Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 27 | a F qiïx x 10-8 C x 104m s T j x i F - x 10-4 N k. b F qV x B F x 10-8 C T x 104m s j x k x 104m s i x F x 10-4 N i x 10-4 N j. Need a force from the magnetic field to balance the downward gravitational force. Its magnitude is mg qvB mg B qv x 10-4 kg m s2 x 10-8 C x 104m s . The right-hand rule requires the magnetic field to be to the east since the velocity is northward the charge is negative and the force is upwards. By the right-hand rule the charge is positive. qv x B F ma qv x B a -------- m x 10-8 C x 104m s T j x i x 10-3 kg . J - m s2 k. See figure on next page. Let F0 qvB then Fa F0 in the - k direction Fb F0 in the j direction Fc 0 since B and velocity are parallel Fd F0 sin 45o in the - j direction Fe F0 in the - j k direction a The smallest possible acceleration is zero when the motion is parallel to the magnetic field. The greatest acceleration is when the velocity and magnetic field are at right angles a JVB d6 x 10- C x 106ms x 10-2 T y 10 .ms . m x 10-31 kg b If a 1 x 1016m s2 qvB sin sin 0 0 . 4 m F F q v B sin tb v -T- ---------- q B sin x 10-15 N x 10 19 C x 10-3 T sin 60 x 106 ms. _ . TT _ _ jv JV _ _ _ _ a F qv x B qBz vx jx k vy j x k vz k x Jt qBz vx - vy i . Set this equal to the given value of F to obtain vx vy Fy - qBz x 10-7 N - - x 10-9 C T -106 ms qBz - x 10-7 N - x 10-9 C T - ms. b The value of vz is indeterminate. -- F F c v F vxFx v F vzFz Fx F 0 9 90 . x x y y z z - qBz x qBz y F qv x B v vyj withvy - x 103ms Fx x 10-3 N Fy 0 and Fz - x 10-3 N Fx q vyBz - vzBy qVyBz Bz Fx qvy x 10-3 N x 10-6 C x 103ms T Fy q vzBx - vxBz 0 which is consistent with F as given in the problem. No force component along the direction of the velocity. Fz q vxBy - vyBx - qvyBx Bx - Fz .
