Halliday- capitulo 28 resolução

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1. (a) The force on the electron is
ˆ i j i FB = qv × B = q vx ˆ + v y ˆ × Bx ˆ + By j = q ( vx By − v y Bx ) k = −1.6 ×10−19 C ⎡ 2.0 × 106 m s ( −0.15 T ) − 3.0 × 106 m s ( 0.030 T ) ⎤ ⎣ ⎦ ˆ = 6.2 × 10−14 N k.

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Thus, the magnitude of FB is 6.2 × 1014 N, and FB points in the positive z direction. (b) This amounts to repeating the above computation with a change in the sign in the charge. Thus, FB has the same magnitude but points in the negative z direction, namely, ˆ F = − 6.2 × 10−14 N k.
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2. (a) We use Eq. 28-3: FB = |q| vB sin φ = (+ 3.2 × 10–19 C) (550 m/s) (0.045 T) (sin 52°) = 6.2 × 10–18 N. (b) a = FB/m = (6.2 × 10– 18 N) / (6.6 × 10– 27 kg) = 9.5 × 108 m/s2. (c) Since it is perpendicular to v , FB does not do any work on the particle. Thus from the work-energy theorem both the kinetic energy and the speed of the particle remain unchanged.

3. (a) Eq. 28-3 leads to v= 6.50 × 10−17 N FB = = 4.00 × 105 m s . −19 −3 eB sin φ . 160 × 10 C 2.60 × 10 T sin 23.0°

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(b) The kinetic energy of the proton is
K=
2 1 2 1 mv = (1.67 ×10−27 kg )( 4.00 × 105 m s ) = 1.34 ×10−16 J , 2 2

which is equivalent to K = (1.34 × 10– 16 J) / (1.60 × 10– 19 J/eV) = 835 eV.

4. The force associated with the magnetic field must point in the j direction in order to

cancel the force of gravity in the − j direction. By the right-hand rule, B points in the

− k direction (since i × − k = j ). Note that the charge is positive; also note that we need to assume By = 0. The magnitude |Bz| is given by Eq. 28-3 (with φ = 90°). Therefore, with m = 1.0 × 10−2 kg , v = 2.0 ×104 m/s and q = 8.0 × 10−5 C , we find ⎛ mg ⎞ ˆ ˆ ˆ B = Bz k = − ⎜ ⎟ k = (−0.061 T)k qv ⎠ ⎝

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5. Using Eq. 28-2 and Eq. 3-30, we obtain
F = q v x By − v y Bx k = q v x 3Bx − v y Bx k

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where we use the fact that By = 3Bx. Since the