The charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field, \[\mathbf{B}={{B}_{0}}\overset\frown{k}\].
A) They have equal z-components of moments.
B) They must have equal charges.
C) They necessarily represent a particle, antiparticle pair.
D) The charge to mass ratio satisfies –
\[{{(\dfrac{{{q}_{1}}}{m})}_{1}}+{{(\dfrac{{{q}_{2}}}{m})}_{2}}=0\]
Answer
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Hint: We are given two charged particles in a magnetic field. Also, it is said that they follow the exact opposite path in the field, i.e., their behaviours in the field are just opposite although their magnitudes of components of pitch remain the same.
Complete answer:
We know any charged particle in a magnetic field with an initial horizontal velocity component exhibits a helical motion in the field. The pitch developed by the helical motion of the charged particle can be given by –
\[p=Tv\cos \theta \]
Where, T is the time period of completing one helical motion,
v is the initial velocity of the charged particle,
\[\theta \] is the angle made by the velocity of the charged particle with the horizontal.
We know the formula which determines the time period as –
\[\tau =\dfrac{2\pi m}{qB}\]
Substituting the above equation in the formula for pitch gives –
\[p=\dfrac{2\pi m}{qB}v\cos \theta \]
Rearranging the terms, so as to get the charge to mass ratio of the particle gives –
\[\dfrac{q}{m}=\dfrac{2\pi v\cos \theta }{pB}\]
We know that for the given two particles, the helical path has an equal magnitude, as a result p is constant. Also, all the other parameters are constant and therefore the charge to mass ratio is a constant.
\[\dfrac{2\pi v\cos \theta }{pB}=\text{constant}\]
Now, we are given that the two particles follow opposite paths. For all these conditions to follow, the two particles should have opposite charges with equal magnitudes. Therefore, the sum of the charge to mass ratio of the two particles will be zero.
This satisfies the equation –
\[{{(\dfrac{{{q}_{1}}}{m})}_{1}}+{{(\dfrac{{{q}_{2}}}{m})}_{2}}=0\]
So, the correct answer is “Option D”.
Note:
The pitch of a helical path followed by a charged particle in a magnetic field is the linear distance covered by the particle along the horizontal in one rotation. For a particle projected vertically, the cosine component becomes zero and the particle doesn’t have a pitch, it undergoes circular motion.
Complete answer:
We know any charged particle in a magnetic field with an initial horizontal velocity component exhibits a helical motion in the field. The pitch developed by the helical motion of the charged particle can be given by –
\[p=Tv\cos \theta \]
Where, T is the time period of completing one helical motion,
v is the initial velocity of the charged particle,
\[\theta \] is the angle made by the velocity of the charged particle with the horizontal.
We know the formula which determines the time period as –
\[\tau =\dfrac{2\pi m}{qB}\]
Substituting the above equation in the formula for pitch gives –
\[p=\dfrac{2\pi m}{qB}v\cos \theta \]
Rearranging the terms, so as to get the charge to mass ratio of the particle gives –
\[\dfrac{q}{m}=\dfrac{2\pi v\cos \theta }{pB}\]
We know that for the given two particles, the helical path has an equal magnitude, as a result p is constant. Also, all the other parameters are constant and therefore the charge to mass ratio is a constant.
\[\dfrac{2\pi v\cos \theta }{pB}=\text{constant}\]
Now, we are given that the two particles follow opposite paths. For all these conditions to follow, the two particles should have opposite charges with equal magnitudes. Therefore, the sum of the charge to mass ratio of the two particles will be zero.
This satisfies the equation –
\[{{(\dfrac{{{q}_{1}}}{m})}_{1}}+{{(\dfrac{{{q}_{2}}}{m})}_{2}}=0\]
So, the correct answer is “Option D”.
Note:
The pitch of a helical path followed by a charged particle in a magnetic field is the linear distance covered by the particle along the horizontal in one rotation. For a particle projected vertically, the cosine component becomes zero and the particle doesn’t have a pitch, it undergoes circular motion.
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