Assertion (A) : An electron and a proton enter with the same momentum $\vec{p}$ in a magnetic field $\vec{B}$ such that $\vec{p} \perp \vec{B}$. Then both describe a circular path of the same radius. Reason (R) : The radius of the circular path described by the charged particle (charge $q$, mass $m$) moving in the magnetic field $\vec{B}$ is given by $r = \dfrac{mv}{qB}$. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false and Reason (R) is also false.

> [!TLDR] > The radius of circular motion in a perpendicular magnetic field depends on momentum, not mass or velocity separately. Since both particles have the same momentum, they trace the same radius — the assertion is true, and the reason correctly explains it. **Why this works — the core idea** When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force $\vec{F} = q(\vec{v} \times \vec{B})$ acts as a centripetal force. The particle is forced into a circular path whose radius depends on how much "oomph" (momentum) it has versus how strongly the field tries to bend it. The key insight: the radius formula $r = \frac{mv}{qB}$ is really $r = \frac{p}{qB}$ — it's momentum that matters, not mass or speed alone. --- 1. **Start with the force balance.** For a particle of charge $q$, mass $m$, and speed $v$ moving perpendicular to $\vec{B}$, the magnetic force provides the centripetal force: $$ qvB = \frac{mv^2}{r} $$ Cancel one $v$ (valid since $v \neq 0$): $$ qB = \frac{mv}{r} \quad \Rightarrow \quad r = \frac{mv}{qB} $$ 2. **Rewrite in terms of momentum.** Linear momentum $p = mv$, so: $$ r = \frac{p}{qB} $$ This is the cleaner, more revealing form. The radius depends only on the magnitude of momentum, the charge magnitude, and the field strength — not on mass or velocity individually. 3. **Apply to the given situation.** Both the electron and the proton have the **same momentum** $\vec{p}$ (same magnitude and direction), and both have the **same magnitude of charge** $|q| = e$ (ignoring sign, which only affects direction of rotation, not radius). They enter the same magnetic field $\vec{B}$ with $\vec{p} \perp \vec{B}$. Therefore: $$ r_{\text{electron}} = \frac{p}{eB} = r_{\text{proton}} $$ Both paths have identical radii. 4. **Check the reason statement.** …

Assertion (A) : An electron and a proton enter with the same momentum vecp in a magnetic field vecB such that

Question

\vec{p}

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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. …

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The radius of circular motion in a perpendicular magnetic field depends on momentum, not mass or velocity separately. Since both particles have the same momentum, they trace the same radius — the assertion is true, and the reason correctly explains it.

Why this works — the core idea

When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force $\vec{F} = q(\vec{v} \times \vec{B})$ acts as a centripetal force. The particle is forced into a circular path whose radius depends on how much "oomph" (momentum) it has versus how strongly the field tries to bend it. The key insight: the radius formula $r = \frac{mv}{qB}$ is really $r = \frac{p}{qB}$ — it's momentum that matters, not mass or speed alone.

Start with the force balance. For a particle of charge $q$ , mass $m$ , and speed $v$ moving perpendicular to $\vec{B}$ , the magnetic force provides the centripetal force:

qvB = \frac{mv^2}{r}

Cancel one $v$ (valid since $v \neq 0$ ):

qB = \frac{mv}{r} \quad \Rightarrow \quad r = \frac{mv}{qB}

Rewrite in terms of momentum. Linear momentum $p = mv$ , so:

r = \frac{p}{qB}

This is the cleaner, more revealing form. The radius depends only on the magnitude of momentum, the charge magnitude, and the field strength — not on mass or velocity individually.

Apply to the given situation. Both the electron and the proton have the same momentum $\vec{p}$ (same magnitude and direction), and both have the same magnitude of charge $|q| = e$ (ignoring sign, which only affects direction of rotation, not radius). They enter the same magnetic field $\vec{B}$ with $\vec{p} \perp \vec{B}$ . Therefore:

r_{\text{electron}} = \frac{p}{eB} = r_{\text{proton}}

Both paths have identical radii.

Check the reason statement. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 13 on this concept.

CBSE 2026Set DS1 mark
$10^{-5}$
›Reveal solutionSolution

Using $r = \dfrac{\sqrt{2mE}}{eB}$ , the radius comes out to about $1.07$ m.

