Q.An electron is released from rest in a region of uniform electric and magnetic fields acting parallel to each other. The electron will (A) move in a straight line. (B) move in a circle. (C) remain stationary. (D) move in a helical path.

Showing the 12 most recent of 23 on this concept.

1. 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

   ✓Final answer

   (a) circle

   When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force $\vec{F} = q\,\vec{v}\times\vec{B}$ is always perpendicular to $\vec{v}$ and of constant magnitude $qvB$. It does no work (speed constant) and acts as a centripetal force, so the particle moves in a circle of radius $r = \dfrac{mv}{qB}$.

2. 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. In the time for one revolution (the period T = 2pim/qB), the particle advances a distance v_parallel * T along B - this distance is defined as the pitch of the helix.

   ✓Final answer

   (a) pitch.

3. 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 velocity but not its magnitude. Since the force stays perpendicular to velocity, the particle moves in a circle of radius $r = \dfrac{mv}{qB}$.

   ✓Final answer

   The path is circular (option d).

4. 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

   Momentum is a vector quantity, $\vec p = m\vec v$. Although the magnitude of $\vec v$ stays fixed, the magnetic force continuously acts perpendicular to $\vec v$, which means it continuously changes the direction of $\vec v$ (this is exactly what produces circular motion in a uniform field — the perpendicular force acts as a centripetal force). So even though $|\vec p|$ is constant, the direction of $\vec p$ keeps changing, meaning the momentum vector $\vec p$ itself is not constant.

   Conclusion

   • Energy (kinetic energy) — constant, since the force does no work.

   • Momentum (a vector) — changes, because its direction is continuously altered even though its magnitude is fixed.

   ✓Final answer

   (b) Energy remains constant, but momentum changes (in direction).

5. 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 uniform circular motion, with

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

   (If the velocity had a component along B, the path would be a helix — but here it enters exactly perpendicular to the field.)

   ✓Final answer

   (a) circular.

6. CBSE 2024Set ANNUAL1 markMCQ
   Q.When a charged particle moves in a uniform magnetic field in a direction perpendicular to the field, then the path of the particle will be -

   (a) Parabolic

   (b) Circular

   (c) Straight line

   (d) Helical

   ›Reveal solutionSolution

   A uniform magnetic force acting always perpendicular to the velocity provides centripetal force, so the particle moves in a circle.

   When a charged particle of charge q moves with velocity v perpendicular to a uniform magnetic field B, it experiences a magnetic force:

   $\vec{F} = q\vec{v} \times \vec{B}$

   This force is always perpendicular to the velocity, so it changes only the direction of the velocity, never its magnitude (speed stays constant). A force of constant magnitude always perpendicular to velocity is exactly the condition for uniform circular motion — it acts as the centripetal force, $qvB = \dfrac{mv^2}{r}$, giving a fixed radius $r = \dfrac{mv}{qB}$.

   (If the particle also has a velocity component along B, the path becomes helical instead — but here the motion is purely perpendicular to B.)

   ✓Final answer

   (b) Circular.

7. CBSE 2024Set ANNUAL1 markMCQ
   Q.A charged particle enters at 30° to the magnetic field. Its path becomes :

   (a) circular

   (b) helical

   (c) elliptical

   (d) straight line

   ›Reveal solutionSolution

   Entering at an oblique angle (neither 0° nor 90°) to B gives a helix — the velocity component along B is unaffected, while the perpendicular component causes circular motion.

   When a charged particle enters a uniform magnetic field at an angle θ (here 30°) to $\vec B$, resolve its velocity into two components:

   • $v_\parallel = v\cos\theta$, along $\vec B$: experiences no magnetic force ($\vec F = q\vec v\times\vec B = 0$ for this component), so the particle drifts uniformly along the field direction.
   • $v_\perp = v\sin\theta$, perpendicular to $\vec B$: experiences a force $qv_\perp B$ that is always perpendicular to this velocity component, producing uniform circular motion in the plane perpendicular to $\vec B$.

   The combination of uniform linear drift along $\vec B$ and circular motion perpendicular to it traces out a helix (a 3-D spiral) — this is exactly what happens for any angle other than 0° (straight line, no perpendicular component) or 90° (pure circle, no parallel component).

   ✓Final answer

   Helical path — option (b).

8. CBSE 2023Set ANNUAL1 markMCQ
   Q.A charged particle enters with some speed at 30 degrees to the magnetic field. Its path will be :

   (a) helical

   (b) circular

   (c) parabolic

   (d) None of these

   ›Reveal solutionSolution

   A charge entering at an angle (other than 0 or 90 degrees) to a magnetic field follows a helical path.

