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QPages 230-234: Q45: HC Verma: Concepts of Physics - Volume 2 Magnetic Field Pages 230-234: Q45 Solution: Two particles, each with mass are placed at a separation in a…

Two particles, each with mass mm are placed at a separation dd in a uniform magnetic field BB, as shown in the figure. They have opposite charges of equal magnitude qq. At time t=0t = 0, the particles are projected towards each other, each with a speed vv. Suppose the Coulomb force between the charges is switched off. a) Find the maximum value vmv_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=vm/2v = v_m/2? c) At what instant will a collision occur between the particles if v=2vmv = 2v_m? d) Suppose v=2vmv = 2v_m and the collision between the particles is completely inelastic. Describe the motion after the collision.

Step-by-Step Verified Solution

**Answer:** (a) vm=qBd2mv_m = \frac{q B d}{2m} (b) Min separation = d/2d/2, Max = dd (c) t=πmqBt = \frac{\pi m}{q B} (d) They move together in a straight line. **Step-by-Step Explanation:** (a) For no collision, the particles must not meet at the midpoint. The condition gives vm=qBd2mv_m = \frac{q B d}{2m}. (b) For v=vm/2v = v_m/2, the minimum separation is d/2d/2 and maximum is dd. (c) For v=2vmv = 2v_m, the collision occurs at time t=πmqBt = \frac{\pi m}{q B}. (d) After inelastic collision, they have zero net charge motion and move in a straight line.

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