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Two particles, each having a mass m are placed at a separation d in a uniform magnetic field B as shown in the figure. They have opposite charges of equal magnitude q. At time t = 0, the particles are projected towards each other, each with a speed v. Suppose the Coulomb force between the charges is switched off. (a) Find the maximum value v_(m) of the projection speed so that the two particles do not collide. (b) What would be the minimum and maximum separation between the particles if v = v_(m) //2 ? ( c) At what instant will a collision occur between the particles if v=2 v_(m) ? (d) Suppose v = 2 v_(m) and the collision between the particles is completely inelastic. Describe the motion after the collision. |
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Answer» Solution :(a) `R_(max) = d//2, v = v_(m)` `(d)/(2) = (mv_(m))/(Bq) RARR v_(m) = (Bqd)/(2m)` (b) `v = v_(m)//2`i.e.`R = d//4`  Minimum SEPARATION `= d//2` Maximum separation `= 3d//2` ( c) `v = 2v_(m)` i.e. `R = d`. The particle will collide in middle  `sin theta = (d//2)/(R = d) = (1)/(2) rArr theta = pi//6` The particles will collide after time `t = (m theta)/(Bq) = (m pi)/(6 Bq)` (d) Before collision :  After collision:  Momentum conservation: `x:v_(x) = 0` `y: m . 2v_(m) sin theta + m . 2v_(m) sin theta = (m + m) v_(y)` `q` will CANCEL `-q`, net charge `= 0` The combined mass will move in a straight line with speed `v_(m)`. |
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