Concept. A charged particle moving perpendicular to a magnetic field goes in a circle whose radius is set by balancing the magnetic force against the centripetal requirement: $qvB = \dfrac{mv^2}{r}$ , giving $r = \dfrac{mv}{qB} = \dfrac{p}{qB}$ . The momentum is found from the kinetic energy, $p = \sqrt{2mE}$ .

Calculation. Kinetic energy $E = 10\text{ eV} = 10\times1.6\times10^{-19} = 1.6\times10^{-18}$ J. …

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CBSE 2026Set A1 markMCQ
$Q. If a charged particle of mass m and charge q enters a uniform magnetic field B at an angle θ in the direction of field with velocity v, then the path of the particle is helical. The radius of circular path of the helix will be (A) mv/qB (B) mv cosθ/qB (C) mv sinθ/qB (D) 2πmv/qB$
›Reveal solutionSolution
The perpendicular component v sinθ gives circular motion; r = m v sinθ / qB.

When a charge enters a magnetic field at angle θ to B, resolve its velocity:

Component along B: $v\cos\theta$ — unaffected by the field, gives uniform motion along the axis (the pitch of the helix).

Component perpendicular to B: $v\sin\theta$ — experiences the magnetic force $qvB\sin\theta$ , producing circular motion. …
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CBSE 2025Set 55/4/11 markMCQ
$+q$
›Reveal solutionSolution

The charge enters with velocity along $+Y$ and the field is into the page ( $-Z$ ), so the magnetic force keeps it in the $XY$ plane — the path is a semicircle in the $XY$ plane. Option (B).

Reading the figure.

Figure — 55/4/1 Q3

The axes are $X$ (right), $Y$ (up) and $Z$ (out of the page, toward the viewer). The $\times$ grid marks a uniform field into the page, i.e. $\vec{B} = -B\,\hat{k}$ , filling the upper region. The charge $+q$ sits on the $+X$ axis and enters moving along $+Y$ , so $\vec{v} = v\,\hat{j}$ .

Force direction.
$\vec{F} = q\,\vec{v}\times\vec{B} = q\,(v\,\hat{j})\times(-B\,\hat{k}) = -qvB\,(\hat{j}\times\hat{k}) = -qvB\,\hat{i}$
The force is along $-X$ , i.e. it lies in the $XY$ plane, perpendicular to $\vec{v}$ . …
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CBSE 2025Set 55/5/11 markMCQ
$10^{6}\ \mathrm{ms^{-1}}$
›Reveal solutionSolution

The radius of cyclotron motion is $r = \frac{mv}{qB}$ . Using the kinetic energy gained from the potential difference, we find the charge-to-mass ratio, then substitute into the radius formula to get 5 cm.

Why this works — the physics of circular motion in a magnetic field

When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force acts as a centripetal force, bending the path into a circle. The key relation comes from equating:
$q v B = \frac{m v^2}{r}$
which simplifies to the cyclotron radius:
$r = \frac{mv}{qB}$
The problem gives us $v = 10^6\ \mathrm{m/s}$ and $B = 0.4\ \mathrm{T}$ , but we don't directly know $m/q$ — the mass-to-charge ratio of the particle. However, we can find it from the acceleration step: the particle was accelerated from rest through a potential difference of $10\ \mathrm{kV} = 10^4\ \mathrm{V}$ .

Step-by-step solution

1. Find the kinetic energy gained

When a charge $q$ is accelerated through a potential difference $V$ , it gains kinetic energy equal to the work done by the electric field:
$\frac{1}{2} m v^2 = qV$
We know $v = 10^6\ \mathrm{m/s}$ and $V = 10^4\ \mathrm{V}$ . This gives us a direct relation between $m$ and $q$ .

2. Extract the charge-to-mass ratio

From $\frac{1}{2} m v^2 = qV$ , rearrange:
$\frac{q}{m} = \frac{v^2}{2V}$
Plug in the numbers:
$\frac{q}{m} = \frac{(10^6)^2}{2 \times 10^4} = \frac{10^{12}}{2 \times 10^4} = \frac{10^8}{2} = 5 \times 10^7\ \mathrm{C/kg}$
Tip

You don't need to identify the particle — the ratio $q/m$ is all that matters for the radius. This is a common exam trick: they give you $v$ and $V$ so you can find $q/m$ without needing the particle's identity.

