A particle of mass m and charge q is accelerated by a potentiol difference V volt and made to enter a magnetic field region at an angle `y` with the field. At the same moment another particle of same mass and charge is projected in the direction of the field from the same point. What should be the speed of the second particle so that the particles meet again and again after regular interval of time which should be minimum.Also find the time interval after which they meet and the distance travelled by the second particle during the interval.

Question

Pratham Ashish · Answer

hi ,

velocity of first one, v1= (2*q*V /m) ^1/2

for first particle velocity parallel to the field v _{1z ^{= v}1 ^{cos y}}

_{^{perpendicular to field , v}1x^{= v1 sin y}}

speed of the second particle so that the particles meet at minimumtime will be equal to the parallel component of v1

so v2 = v1 cos y

time interval after they meet will be the time taken in one ratation of particle 1,

t = 2*pi* r / v1 siny

where r = m v1 sin y /q B

hence t = 2 m pi / q B

distance travelled by the second one durin the interval

d = v2 * t

= 2 m* v1 cos y* pi / q* B

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