Q.An electron enters a uniform magnetic field with speed $v$. It describes a semicircular path and comes out of the field. The final speed of the electron is : (A) Zero (B) $v$ (C) $\dfrac{v}{2}$ (D) $2v$

Showing the 12 most recent of 13 on this concept.

1. CBSE 2026Set DS1 mark
   Q.An electron of energy 10 eV is revolving round a circular path in a uniform magnetic field of $10^{-5}$ tesla. Determine the radius of the circular path.

   ›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.

   $p = \sqrt{2mE} = \sqrt{2(9.1\times10^{-31})(1.6\times10^{-18})} = \sqrt{2.912\times10^{-48}} = 1.71\times10^{-24}\ \text{kg·m/s}.$

   $r = \frac{p}{eB} = \frac{1.71\times10^{-24}}{(1.6\times10^{-19})(10^{-5})} = \frac{1.71\times10^{-24}}{1.6\times10^{-24}} \approx 1.07\ \text{m}.$

   ✓Final answer

   $r \approx 1.07$ m.

2. 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.

   Setting magnetic force = centripetal force for the perpendicular component:

   $q(v\sin\theta)B = \frac{m(v\sin\theta)^2}{r} \Rightarrow r = \frac{m\,v\sin\theta}{qB}.$

   ✓Final answer

   (C) mv sinθ/qB.

3. CBSE 2025Set 55/4/11 markMCQ
   Q.A particle having charge $+q$ enters a uniform magnetic field $\vec{B}$ as shown in the figure. The particle will describe: (A) a circular path in the XZ plane (B) a semicircular path in the XY plane (C) a helical path with its axis parallel to the Y-axis (D) a semicircular path in the YZ plane

   ›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.

   

   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 $\vec{v}$.

   Nature of the path. Since $\vec{v}\perp\vec{B}$, the entire velocity is perpendicular to the field, so the motion is a circle of radius $r = \dfrac{mv}{qB}$ lying in the plane perpendicular to $\vec{B}$ — that is the $XY$ plane. As the field fills only the upper region, the particle completes half the circle inside the field and exits, tracing a semicircle in the $XY$ plane.

   ✓Final answer

   (B) a semicircular path in the $XY$ plane.

4. CBSE 2025Set 55/5/11 markMCQ
   Q.A charged particle gains a speed of $10^{6}\ \mathrm{ms^{-1}}$ when accelerated from rest through a potential difference of 10 kV. It enters a region of magnetic field $0.4$ T such that its velocity is perpendicular to the field. The radius of the circular path described by it is: (A) 2.5 cm (B) 5 cm (C) 8 cm (D) 10 cm

   ›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:

   $r = \frac{v}{B} \cdot \frac{m}{q} = \frac{v}{B} \cdot \frac{1}{(q/m)}$

   This is cleaner because we have $q/m$ directly.

   4. Substitute and calculate

   $r = \frac{10^6}{0.4} \times \frac{1}{5 \times 10^7}$

   First, $\frac{10^6}{0.4} = \frac{10^6}{4/10} = 10^6 \times \frac{10}{4} = \frac{10^7}{4} = 2.5 \times 10^6$

   Then:

   $r = (2.5 \times 10^6) \times \frac{1}{5 \times 10^7} = \frac{2.5}{5} \times 10^{-1} = 0.5 \times 0.1 = 0.05\ \mathrm{m}$

   Convert to centimetres: $0.05\ \mathrm{m} = 5\ \mathrm{cm}$.

   Watch out

   A common mistake is to forget that $V = 10\ \mathrm{kV}$ means $10^4\ \mathrm{V}$, not $10^3$. Also, watch the unit conversion from metres to centimetres at the end — the options are in cm.

   ---

   ✓Final answer

   The radius of the circular path is 5 cm, which corresponds to option (B).