3. Write the radius formula in terms of known quantities

From $r = \frac{mv}{qB}$ , we can write:

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CBSE 2025Set 55/6/11 markMCQ
$\alpha$
›Reveal solutionSolution
When charged particles enter perpendicular to a magnetic field, the radius depends on momentum and charge: $r = \frac{mv}{qB}$ . Since the proton and α-particle have the same velocity but different mass-to-charge ratios, we find $r_p : r_\alpha = 1 : 2$ .

Why the radius depends on mass and charge

When a charged particle moves perpendicular to a magnetic field, the Lorentz force acts as a centripetal force, bending the particle into a circular path. The magnetic force $qvB$ must equal the centripetal force $\frac{mv^2}{r}$ , which immediately tells us that heavier particles or those with less charge will trace larger circles.

The key insight: the radius is proportional to the particle's momentum-to-charge ratio. Two particles with the same velocity will have radii in the ratio of their $\frac{m}{q}$ values.

Step-by-step calculation

Write the force balance equation The magnetic force provides the centripetal acceleration:

$qvB = \frac{mv^2}{r}$

Solve for the radius Canceling one factor of $v$ from both sides:

$r = \frac{mv}{qB}$ $r = \frac{mv}{qB}$

Identify the particle properties

Proton: mass $m_p$ , charge $q_p = e$

α-particle (helium nucleus): mass $m_\alpha = 4m_p$ , charge $q_\alpha = 2e$

Write the radius for each particle

For the proton:

$r_p = \frac{m_p v}{eB}$
For the α-particle: …
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CBSE 2025Set IMPROVEMENT1 markMCQ
$Q. If the velocity of a charged particle moving perpendicular to the direction of a uniform magnetic field is doubled and the value of the magnetic field is halved, then the radius of the path of the charged particle will become:$
(a) 8 times
(b) Double
(c) 4 times
(d) 3 times
›Reveal solutionSolution

The radius of a charged particle's circular path in a magnetic field is $r = \dfrac{mv}{qB}$ ; doubling $v$ and halving $B$ makes $r$ four times larger.

For a charged particle moving perpendicular to a uniform magnetic field, the magnetic force provides the centripetal force: $qvB = \dfrac{mv^2}{r} \implies r = \dfrac{mv}{qB}$ .

…

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CBSE 2024Set 55/5/11 markMCQ
$m$
›Reveal solutionSolution
The radius of cyclotron motion depends on the ratio $m/q$ . When mass doubles and charge halves, the ratio quadruples, so the new radius is $4R$ .

The key idea here is that a charged particle moving perpendicular to a uniform magnetic field experiences a centripetal force provided by the magnetic Lorentz force. The radius of the circular path is not an independent quantity — it emerges from balancing these two forces.

For any such problem, always start from the force balance equation. The magnetic force is $qvB$ , and the centripetal force required for circular motion is $mv^2/R$ . Setting them equal gives the radius directly.

Write the force balance The magnetic force provides the centripetal force:

$qvB = \frac{mv^2}{R}$
Cancel one factor of $v$ (assuming $v \neq 0$ ):
$qB = \frac{mv}{R}$

Solve for the radius Rearranging:

$R = \frac{mv}{qB}$
This is the standard formula for the cyclotron radius (also called the gyroradius or Larmor radius). Notice that $R$ depends on the ratio $m/q$ , not on $m$ or $q$ individually.
$R = \frac{mv}{qB}$

Identify what changes

The problem states:

New mass: $m' = 2m$

New charge: $q' = q/2$ The magnetic field $B$ and the speed $v$ are not mentioned as changing, so we assume they remain the same. (This is a standard assumption in such problems unless stated otherwise.)

Find the new radius

Substitute the new values into the formula: …
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CBSE 2024Set 55/2/11 markMCQ
$\vec{p}$
›Reveal solutionSolution
The radius of circular motion in a perpendicular magnetic field depends on momentum, not mass or velocity separately. Since both particles have the same momentum, they trace the same radius — the assertion is true, and the reason correctly explains it.

Why this works — the core idea

When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force $\vec{F} = q(\vec{v} \times \vec{B})$ acts as a centripetal force. The particle is forced into a circular path whose radius depends on how much "oomph" (momentum) it has versus how strongly the field tries to bend it. The key insight: the radius formula $r = \frac{mv}{qB}$ is really $r = \frac{p}{qB}$ — it's momentum that matters, not mass or speed alone.

