Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic fi

NCERT Exemplar · Q1

\vec{B} = B_0\hat{k}

(a) They have equal z-components of momenta.

(b) They must have equal charges.

(d) The charge to mass ratio satisfy:

\left(\dfrac{e}{m}\right)_1 + \left(\dfrac{e}{m}\right)_2 = 0

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Concept understanding — Cyclotron Motion Radius

Cyclotron Motion Radius – From Intuition to Formula

Imagine you're pushing a charged ball on a frictionless table, and there's a giant magnet underneath. The moment that ball starts moving, the magnet doesn't pull it or push it forward — it turns it. The force from the magnet always acts sideways, perpendicular to the ball's velocity. So the ball never speeds up or slows down; it just keeps changing direction. If the magnetic field is uniform and the ball keeps moving, it will trace out a perfect circle.

That circle is called cyclotron motion, and the radius of that circle is what we're after.

Why does it curve at all?

The magnetic force on a moving charge is given by:

\vec{F} = q (\vec{v} \times \vec{B})

The cross product means the force is always perpendicular to both the velocity $\vec{v}$ and the magnetic field $\vec{B}$ . For a charge moving perpendicular to a uniform field, this force acts as a centripetal force — it constantly pulls the charge toward the centre of a circle, without doing any work (since force is perpendicular to displacement).

So the charge moves in uniform circular motion. The magnetic force provides the necessary centripetal acceleration.

Deriving the radius

For circular motion, the centripetal force required is:

F_{\text{centripetal}} = \frac{m v^2}{r}

where $m$ is the mass of the particle, $v$ is its speed, and $r$ is the radius of the circle.

The magnetic force (for $\vec{v} \perp \vec{B}$ ) has magnitude:

F_B = |q| v B

Set them equal:

|q| v B = \frac{m v^2}{r}

Cancel one factor of $v$ (assuming $v \neq 0$ ):

|q| B = \frac{m v}{r}

Solve for $r$ :

r = \frac{m v}{|q| B}

That's the cyclotron motion radius (also called the Larmor radius or gyroradius).

What the formula tells you

Faster particle → larger radius (it's harder to turn something moving fast).
Heavier particle → larger radius (more inertia resists the turn).
Stronger magnetic field → smaller radius (the turning force is stronger).
Larger charge → smaller radius (more force for the same field).

Tip

If the particle's velocity has a component parallel to $\vec{B}$ , it doesn't feel any magnetic force in that direction. So the particle moves in a helix — circular motion in the plane perpendicular to $\vec{B}$ , plus constant speed along $\vec{B}$ . The radius formula above still applies using only the perpendicular component of velocity, $v_\perp$ .

A quick example

A proton ( $m = 1.67 \times 10^{-27}$ kg, $q = 1.6 \times 10^{-19}$ C) moves at $2.0 \times 10^6$ m/s perpendicular to a 0.50 T magnetic field.

r = \frac{(1.67 \times 10^{-27})(2.0 \times 10^6)}{(1.6 \times 10^{-19})(0.50)} = \frac{3.34 \times 10^{-21}}{8.0 \times 10^{-20}} = 0.042 \text{ m}

So the proton circles with a radius of about 4.2 cm.

Common mistake to avoid

Watch out

The formula $r = \frac{mv}{|q|B}$ uses the speed $v$ , not velocity. And it assumes the velocity is perpendicular to $\vec{B}$ . If there's a parallel component, use only $v_\perp$ in the numerator. The parallel component doesn't affect the radius — it just carries the particle along the field line.

Why "cyclotron"?

The name comes from the cyclotron, a particle accelerator that uses this exact principle. Particles spiral outward in a magnetic field, gaining energy from an alternating electric field each half-turn. The radius increases as the particle speeds up — exactly what the formula predicts.

Queries like "cyclotron radius formula derivation" and "moving charges and magnetism class 12 physics" are common, since this result comes straight from the Moving Charges and Magnetism chapter of the NCERT/CBSE Class 12 Physics curriculum. It's also a standard numerical-question type in JEE Main and NEET on charged-particle motion in magnetic fields.

Matching helical paths in opposite sense forces the two particles' charge-to-mass ratios to be equal in magnitude but opposite in sign - this is exactly option (d): $(e/m)_1 + (e/m)_2 = 0$ .