5. CBSE 2025Set 55/6/11 markMCQ
   Q.A proton and an $\alpha$-particle enter with the same velocity $\vec{v}$ in a uniform magnetic field $\vec{B}$ (with $\vec{v} \perp \vec{B}$). The ratio of the radii of their paths ($r_p : r_\alpha$) is: (A) 2 (B) $\dfrac{1}{2}$ (C) $\dfrac{1}{4}$ (D) 4

   ›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

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

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

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

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

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

   3. Identify the particle properties

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

   4. Write the radius for each particle

      For the proton:

   $r_p = \frac{m_p v}{eB}$

   For the α-particle:

   $r_\alpha = \frac{4m_p v}{2eB} = \frac{2m_p v}{eB}$

   5. Form the ratio

   $\frac{r_p}{r_\alpha} = \frac{\frac{m_p v}{eB}}{\frac{2m_p v}{eB}} = \frac{m_p v}{eB} \times \frac{eB}{2m_p v} = \frac{1}{2}$

   The α-particle, despite being four times heavier, also carries twice the charge. The net effect is that it curves in a circle twice as large as the proton's path.

   Watch out

   A common mistake is to think "heavier means larger radius" without accounting for charge. The α-particle's double charge partially compensates for its quadruple mass, so the radius ratio is not 1:4 but 1:2.

   ✓Final answer

   The correct option is (B) $\dfrac{1}{2}$.

6. 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: $qvB = \dfrac{mv^2}{r} \implies r = \dfrac{mv}{qB}$.

   Let the new velocity be $v' = 2v$ and new field $B' = B/2$. Then:

   $r' = \dfrac{m v'}{qB'} = \dfrac{m(2v)}{q(B/2)} = 4\left(\dfrac{mv}{qB}\right) = 4r$

   ✓Final answer

   (c) 4 times — the new radius becomes 4 times the original radius.

7. CBSE 2024Set 55/5/11 markMCQ
   Q.A particle of mass $m$ and charge $q$ describes a circular path of radius $R$ in a magnetic field. If its mass and charge were $2m$ and $q/2$ respectively, the radius of its path would be ______. (A) $R/4$ (B) $R/2$ (C) $2R$ (D) $4R$

   ›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.

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

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

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

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

   2. 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.

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

   3. 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.)

   4. Find the new radius

      Substitute the new values into the formula:

   $R' = \frac{m' v}{q' B} = \frac{(2m) v}{(q/2) B} = \frac{2m v}{q B/2} = \frac{2m v}{q B} \times 2 = \frac{4 m v}{q B}$

   But $\frac{m v}{q B} = R$, so:

   $R' = 4R$

   Watch out

   A common mistake is to treat mass and charge changes independently and guess $R' = (2)/(1/2) R = R$ or something similar. Always combine them as a ratio: $R \propto m/q$, so the factor is $(m'/q')/(m/q) = (2m)/(q/2) \div (m/q) = 4$.

   Tip

   You can think of it this way: doubling mass alone would double the radius (more inertia, harder to turn), but halving the charge also doubles the radius (weaker magnetic force). Two doublings multiply to give $4R$.

   ✓Final answer

   The radius becomes $4R$, so the correct option is (D).

8. CBSE 2024Set 55/2/11 markMCQ
   Q.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.

   ›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.

   ---

   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. The reason gives $r = \frac{mv}{qB}$, which is correct. It is the standard formula derived from Newton's second law. Since $mv = p$, this formula directly leads to the conclusion in the assertion. The reason is true and it correctly explains why the assertion holds.

   Watch out

   A common mistake is to think radius depends on mass — "proton is heavier, so it must have a larger radius." But here momentum is fixed, not velocity. The heavier proton moves slower, the lighter electron moves faster, and the product $mv$ stays the same. Radius cares about $mv$, not $m$ alone.

   Tip

   Whenever you see "same momentum" in a magnetic field problem, immediately write $r = p/qB$. It cuts through the clutter — mass and velocity differences cancel out.

   ✓Final answer

   Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A). The correct option is (A).

9. CBSE 2023Set 55/3/11 markMCQ
   Q.A particle of mass $m$ and charge $q$ moving with a uniform velocity $\vec v = v_{0x}\hat{i} + v_{0y}\hat{j}$ enters a region with a magnetic field $\vec B = B_0\hat{j}$. After some time, an electric field $\vec E = E_0\hat{j}$ is also switched on in the region. The resulting path described by the particle will be :

   (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).