Start with the force balance. For a particle of charge $q$ , mass $m$ , and speed $v$ moving perpendicular to $\vec{B}$ , the magnetic force provides the centripetal force:

$qvB = \frac{mv^2}{r}$
Cancel one $v$ (valid since $v \neq 0$ ):
$qB = \frac{mv}{r} \quad \Rightarrow \quad r = \frac{mv}{qB}$

Rewrite in terms of momentum. Linear momentum $p = mv$ , so:

$r = \frac{p}{qB}$
This is the cleaner, more revealing form. The radius depends only on the magnitude of momentum, the charge magnitude, and the field strength — not on mass or velocity individually.

Apply to the given situation. Both the electron and the proton have the same momentum $\vec{p}$ (same magnitude and direction), and both have the same magnitude of charge $|q| = e$ (ignoring sign, which only affects direction of rotation, not radius). They enter the same magnetic field $\vec{B}$ with $\vec{p} \perp \vec{B}$ . Therefore:

$r_{\text{electron}} = \frac{p}{eB} = r_{\text{proton}}$
Both paths have identical radii.

Check the reason statement. …
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CBSE 2023Set 55/3/11 markMCQ
$m$
(a) a circle in $x$ - $z$ plane
(b) a parabola in $x$ - $y$ plane
(c) a helix with constant pitch
(d) a helix with increasing pitch
›Reveal solutionSolution
The particle initially undergoes circular motion in the $x$ - $z$ plane due to the magnetic field. When the electric field is switched on along $y$ , it adds a constant acceleration in that direction, turning the circle into a helix whose pitch increases quadratically with time — so the correct option is (d).

Concept and Intuition

This problem is about the superposition of two well-known motions: cyclotron motion (from a magnetic field perpendicular to velocity) and uniform acceleration (from an electric field parallel to the magnetic field). The key is to see that the magnetic force depends only on the velocity components perpendicular to $\vec B$ , while the electric force acts along $\vec B$ itself. Once you separate the motion into these independent directions, the path becomes clear.

Let’s break it down.

Step-by-step Reasoning

1. Identify the force directions

The magnetic field is $\vec B = B_0 \hat{j}$ — it points along the $y$ -axis. The particle enters with velocity $\vec v = v_{0x}\hat{i} + v_{0y}\hat{j}$ . The magnetic force is
$\vec F_B = q(\vec v \times \vec B) = q\big[(v_{0x}\hat{i} + v_{0y}\hat{j}) \times (B_0\hat{j})\big].$
Since $\hat{i} \times \hat{j} = \hat{k}$ and $\hat{j} \times \hat{j} = 0$ , we get
$\vec F_B = q v_{0x} B_0 \, \hat{k}.$
So the magnetic force acts purely along the $z$ -axis — it has no component along $y$ (the direction of $\vec B$ ). This means the $y$ -component of velocity is unaffected by the magnetic field.

2. Motion before the electric field is switched on

Initially, only $\vec B$ is present. The particle has:

$v_y = v_{0y}$ constant (no force along $y$ ).

$v_x = v_{0x}$ and $v_z$ varying due to the magnetic force.

The magnetic force $q v_x B_0$ in the $z$ -direction, together with the perpendicular component of velocity, produces uniform circular motion in the $x$ - $z$ plane. The radius (cyclotron radius) is
$r = \frac{m v_{\perp}}{|q| B_0},$
where $v_{\perp} = v_{0x}$ (since the $y$ -component is parallel to $\vec B$ and doesn't contribute to the circular motion). The angular frequency (cyclotron frequency) is
$\omega = \frac{|q| B_0}{m}.$
So the particle moves in a circle in the $x$ - $z$ plane while drifting with constant speed $v_{0y}$ along $y$ . That combination — circular motion in a plane plus uniform motion perpendicular to that plane — is a helix of constant pitch.

Watch out

A common mistake is to think the initial motion is a circle. It is actually a helix from the very beginning, because the particle already has a constant $v_y$ along the magnetic field. The circle is only the projection onto the $x$ - $z$ plane.

3. What happens when the electric field is switched on?

The electric field $\vec E = E_0 \hat{j}$ is along the same direction as $\vec B$ . It exerts a force
$\vec F_E = q E_0 \hat{j}.$
This force is parallel to the magnetic field, so it does not affect the circular motion in the $x$ - $z$ plane. It only accelerates the particle along $y$ .

