A charged particle +q in an electric field vecE experiences a force in the direction of the field, and its kin

Question

+q

(i) The ratio of their kinetic energies

\dfrac{K_1}{K_2}

is : (A)

\dfrac{1}{2}

(B)

\dfrac{1}{4}

(C)

4

(D)

1

(ii) The ratio of the radii of the circular paths described by them

\dfrac{r_1}{r_2}

is : (A)

\dfrac{1}{2}

(B)

\sqrt{2}

(C)

\dfrac{1}{\sqrt{2}}

(D)

2

(iii) Suppose particles 1 and 2 enter the magnetic field

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

with velocities

\vec{v_1} = v_1\hat{i}

and

\vec{v_2} = v_2\hat{i}

. Then : (A) both particles revolve clockwise (B) both particles revolve anticlockwise (C) particle 1 revolves clockwise while particle 2 revolves anticlockwise (D) particle 1 revolves anticlockwise while particle 2 revolves clockwise (iv)(a) If the period of revolution for particle 1 is

4

s, then for particle 2 the period will be : (A)

1

s (B)

2

s (C)

4

s (D)

8

(OR)

(iv)(b) If the values of momentum for particles 1 and 2 are

p_1

and

p_2

, then : (A)

p_1 = \dfrac{p_2}{2}

(B)

p_1 = p_2

(C)

p_1 = 2p_2

(D)

p_1 = 4p_2

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Part (a)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. …

Part (b)Concept understanding — Charged Particle in Magnetic Field

Charged Particle in a Magnetic Field

When a charged particle moves through a magnetic field, the field grabs it sideways. Unlike an electric field, which can speed a charge up or slow it down, a magnetic field only bends the path — it never changes the particle's speed. Understanding why leads directly to circular and helical motion, the basis of cyclotrons, mass spectrometers and the aurora.

The force: always sideways

A particle of charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ feels the magnetic (Lorentz) force:

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

Because of the cross product, $\vec{F}$ is perpendicular to both $\vec{v}$ and $\vec{B}$ . Its magnitude is

F = |q|\,v\,B\sin\theta

where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$ .

Important

Since $\vec{F} \perp \vec{v}$ , the force does no work: $\vec{F}\cdot\vec{v} = 0$ . Therefore the kinetic energy and the speed stay constant — the field only changes the direction of motion, never the magnitude.

Case 1: velocity perpendicular to the field → a circle

If $\vec{v} \perp \vec{B}$ ( $\theta = 90^\circ$ ), the force $F = qvB$ stays constant in size and always points toward one central point. That is exactly the condition for uniform circular motion, with the magnetic force acting as the centripetal force:

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

Solving for the radius:

r = \frac{mv}{qB}

The time period of one revolution is

T = \frac{2\pi r}{v} = \frac{2\pi m}{qB}

Note

The period $T$ (and the frequency $f = qB/2\pi m$ , the cyclotron frequency) does not depend on the speed or the radius. A faster particle traces a bigger circle but takes exactly the same time per loop. This speed-independence is what makes the cyclotron work.

Case 2: velocity at an angle → a helix

If $\vec{v}$ makes an angle $\theta$ with $\vec{B}$ , split it into two parts:

Perpendicular component $v_\perp = v\sin\theta$ — feels the magnetic force and drives circular motion of radius $r = \dfrac{mv_\perp}{qB}$ .
Parallel component $v_\parallel = v\cos\theta$ — feels no force (since $\vec{v_\parallel}\times\vec{B}=0$ ) and carries the particle steadily along the field line.

Combining a circle with a steady drift gives a helix. The distance advanced along the field in one full turn is the pitch:

p = v_\parallel\,T = v\cos\theta \cdot \frac{2\pi m}{qB}

A quick example

An electron ( $m = 9.1\times10^{-31}\ \text{kg}$ , $q = 1.6\times10^{-19}\ \text{C}$ ) enters a $0.02\ \text{T}$ field at $10^{6}\ \text{m/s}$ , perpendicular to $\vec{B}$ :

$r = \frac{mv}{qB} = \frac{(9.1\times10^{-31})(10^{6})}{(1.6\times10^{-19})(0.02)} \approx 2.8\times10^{-4}\ \text{m}$ …

Why this formula?