   The pitch of a helix is the distance the particle moves along the axis ($y$) in one complete revolution. One revolution takes time $T = 2\pi / \omega$. The distance travelled along $y$ in that time is

   $\Delta y = v_y(t) \cdot T,$

   but since $v_y$ is changing, the distance covered in each successive revolution is different. Specifically, during the $n$-th revolution, the average $v_y$ is larger (or smaller) than during the previous one, so the pitch increases (or decreases) with time.

   Tip

   If the electric field were zero, $v_y$ would be constant and the pitch would be constant — a regular helix. The electric field adds a linear change to $v_y$, so the pitch changes linearly with time, meaning the $y$-displacement grows quadratically. That’s the signature of an increasing pitch helix.

   5. Eliminate the other options

   • (a) A circle in the $x$-$z$ plane — This would require $v_y = 0$ always. But $v_y$ is not zero initially, and even if it were, the electric field would accelerate it along $y$, so the motion is never confined to a plane.
   • (b) A parabola in the $x$-$y$ plane — The magnetic force has a $z$-component, so the motion is three-dimensional. A parabola would require no force out of the $x$-$y$ plane, which is false here.
   • (c) A helix with constant pitch — This would require $v_y$ constant, which is false once the electric field is on.
   • (d) A helix with increasing pitch — This matches our analysis exactly.

   ---

   ✓Final answer

   The correct option is (d) a helix with increasing pitch.

10. CBSE 2023Set 55/4/11 markMCQ
    Q.An electron enters a uniform magnetic field with speed $v$. It describes a semicircular path and comes out of the field. The final speed of the electron is : (A) Zero (B) $v$ (C) $\dfrac{v}{2}$ (D) $2v$

    ›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.

    1. 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.

    2. 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.

    3. 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.

    4. The final speed

       The electron enters at speed $v$ and exits at speed $v$. No option other than $v$ is consistent with the work-energy theorem applied to a purely magnetic force.

    Watch out

    A common mistake is to think that because the path curves, the particle must slow down or that "coming out" implies some loss of energy. But curvature alone does not imply a change in speed — only a change in direction. The magnetic force is a deflecting force, not an accelerating one in the sense of changing speed.

    Tip

    If you ever see a problem where a charged particle moves through a uniform magnetic field and the only force is magnetic, the speed is always conserved. The radius of the path changes with speed, but the speed itself does not change unless an electric field or some other force is present.

    ✓Final answer

    The final speed of the electron is $v$, so the correct option is (B).

11. CBSE 2019Set 55/2/11 mark
    Q.A proton is accelerated through a potential difference $V$, subjected to a uniform magnetic field acting normal to the velocity of the proton. If the potential difference is doubled, how will the radius of the circular path described by the proton in the magnetic field change ?

    ›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

    1. 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:

    $v = \sqrt{\frac{2qV}{m}}$

    2. Express the initial radius $r_1$

       Substituting this velocity into the radius formula:

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

    3. Find the new radius $r_2$ when potential is doubled

       When the potential difference becomes $2V$, the new velocity is:

    $v' = \sqrt{\frac{2q(2V)}{m}} = \sqrt{2} \cdot \sqrt{\frac{2qV}{m}} = \sqrt{2} \cdot v$

    The new radius becomes:

    $r_2 = \frac{mv'}{qB} = \frac{m(\sqrt{2} \cdot v)}{qB} = \sqrt{2} \cdot r_1$

    4. Calculate the ratio

    $\frac{r_2}{r_1} = \sqrt{2}$

    $r \propto \sqrt{V}$

    The cyclotron radius is proportional to the square root of the accelerating potential.

    Watch out

    A common mistake is to assume the radius doubles when the voltage doubles. Remember that kinetic energy is proportional to $v^2$, so doubling the energy only increases velocity—and hence radius—by $\sqrt{2}$.

    ✓Final answer

    The radius of the circular path increases by a factor of $\boxed{\sqrt{2}}$ when the potential difference is doubled.

12. CBSE 2019Set ANNUAL1 markMCQ
    Q.A charged particle of mass $m$ and charge $q$ moves along a circular path with a velocity $v$ perpendicular to a magnetic field $\vec{B}$. The radius of the circular path is

    (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

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

    Step 3 — Solve for $r$

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

    ✓Final answer

    $r = \dfrac{mv}{qB}$ — option (a).