   Resolve the velocity into a component parallel to B and one perpendicular to B. The parallel component feels no force and gives steady straight-line drift; the perpendicular component feels qvB and gives uniform circular motion. Their superposition is a helix (a circle that advances along B). This case from the NCERT/CBSE Class 12 chapter applies for any angle between 0 and 90 degrees, including 30 degrees.

   ✓Final answer

   (a) helical.

9. CBSE 2022Set ANNUAL1 markMCQ
   Q.The force ($\vec{F}$) acting on a particle of charge q moving with velocity ($\vec{v}$) in magnetic field ($\vec{B}$) is:

   (a) $\dfrac{q}{\vec{v}\times\vec{B}}$

   (b) $\dfrac{\vec{v}\times\vec{B}}{q}$

   (c) $q(\vec{v}\times\vec{B})$

   (d) $\vec{v}\times\vec{q}\times\vec{B}$

   ›Reveal solutionSolution

   The magnetic force on a charge is $\vec F = q(\vec v \times \vec B)$: option (C).

   A charge $q$ moving with velocity $\vec v$ in a magnetic field $\vec B$ experiences the magnetic part of the Lorentz force:

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

   • Its magnitude is $F = qvB\sin\theta$, where $\theta$ is the angle between $\vec v$ and $\vec B$.
   • The force is perpendicular to both $\vec v$ and $\vec B$ (given by the right-hand rule).

   Dimensionally the other options are wrong (dividing by a vector is meaningless).

   ✓Final answer

   (C) $q(\vec{v}\times\vec{B})$.

10. CBSE 2022Set ANNUAL1 markMCQ
    Q.A proton enters into a uniform magnetic field perpendicularly to it. The path of the proton would be:

    (a) Elliptical

    (b) Circular

    (c) Parabolic

    (d) Linear

    ›Reveal solutionSolution

    A charge entering $\perp$ to a uniform $\vec B$ moves in a circle, since the magnetic force provides constant centripetal force: option (B).

    The magnetic force $\vec F = q(\vec v \times \vec B)$ is always perpendicular to the velocity, so it does no work and the speed stays constant.

    When the proton enters perpendicular to $\vec B$, this constant-magnitude force acts as a centripetal force, bending the path into a circle of radius

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

    So the path is circular. (If the velocity had a component along $\vec B$, the path would be helical; here it is purely perpendicular, giving a circle.)

    ✓Final answer

    (B) Circular.

11. CBSE 2020Set 55/2/11 markMCQ
    Q.An electron is released from rest in a region of uniform electric and magnetic fields acting parallel to each other. The electron will (A) move in a straight line. (B) move in a circle. (C) remain stationary. (D) move in a helical path.

    ›Reveal solutionSolution

    When electric and magnetic fields are parallel and a charged particle starts from rest, only the electric field exerts a force initially; the particle accelerates along the field direction in a straight line.

    Understanding Forces on a Charged Particle

    The motion of a charged particle in electromagnetic fields depends on two forces: the electric force $\vec{F}_E = q\vec{E}$ and the magnetic force $\vec{F}_B = q(\vec{v} \times \vec{B})$. The key insight here is that the magnetic force has a peculiar property—it acts only when the particle has a velocity component perpendicular to the magnetic field.

    When the electric and magnetic fields are parallel to each other, we need to think carefully about what happens at each instant.

    Step-by-Step Analysis

    1. Initial condition: the electron at rest

    The electron starts from rest, so $\vec{v} = 0$ initially. The magnetic force is $\vec{F}_B = q(\vec{v} \times \vec{B}) = 0$ because the velocity is zero. Only the electric force acts:

    $\vec{F} = -e\vec{E}$

    The electron experiences a force opposite to the electric field direction (since its charge is negative).

    2. The electron begins to accelerate

    Under the electric force alone, the electron accelerates in a direction parallel to $\vec{E}$ (but opposite in sense). Since $\vec{E} \parallel \vec{B}$, the velocity that develops is also parallel to $\vec{B}$.

    3. What about the magnetic force as the electron moves?

    Once the electron has velocity $\vec{v}$, we check the magnetic force:

    $\vec{F}_B = -e(\vec{v} \times \vec{B})$

    But here's the crucial point: $\vec{v}$ is parallel to $\vec{B}$ (both along the same line). The cross product of two parallel vectors is zero:

    $\vec{v} \times \vec{B} = 0 \quad \text{when } \vec{v} \parallel \vec{B}$

    So the magnetic force remains zero throughout the motion.