The $y$ -component of motion now becomes uniformly accelerated:
$a_y = \frac{q E_0}{m}, \quad v_y(t) = v_{0y} + \frac{q E_0}{m} t, \quad y(t) = v_{0y} t + \frac{1}{2} \frac{q E_0}{m} t^2.$
4. Combine the motions

In the $x$ - $z$ plane, the particle continues its circular motion with constant angular speed $\omega$ and radius $r$ — unchanged by the electric field. Along $y$ , it now has a velocity that increases linearly with time (if $q$ and $E_0$ have the same sign) or decreases (if opposite signs). …
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CBSE 2023Set 55/4/11 markMCQ
$v$
›Reveal solutionSolution
The magnetic force on a moving charge is always perpendicular to its velocity, so it does no work and cannot change the particle's speed. The electron's final speed remains $v$ .

The key insight here is about the nature of the magnetic force. When a charged particle moves through a uniform magnetic field, the force it experiences is given by $\vec{F} = q(\vec{v} \times \vec{B})$ . This cross product means the force is always perpendicular to both the velocity and the magnetic field.

Because the force is perpendicular to the velocity at every instant, it can only change the direction of motion — never the magnitude of the velocity. Work done by a force is $W = \vec{F} \cdot \vec{d}$ , and since $\vec{F} \perp \vec{v}$ at all times, the dot product is zero. No work means no change in kinetic energy, and therefore no change in speed.

Let's walk through the problem step by step.

What the magnetic force does

The electron enters the field with speed $v$ . The magnetic force acts as a centripetal force, bending the electron's path into a circle. For a semicircular path, the electron simply traces half of that circle before exiting. The force is always directed toward the centre of the circle, perpendicular to the instantaneous velocity.

Why speed stays constant

Since $\vec{F} \perp \vec{v}$ , the power delivered by the magnetic force is $P = \vec{F} \cdot \vec{v} = 0$ . No power means no change in kinetic energy: $\Delta K = 0$ . The electron's kinetic energy $\frac{1}{2} m v^2$ remains the same throughout the motion.

What the semicircular path tells us

The fact that the path is a semicircle (rather than a full circle or some other arc) only affects the geometry of the exit point and the time spent inside the field. It has no bearing on the speed. Whether the particle goes through a quarter-circle, a semicircle, or a full circle, the speed is unchanged as long as the field is uniform and the particle doesn't lose energy through collisions or radiation.

The final speed …
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CBSE 2019Set 55/2/11 mark
$V$
›Reveal solutionSolution
A charged particle's cyclotron radius depends on its momentum. Doubling the accelerating voltage increases kinetic energy (and hence momentum) by $\sqrt{2}$ , so the radius increases by a factor of $\sqrt{2}$ .

When a charged particle moves through a magnetic field perpendicular to its velocity, the Lorentz force provides the centripetal acceleration needed for circular motion. The radius of this path—called the cyclotron radius—depends on how fast the particle is moving. The key insight is that the accelerating potential difference determines the particle's speed through energy conservation.

The magnetic force $qvB$ acts as the centripetal force, giving us:
$qvB = \frac{mv^2}{r}$
Rearranging for the radius:
$r = \frac{mv}{qB}$
This tells us the radius is proportional to the particle's momentum $mv$ . Now we need to connect the velocity to the potential difference.

Step-by-step solution

Find the velocity after acceleration through potential $V$

When the proton (charge $q$ , mass $m$ ) is accelerated through potential difference $V$ , it gains kinetic energy equal to the work done by the electric field:

$qV = \frac{1}{2}mv^2$
Solving for velocity:
$v = \sqrt{\frac{2qV}{m}}$

Express the initial radius $r_1$

Substituting this velocity into the radius formula:

$r_1 = \frac{mv}{qB} = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{1}{B}\sqrt{\frac{2mV}{q}}$

Find the new radius $r_2$ when potential is doubled

When the potential difference becomes $2V$ , the new velocity is: …
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CBSE 2019Set ANNUAL1 markMCQ
$m$
(a) $\dfrac{mv}{qB}$
(b) $\dfrac{mB}{qv}$
(c) $\dfrac{mq}{vB}$
(d) $\dfrac{qv}{mB}$
›Reveal solutionSolution

Set the magnetic force equal to the required centripetal force for circular motion and solve for the radius.

Step 1 — Magnetic force (perpendicular $v$ , $B$ )

$F = qvB$

Step 2 — Centripetal force requirement

…

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