A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters the cube across one of its faces with velocity $\vec{v}$ and a positron enters via opposite face with velocity $-\vec{v}$. At this instant, (a) the electric forces on both the particles cause identical accelerations. (b) the magnetic forces on both the particles cause equal accelerations. (c) both particles gain or loose energy at the same rate. (d) the motion of the centre of mass (CM) is determined by B alone.

> [!TLDR] > The magnetic forces on the two particles are equal, they exchange energy at the same rate, and the centre-of-mass motion is fixed by $\vec{B}$ alone -- so statements (b), (c) and (d) are true; the electric-force statement (a) is false. ## Forces on each particle Both sit in the same uniform $\vec{E}$ and $\vec{B}$. Electron: charge $-e$, velocity $\vec{v}$. Positron: charge $+e$, velocity $-\vec{v}$. **(a) Electric force -- NOT identical.** $$\vec{F}_{E,e}=-e\vec{E},\qquad \vec{F}_{E,p}=+e\vec{E}.$$ Equal and opposite, so they give opposite accelerations, not identical ones. Statement (a) is false. **(b) Magnetic force -- equal.** $$\vec{F}_{B,e}=-e(\vec{v}\times\vec{B}),\qquad \vec{F}_{B,p}=+e\big((-\vec{v})\times\vec{B}\big)=-e(\vec{v}\times\vec{B}).$$ Both equal $-e(\vec{v}\times\vec{B})$ -- the magnetic forces are the same. Statement (b) is true. **(c) Rate of energy change -- equal.** Only the electric force does work (the magnetic force is $\perp\vec{v}$). Power $P=\vec{F}_E\cdot\vec{v}$: …

A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters the cub

NCERT Exemplar · Q9

\vec{v}

(a) the electric forces on both the particles cause identical accelerations.

(b) the magnetic forces on both the particles cause equal accelerations.

(d) the motion of the centre of mass (CM) is determined by B alone.

Dnh Dd CbseMCQ· 1mImportance★★★★★

62% · 34/55 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

The magnetic forces on the two particles are equal, they exchange energy at the same rate, and the centre-of-mass motion is fixed by $\vec{B}$ alone -- so statements (b), (c) and (d) are true; the electric-force statement (a) is false.

Forces on each particle

Both sit in the same uniform $\vec{E}$ and $\vec{B}$ . Electron: charge $-e$ , velocity $\vec{v}$ . Positron: charge $+e$ , velocity $-\vec{v}$ .

Electric force -- NOT identical. $\vec{F}_{E,e}=-e\vec{E},\qquad \vec{F}_{E,p}=+e\vec{E}.$ Equal and opposite, so they give opposite accelerations, not identical ones. Statement (a) is false.
Magnetic force -- equal. $\vec{F}_{B,e}=-e(\vec{v}\times\vec{B}),\qquad \vec{F}_{B,p}=+e\big((-\vec{v})\times\vec{B}\big)=-e(\vec{v}\times\vec{B}).$ Both equal $-e(\vec{v}\times\vec{B})$ -- the magnetic forces are the same. Statement (b) is true.
Rate of energy change -- equal. Only the electric force does work (the magnetic force is $\perp\vec{v}$ ). Power $P=\vec{F}_E\cdot\vec{v}$ : …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ8 — Two identical current carrying coaxial loops, carry current I in an op…→Next questionQ10 — A charged particle would continue to move with a constant velocity in …

Q.A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters the cube across one of its faces with velocity $\vec{v}$ and a positron enters via opposite face with velocity $-\vec{v}$ . At this instant,

Concept understanding — Charged Particle in Magnetic Field

Charged Particle in a Magnetic Field

The force: always sideways

Case 1: velocity perpendicular to the field → a circle

Case 2: velocity at an angle → a helix

A quick example

Why this formula?

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

1. The Fundamental Force Law: Lorentz Force

Why this form?

2. Circular Motion in a Uniform Magnetic Field

Step 1: Force provides centripetal acceleration

Step 2: Equate and solve for $r$

Why this makes sense:

3. Angular Frequency (Cyclotron Frequency)

Why this is remarkable:

4. General Motion: Helical Path …

Forces on each particle

Method: Comparing Forces on Two Oppositely-Charged, Oppositely-Moving Particles in the Same Field

Steps

Q.A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters the cube across one of its faces with velocity v⃗\vec{v}v and a positron enters via opposite face with velocity −v⃗-\vec{v}−v. At this instant,

Forces on each particle

🎓Unlock everything free for 14 days

Q.A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters the cube across one of its faces with velocity $\vec{v}$ and a positron enters via opposite face with velocity $-\vec{v}$ . At this instant,

Unlock everything free for 14 days