    Tip

    The magnetic force $q(\vec{v} \times \vec{B})$ vanishes whenever velocity is parallel (or antiparallel) to the magnetic field. This is why particles can stream freely along magnetic field lines.

    4. The resulting motion

    With only the electric force acting continuously in one direction, the electron undergoes uniformly accelerated motion along a straight line parallel to the fields. The magnetic field, despite being present, plays no role in the dynamics.

    Watch out

    A common mistake is thinking "magnetic field present → curved path." This is only true when the velocity has a component perpendicular to $\vec{B}$. Parallel motion is unaffected by the magnetic field.

    5. Why not the other options?

    • (B) Circle: Requires $\vec{v} \perp \vec{B}$, which never develops here.
    • (C) Stationary: The electric force accelerates the particle; it cannot remain at rest.
    • (D) Helix: Requires both parallel and perpendicular velocity components relative to $\vec{B}$; starting from rest with parallel fields, no perpendicular component ever arises.

    ✓Final answer

    The correct option is (A) — the electron moves in a straight line.

12. CBSE 2020Set 55/3/11 markMCQ
    Q.A region has a uniform magnetic field in it. A proton enters into the region with velocity making an angle of $45°$ with the direction of the magnetic field. In this region the proton will move on a path having the shape of a (A) straight line (B) circle (C) spiral (D) helix

    ›Reveal solutionSolution

    A charged particle entering a magnetic field at an angle splits its velocity into parallel (unaffected) and perpendicular (circular motion) components; the combination produces helical motion. The answer is (D).

    When a charged particle moves through a magnetic field, the Lorentz force acts on it. The key insight is that the magnetic force $\vec{F} = q(\vec{v} \times \vec{B})$ is always perpendicular to both the velocity and the field. This means the force can only act on the component of velocity that is perpendicular to $\vec{B}$ — the parallel component remains completely unaffected.

    Think of it this way: the magnetic field "sees" only the perpendicular part of the velocity and bends it into circular motion, while the parallel part continues undisturbed, carrying the particle forward along the field direction. The result is a corkscrew path.

    Let me break down what happens to our proton entering at $45°$:

    1. Decompose the velocity into components If the proton enters with speed $v$ at $45°$ to $\vec{B}$, we can write:

    $v_\parallel = v \cos 45° = \frac{v}{\sqrt{2}}$

    $v_\perp = v \sin 45° = \frac{v}{\sqrt{2}}$

    The parallel component is along $\vec{B}$, the perpendicular component lies in the plane normal to $\vec{B}$.

    2. Analyze the motion due to each component

       The magnetic force is $\vec{F} = q\vec{v} \times \vec{B}$. Since $\vec{v}_\parallel$ is parallel to $\vec{B}$, their cross product vanishes — no force acts on this component. The proton continues with constant velocity $v_\parallel$ along the field direction.

       The perpendicular component $v_\perp$ experiences the full magnetic force, which is always perpendicular to $v_\perp$ itself. This is the classic setup for uniform circular motion in the plane perpendicular to $\vec{B}$.

    3. Combine the two motions

       In the plane perpendicular to $\vec{B}$: circular motion with radius $r = \frac{m v_\perp}{qB}$.

       Along the direction of $\vec{B}$: uniform motion with velocity $v_\parallel$.

       Superimposing these gives a helix — the particle traces out circles while simultaneously advancing along the field direction.

    4. Pitch of the helix

       The distance the particle advances along $\vec{B}$ during one complete circular orbit (period $T = \frac{2\pi m}{qB}$) is called the pitch:

    $p = v_\parallel \cdot T = \frac{v \cos 45° \cdot 2\pi m}{qB} = \frac{\sqrt{2}\pi m v}{qB}$

    Tip

    Special cases help build intuition:

    • If the particle enters parallel to $\vec{B}$ ($\theta = 0°$), then $v_\perp = 0$ and it moves in a straight line.

    • If it enters perpendicular ($\theta = 90°$), then $v_\parallel = 0$ and it moves in a circle.

    • Any angle in between produces a helix.

    Watch out

    A spiral (option C) would require the radius to change continuously — either growing or shrinking. But the magnetic force does no work (always perpendicular to velocity), so the speed remains constant and the radius stays fixed. The path is a helix, not a spiral.

    ✓Final answer

    The correct option is (D) helix.