Q.An electron and a positron are released from $(0, 0, 0)$ and $(0, 0, 1.5R)$ respectively, in a uniform magnetic field $\vec{B} = B_0\hat{i}$, each with an equal momentum of magnitude $p = eBR$. Under what conditions on the direction of momentum will the orbits be non-intersecting circles?

Showing the 12 most recent of 15 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.A proton and a deuteron, both having the same kinetic energy, enter perpendicularly into a uniform magnetic field B. For motion of proton and deuteron on circular path of radius $R_p$ and $R_d$ respectively, the correct statement is (A) $R_d = \sqrt{2}R_p$ (B) $R_d = \dfrac{R_p}{\sqrt{2}}$ (C) $R_d = R_p$ (D) $R_d = 2R_p$

   ›Reveal solutionSolution

   Same kinetic energy, same charge, different mass: the circular-motion radius scales as $\sqrt{m}$, so the deuteron's radius is $\sqrt2$ times the proton's.

   Concept and Intuition

   A charged particle moving perpendicular to a magnetic field $B$ travels in a circle of radius $R = \dfrac{mv}{qB}$, from balancing the magnetic force against the centripetal requirement ($qvB = mv^2/R$). When particles share the same kinetic energy $K=\tfrac12 mv^2$ rather than the same speed, it's more useful to write momentum as $p = mv = \sqrt{2mK}$, giving

   $R = \frac{\sqrt{2mK}}{qB}.$

   So for fixed $K$ and $B$, $R \propto \sqrt{m}/q$. The deuteron (one proton + one neutron) has essentially twice the proton's mass but the same charge magnitude $+e$, so its radius scales up by $\sqrt{2}$ relative to the proton's.

   Step-by-Step Solution

   1. $R_p = \dfrac{\sqrt{2m_pK}}{eB}$, $R_d = \dfrac{\sqrt{2m_dK}}{eB}$ (same $K$, same $B$, same charge magnitude $e$ for both).
   2. Ratio: $\dfrac{R_d}{R_p} = \sqrt{\dfrac{m_d}{m_p}}$. …

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2. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.A particle having charge 'q' enters a uniform transverse magnetic field $\vec{B}$. It is deflected through a distance 'x' while travelling a distance 'y' as shown in figure. The magnitude of the momentum of the particle is [FIGURE: a particle enters horizontally from the left travelling distance y, then curves upward on a parabolic-like path, being deflected a vertical distance x] (A) $\dfrac{qB}{2}[y^2+x^2]$ (B) $\dfrac{qBy^2}{x}$ (C) $\dfrac{qB}{2}\left[\dfrac{y^2}{x}+x\right]$ (D) $\dfrac{qBy^2}{2x}$

   ›Reveal solutionSolution

   A charged particle deflects along a circular arc in a transverse field; eliminating the turn-angle between the deflection $x$ and the along-path distance $y$ gives the orbit radius, hence the momentum. Answer: (C).

   Concept and Intuition

   A charged particle moving perpendicular to a uniform $\vec B$ travels on a circular arc of radius $R=\dfrac{mv}{qB}=\dfrac{p}{qB}$. The two measured quantities $x$ (deflection) and $y$ (distance travelled before that deflection) are just the sagitta and chord-projection of this same circle, so geometry alone fixes $R$ in terms of $x$ and $y$ — no need to assume $x\ll y$.

   Step-by-Step Solution

   1. Put the entry point at the origin with the initial velocity along the $+y'$ (horizontal) direction. The particle's path is a circular arc of radius $R$.
   2. After the velocity vector has turned through angle $\theta$, the coordinates on the circle are $y=R\sin\theta$ (distance travelled along the original direction) and $x=R(1-\cos\theta)$ (perpendicular deflection).
   3. From these: $R\cos\theta = R-x$ and $R\sin\theta=y$. Squaring and adding: $y^2+(R-x)^2=R^2$. …

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3. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.If a charged particle enters a uniform magnetic field normally with certain velocity, then the time period of revolution of the particle (A) decreases with increase of velocity of the particle (B) increases with increase of radius of the orbit (C) increases with increase of magnetic field (D) decreases with increase of specific charge of the particle

   ›Reveal solutionSolution

   The cyclotron period $T=2\pi m/(qB)$ depends only on the specific charge $q/m$ and the field $B$ — never on the particle's speed or the orbit radius.

   Concept and Intuition

   When a charged particle enters a magnetic field perpendicular to its velocity, the magnetic force provides centripetal force: $qvB = \dfrac{mv^2}{r}$, giving $r = \dfrac{mv}{qB}$. The period is $T=\dfrac{2\pi r}{v} = \dfrac{2\pi m}{qB}$. Notice $v$ cancels out completely — a faster particle moves in a larger circle, and the two effects exactly compensate so that the time to complete one revolution stays the same. This is the basis of the cyclotron's fixed-frequency operation.