Charged Particle in a Magnetic Field — Why the Key Formulas Hold

Let's build this from first principles. The core idea is that a magnetic field exerts a force only on a moving charge, and that force is always perpendicular to both the velocity and the field.

1. The Fundamental Force Law: Lorentz Force

The starting point is the Lorentz force for a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ :

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

Why this form?

Cross product $\vec{v} \times \vec{B}$ means the force is perpendicular to both $\vec{v}$ and $\vec{B}$ .
Magnitude: $F_m = |q| v B \sin\theta$ , where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$ .
Direction: given by the right-hand rule (for positive $q$ ).

Key insight: Because $\vec{F}_m \perp \vec{v}$ , the magnetic force does no work — it changes only the direction of velocity, not its speed.

2. Circular Motion in a Uniform Magnetic Field

Consider a charge $q$ moving with speed $v$ perpendicular to a uniform $\vec{B}$ (so $\theta = 90^\circ$ , $\sin\theta = 1$ ).

Step 1: Force provides centripetal acceleration

The magnetic force is the only radial force:

F_m = q v B

This must equal the centripetal force required for circular motion:

F_c = \frac{m v^2}{r}

Step 2: Equate and solve for $r$

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

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

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

Thus:

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

This is the radius of the circular path (cyclotron radius).

Why this makes sense:

Larger mass $m$ → harder to turn → larger $r$
Larger charge $q$ or stronger $B$ → stronger force → tighter turn → smaller $r$
Faster speed $v$ → more momentum → larger $r$

3. Angular Frequency (Cyclotron Frequency)

From the circular motion relation:

\omega = \frac{v}{r}

Substitute $r = \frac{m v}{q B}$ :

\omega = \frac{v}{\frac{m v}{q B}} = \frac{q B}{m}

Thus:

\boxed{\omega_c = \frac{q B}{m}}

Why this is remarkable:

$\omega_c$ is independent of speed $v$ — all particles with same $q/m$ have the same angular frequency, regardless of how fast they move.
This is the principle behind cyclotrons (particle accelerators).

4. General Motion: Helical Path …

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(i) (A) $K_1/K_2=\tfrac12$ ; (ii) (D) $r_1/r_2=2$ ; (iii) (D) particle 1 anticlockwise, particle 2 clockwise; (iv)(a) (A) $T_2=1$ s; and the OR alternative (iv)(b) (B) $p_1=p_2$ .

Part (a) — Case study parts (i)–(iii) and (iv)(a)

Data: particle 1 has mass $m$ , charge $-q$ ; particle 2 has mass $m/2$ , charge $+2q$ . Both start from rest and are accelerated through the same potential difference $V$ .

Key relations: kinetic energy gained $K=|q|V$ (mass-independent); radius $r=\dfrac{mv}{|q|B}=\dfrac{\sqrt{2mK}}{|q|B}$ ; period $T=\dfrac{2\pi m}{|q|B}$ .

Kinetic-energy ratio. $K_1=|{-q}|V=qV,\qquad K_2=|{+2q}|V=2qV\ \Rightarrow\ \frac{K_1}{K_2}=\frac{qV}{2qV}=\frac12.\quad\textbf{(A)}$
Radius ratio. First the momenta: $p_1=\sqrt{2m\,K_1}=\sqrt{2m\,qV},\qquad p_2=\sqrt{2\cdot\tfrac{m}{2}\cdot K_2}=\sqrt{2\cdot\tfrac{m}{2}\cdot 2qV}=\sqrt{2mqV}.$ So $p_1=p_2$ . Then $r_1=\frac{p_1}{|{-q}|B}=\frac{\sqrt{2mqV}}{qB},\qquad r_2=\frac{p_2}{|{+2q}|B}=\frac{\sqrt{2mqV}}{2qB}\ \Rightarrow\ \frac{r_1}{r_2}=2.\quad\textbf{(D)}$
Sense of revolution. With $\vec B=B_0\hat k$ and $\vec v=v\hat i$ (note $\hat i\times\hat k=-\hat j$ ):

Particle 1 ( $-q$ ): $\vec F=(-q)(v\hat i\times B_0\hat k)=(-q)(vB_0)(-\hat j)=+qvB_0\,\hat j$ — force toward $+y$ . Moving $+x$ with a leftward (up) push $\Rightarrow$ anticlockwise.
Particle 2 ( $+2q$ ): $\vec F=(+2q)(v\hat i\times B_0\hat k)=-2qvB_0\,\hat j$ — force toward $-y$ . Moving $+x$ with a downward push $\Rightarrow$ clockwise.

