnetic field is uniform over each face of the box, the total magnetic flux around the box is (1 cm x 2 cm)(-2 T- 2 T + 1 T + 5 T) -t (1 cm x 1 cm)(-2 T)+ Qunknown = 0 j j E-,, . arli = -0.0002 T m2 3 B~,,,,~ COS@ = -2.0 T The angle 8 must be 180". Because 8is the angle between $ and the outward normal of di, the field B is directed into the face. 34.2. Model: Visualize: magnetic field is in the direction of the negative z-axis. Solve: Electric and magnetic fields exert forces on charged particles. Assume the fields are uniform. Please refer to Figure Ex34.2. The electric field is in the direction of the positive y-axis and the Substituting into the Lorentz force law, V m F, = q( E + C x E) = (1.6 x C) 1.0 x 106j - + (-2.0 x 1073 m / s) x (-0.10i T)) = (3.2; + 1.6j) x N 3 = dm x N = 3.58 x N 8 = tad( N = 26.6" 1.6 x 3.2 x 10-1~ N 1 The angle 8 is above the positive x-axis. 34.3. Model: Solve: Electric and magnetic fields exert forces on charged particles. Assume the fields are uniform. Substituting into the Lorentz force law, I$,,, = q(E+ G x = (-1.6 x C) 5.0 x IO6; m / x (-0.lOi T = -(3.2 x 10-14)(1 -3) N - (8.0 x 10-l.S) N= (-3.2 x - 4.8 x 10-l4j) N 34.4. Model: Visualize: Please refer to Figure Ex34.4. The magnetic force on the negative electron by the right-hand rule directed downward. So that the electron is undeflected, we must apply an electric field to cause an electric force directed upward. That is, the electric field must point downward. Solve: Assume that the electric and magnetic fields are uniform fields. For the electron to not deflect, F;, = 4 a clC x 4 = eE =j E = vBsin90" = (2.0 x 10' m / s)(O.OIO T) = 2.0 x 10' V / m 34.5. Model: Assume the electric and magnetic fields are uniform. Visualize: Solve: The force on the proton. which is the sum of the electric and magnetic forces, is Please refer to Figure Ex34.5. F = + = -Fcos30°f + Fsin30'3 = (-2.77; +1.603)x N
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34-2 Chapter 34 Since v’ points out of the page, the magnetic force is F, = e< x B = 1.66 x lO-I33 N.Thus ---. =eE=F-F,=-2.77~10-‘~~N~~=~,/e=-1.73x10~~ V/m That is, the electric field is = (1.73~10~ V/m, left). 34.6. Model: Solve: Assume that the electric and magnetic fields are uniform fields. Substituting into the Lorentz force law, = q(E+ x B) 3 (9.6 x N)(; - i) = -(1.60 x .)[E+ (5.0 x lo6; m / s) x (0.103 T)] E = -(6.0 x io5 N / c)(; - - x c)L = (-6.0; + i.oL ) x c 34.7. Model: Apply the Galilean transformation of velocity. Solve: (a) In the laboratory frame S, the speed of the proton is v = j(1.41 x lo6 m / s)’ +(1.41 x / s)’ = 2.0 x lo6 m / s The angle the velocity vector makes with the positive y-axis is = 45” 1 1.41~10~ m/s e = tan-’ (b) In the rocket frame S’, we need to fist determine the vector V’. Equation 34.5 yields: V’ = V - v = (1.41 x lo6; + 1.41 x 106y) m / s -(1.0 x lo6;) m / s = (0.41 x lo6; + 1.41 x 106j) m / s The speed of the proton is v’=4(O.41x1O6 m/s)’ +(1.41x106 m/s)’ =1.47x106 m/s The angle the velocity vector makes with the positive y’-axis is = 16.2” 1 0.41~10~ 8’ = tan-’ 34.8. Model: Apply the Galilean transformation of fields.
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Ch 34 Physics for Scientists and Engineers -...

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