Question

A long, horizontal wire is carrying a current 1. A particle of mass m and charge q is fired horizontally at a speed vo. The initial velocity of the particle is parallel to the wire, and its initial position is a distancer directly below the wire, as shown in (Figure 1). What initial speed must the particle have for it to travel in a straight line? Do not ignore gravity. Express your answer in terms of l, m, q, r, uo, and the acceleration of gravity g.

Answer 1

Magnetic force at distance r due to long wire carrying current I is given by

B=($\mu$₀I)/(2$\pi$r)

Magnetic force on particle of charge q moving with velocity v₀ is

F_B=qv₀B=(qv₀$\mu$₀I)/(2$\pi$r)

Direction of this force will be in upward direction(given by right hand thumb rule)

To travel in straight line this magnetic force should be equal to the gravitation force.

So

(qv₀$\mu$₀I)/(2$\pi$r)=mg

Initial speed,v₀=(2$\pi$mgr)/(qI$\mu$₀)