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  • Two charged particles traverse identical helical paths in an opposite sense in a uniform magnetic field

Two charged particles traverse identical helical paths in an opposite sense in a uniform magnetic field.

B=B_{0}\widehat{k}

a) they have equal z-components of momenta

b) they must have equal charges

c) they necessarily represent a particle-antiparticle pair

d) the charge to mass ratio satisfy: \left (\frac{e}{m} \right )_{1}+\left (\frac{e}{m} \right )_{2}=0

Answers (1)

The answer is the option (d) As shown in the diagram if the particle is drawn in the x-y plane at an angle of θ and moving with a velocity v, so as to make one component along the field (v \cos\theta ) and the other perpendicular(v \sin \theta ), we need to resolve the velocity in rectangular components. After doing so, the particle gains a constant velocity along the field of (v \cos\theta ). The distance travelled by the particle is known as pitch.

The pitch of the helix (i.e, linear distance travelled in one rotation ) will be given by p = T(v\cos \theta )=2p\frac{m}{qB}(v \cos \theta)

For given pitch p corresponds to change particle, we have

\frac{q}{m}=\frac{2 \pi v\cos \theta}{qB}

In this particular case, the path covered by the particles is identical and helical in a completely opposite manner in the presence of a uniform magnetic field, B. This indicates their LHS is the same and has the opposite sign. Therefore, \left (\frac{e}{m} \right )_{1}+\left (\frac{e}{m} \right )_{2}=0

 

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