Setting up the helix

For a charged particle of mass $m$ , charge $q$ , moving in $\vec{B}=B_0\hat{k}$ , split the velocity into $v_\perp$ (perpendicular to $\vec{B}$ ) and $v_\parallel$ (along $\vec{B}$ ). The perpendicular part gives circular motion:

r = \frac{mv_\perp}{|q|B_0}, \qquad T = \frac{2\pi m}{|q|B_0}

and the parallel part carries the particle steadily along the field, giving a helix of pitch

p = v_\parallel T = \frac{2\pi m v_\parallel}{|q|B_0}.

Same shape, opposite sense

"Identical helical paths" means the two particles trace the same radius $r$ and the same pitch $p$ . The sense of rotation (clockwise or anticlockwise, viewed along $\vec{B}$ ) is fixed entirely by the sign of the charge - a positive charge circulates one way, a negative charge the other way, for the same $\vec{B}$ . "Completely opposite sense" therefore means the two charges have opposite sign.

Both $r$ and $T$ (and hence $p$ ) are governed by the single combination $|q|/m$ , through the cyclotron angular frequency $\omega = |q|B_0/m$ . For the two particles to trace geometrically identical helices in this same field, this angular frequency must be the same for both:

\frac{|q_1|}{m_1} = \frac{|q_2|}{m_2}.

Combine this with the opposite sign of the charges: writing $e/m$ for the signed charge-to-mass ratio of each particle,

\frac{q_1}{m_1} = -\frac{q_2}{m_2} \quad\Longrightarrow\quad \left(\frac{e}{m}\right)_1 + \left(\frac{e}{m}\right)_2 = 0.

This is exactly stem option (d).

Why the other options are not forced

(a) equal $z$ -components of momenta: the pitch condition fixes $v_\parallel$ to be the same for both particles once $|q|/m$ is matched, but the mass $m$ need not be the same - so $p_z = mv_\parallel$ need not be equal.
(b) equal charges: the charges must be opposite in sign, so, other than the trivial case $q=0$ , they cannot be equal.
(c) a particle-antiparticle pair: this would additionally require the two masses to be exactly equal, which is not implied - any two species with the same $|q|/m$ magnitude and opposite charge sign satisfy the condition, not only a particle and its antiparticle.

✓Final answer

\left(\dfrac{e}{m}\right)_1 + \left(\dfrac{e}{m}\right)_2 = 0

Method: Comparing Two Charged Particles' Helical Paths in a Uniform Field

Whenever a problem describes two (or more) charged particles tracing helices in the same field and asks what must be true of their charges/masses, the technique is to write each particle's path parameters symbolically and match them term by term.

Steps

Step 1: Write the three helix parameters for a general particle

For a particle of mass $m$ , charge magnitude $|q|$ , in field $\vec B=B_0\hat k$ , split velocity into $v_\perp$ and $v_\parallel$ :

r = \frac{mv_\perp}{|q|B_0}, \qquad T = \frac{2\pi m}{|q|B_0}, \qquad p = v_\parallel T

Notice $r$ and $T$ (and therefore $p$ ) depend on the particle's charge and mass only through the single combination $|q|/m$ (equivalently, the cyclotron frequency $\omega=|q|B_0/m$ ).

Step 2: Translate "identical path shape" into an equation

"Identical helices" means both particles have the same $r$ , $T$ , and $p$ in the same $\vec B$ . Since these depend only on $|q|/m$ (and on $v_\perp, v_\parallel$ , which the problem may or may not force equal), matching path shape between particle 1 and particle 2 gives:

\frac{|q_1|}{m_1} = \frac{|q_2|}{m_2}

Step 3: Translate "opposite sense" into a sign condition

The rotational sense (clockwise vs anticlockwise, viewed along $\vec B$ ) is fixed purely by the sign of the charge -- it does not depend on mass, speed, or radius. "Completely opposite sense" therefore forces the two charges to have opposite sign.

Applying to this problem: combine the magnitude condition from Step 2 with the opposite-sign condition from Step 3 -- a same-magnitude, opposite-sign charge-to-mass ratio is exactly $(e/m)_1+(e/m)_2=0$ . Check each remaining option (equal momenta, equal charges, particle-antiparticle pair) against only what Steps 2-3 actually proved, not against extra assumptions the problem never stated.

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Q.Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field B⃗=B0k^\vec{B} = B_0\hat{k}B=B0​k^.

Setting up the helix

Same shape, opposite sense

Why the other options are not forced

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Q.Two charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field $\vec{B} = B_0\hat{k}$ .

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