So particle 1 revolves anticlockwise, particle 2 clockwise. (D)

(iv)(a) Period of particle 2. Since $T=\dfrac{2\pi m}{|q|B}\propto\dfrac{m}{|q|}$ , …

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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 36 on this concept.

CBSE 2026Set V11 markMCQ
$Q. The path traced by a charged particle moving perpendicular to a uniform magnetic field is :$
(a) circle
(b) straight line
(c) helix
(d) ellipse
›Reveal solutionSolution

(a) circle …

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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 2026Set ANNUAL1 markMCQ
$Q. The distance travelled by a charged particle in one rotation along the magnetic field is called$
(a) pitch
(b) angular frequency
(c) radius of helix
(d) angular displacement
›Reveal solutionSolution

A charged particle entering a magnetic field with velocity components both along and perpendicular to B moves in a helix; the axial distance covered in one full turn is called the pitch.

The component of velocity perpendicular to B (v_perp) causes circular motion (radius r = m v_perp / qB), while the component along B (v_parallel) is unaffected and produces uniform linear motion along the field direction. The combination is a helix. I …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. A positively charged particle enters in a perpendicular uniform magnetic field, its path will be:$
(a) Elliptical
(b) Parabolic
(c) Linear
(d) Circular
›Reveal solutionSolution

A charged particle moving perpendicular to a uniform magnetic field traces a circle because the magnetic force is always perpendicular to velocity.

When a positive charge $q$ moves with speed $v$ perpendicular to a uniform field $B$ , it experiences a force $F = qvB$ directed perpendicular to $v$ (by $\vec F = q\vec v \times \vec B$ ). This force acts as a centripetal force, constantly changing the direction of veloci …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If a charged particle enters perpendicularly into a uniform magnetic field, then which of the following statements is true?$
(a) Both energy and momentum remain constant.
(b) Energy remains constant, but momentum changes.
(c) Both energy and momentum change.
(d) Energy changes but momentum remains constant.
›Reveal solutionSolution

The magnetic force is always perpendicular to the velocity, so it can never do work on the charge — kinetic energy (and hence speed) stays exactly constant. But the force continuously deflects the particle into a circular path, constantly changing the direction of its momentum vector even while its magnitude is unchanged.

The magnetic force and work done

The force on a charge $q$ moving with velocity $\vec v$ in a magnetic field $\vec B$ is the Lorentz (magnetic) force:

$\vec F=q\,\vec v\times\vec B$

By the definition of the cross product, $\vec F$ is always perpendicular to $\vec v$ . The (infinitesimal) work done by this force over a displacement $d\vec s = \vec v\,dt$ is

$dW=\vec F\cdot d\vec s = (q\,\vec v\times\vec B)\cdot(\vec v\,dt)=0$

because $\vec v\times\vec B$ is perpendicular to $\vec v$ , so its dot product with $\vec v$ is zero. Since $dW=0$ at every instant, the total work done by the magnetic force is always zero.

By the work–energy theorem, since no work is done, the kinetic energy — and hence the speed $|\vec v|$ and hence the magnitude of momentum $|\vec p|=m|\vec v|$ — of the particle remains constant.

Why momentum itself still changes

…

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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 2025Set ANNUAL1 markMCQ
$Q. A charged particle enter in a magnetic field perpendicular to the magnetic lines of forces. The path of the charged particle is :$
(a) circular
(b) ellipse
(c) straight line
(d) helical
›Reveal solutionSolution

A charge moving perpendicular to a uniform magnetic field traces a circle, because the magnetic force always acts as a centripetal force.

The magnetic force on a moving charge is $\vec{F} = q\vec{v}\times\vec{B}$ . When $\vec{v} \perp \vec{B}$ , this force has constant magnitude $qvB$ and is always directed perpendicular to $\vec{v}$ — i.e., it always points toward a fixed centre. A force of constant magnitude always perpendicular to velocity is exactly the condition for unifo …

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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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Part (a) — Case study parts (i)–(iii) and (iv)(a)

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