   Step-by-Step Solution

   1. Write $T = \dfrac{2\pi m}{qB} = \dfrac{2\pi}{(q/m)\,B}$.
   2. Velocity dependence: $T$ has no $v$ in it at all, so option (A) — "decreases with increase of velocity" — is false; $T$ is unchanged.
   3. Radius dependence: since $r=mv/(qB)$, a larger radius (at fixed $B$, $q/m$) simply means a proportionally larger $v$; $T$ still stays constant. Option (B) is false. …

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4. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.If the operating magnetic field in a cyclotron for accelerating protons is 668 mT, then the angular frequency of the oscillator of the cyclotron is (Charge of proton $=1.6\times10^{-19}$ C and mass of proton $=1.67\times10^{-27}$ kg) (A) $9.6\times10^7$ rad s$^{-1}$ (B) $3.2\times10^7$ rad s$^{-1}$ (C) $12.8\times10^7$ rad s$^{-1}$ (D) $6.4\times10^7$ rad s$^{-1}$

   ›Reveal solutionSolution

   The cyclotron's operating angular frequency is simply the charged particle's cyclotron frequency $\omega=qB/m$, which comes out to $6.4\times10^7$ rad/s for a proton in a 668 mT field.

   Concept and Intuition

   Inside a cyclotron, a charged particle moving in a magnetic field experiences the magnetic force as the centripetal force for circular motion: $qvB = \dfrac{mv^2}{r}$, which rearranges to $v/r = \omega = qB/m$. Remarkably, this angular frequency (the "cyclotron frequency") doesn't depend on the particle's speed or the radius of its path — that's exactly what makes the cyclotron work, since the oscillator driving the accelerating voltage can run at a single fixed frequency throughout the whole spiral.

   Step-by-Step Solution

   1. $\omega = \dfrac{qB}{m}$, with $q=1.6\times10^{-19}$ C, $B=668\text{ mT}=0.668$ T, $m=1.67\times10^{-27}$ kg.
   2. Numerator: $qB = 1.6\times10^{-19}\times0.668 = 1.0688\times10^{-19}$. …

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5. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.If an electron and a proton enter normally into a uniform magnetic field with equal kinetic energies, then (A) the electron travels in circular path of less radius. (B) the proton travels in circular path of less radius. (C) both electron and proton travel in circular paths of same radius. (D) both electron and proton travel in straight line path.

   ›Reveal solutionSolution

   With equal kinetic energy, the lighter particle (electron) has a smaller circular radius in a magnetic field because radius scales with $\sqrt{m}$.

   Concept and Intuition

   The magnetic force provides centripetal force: $qvB=\dfrac{mv^2}{r}\Rightarrow r=\dfrac{mv}{qB}=\dfrac{p}{qB}$. Expressing momentum in terms of kinetic energy, $p=\sqrt{2mKE}$, so $r=\dfrac{\sqrt{2mKE}}{qB}$. For the same KE and charge magnitude, $r\propto\sqrt m$.

   Step-by-Step Solution

   1. $r=\dfrac{\sqrt{2mKE}}{qB}$.
   2. For equal KE and equal $|q|$ (electron and proton have the same charge magnitude) and the same B: $r\propto\sqrt{m}$.
   3. $m_{electron}\ll m_{proton}$, so $r_{electron}\ll r_{proton}$.
   4. Hence the electron travels in a circle of smaller radius. …

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6. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.Two charged particles of specific charges in the ratio $2:1$ and masses in the ratio $1:4$ moving with same kinetic energy enter a uniform magnetic field at right angles to the direction of the field. The ratio of the radii of the circular paths in which the particles move under the influence of the magnetic field is (A) $2:1$ (B) $1:1$ (C) $4:1$ (D) $8:1$

   ›Reveal solutionSolution

   Expressing the cyclotron radius in terms of specific charge and mass for equal kinetic energies shows the two particles trace circles of exactly equal radius, i.e. ratio $1:1$.

   Concept and Intuition

   For a charged particle moving in a uniform magnetic field, the radius of its circular path is $r=\dfrac{mv}{qB}$. Since kinetic energy (not speed) is given as equal here, it's cleanest to eliminate $v$ using $KE=\frac12 mv^2 \Rightarrow v=\sqrt{2KE/m}$, so $r$ ends up depending on mass and specific charge $q/m$ in a particular combined way — not simply on mass or charge alone.

   Step-by-Step Solution

   1. Write $r = \dfrac{mv}{qB}$ and substitute $v=\sqrt{\dfrac{2\,KE}{m}}$: $r = \dfrac{m}{qB}\sqrt{\dfrac{2KE}{m}} = \dfrac{\sqrt{2\,KE\,m}}{qB}$.
   2. Since $KE$ and $B$ are the same for both particles, $r \propto \dfrac{\sqrt m}{q}$.
   3. Rewrite using specific charge $s=q/m$: $q = sm$, so $r \propto \dfrac{\sqrt m}{sm} = \dfrac{1}{s\sqrt m}$.
   4. $\dfrac{r_1}{r_2} = \dfrac{1/(s_1\sqrt{m_1})}{1/(s_2\sqrt{m_2})} = \dfrac{s_2\sqrt{m_2}}{s_1\sqrt{m_1}}$. …

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7. AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
   Q.Two charged particles of same charge but different masses $m_1$ and $m_2$ are projected with the same velocity V normally into the uniform magnetic field $\vec{B}$ as shown in figure. The maximum separation of the particles is [FIGURE] (a region of uniform magnetic field shown as dots coming out of the page (field $\vec{B}$); two charged particles enter from the bottom with velocity V directed upward and follow two different semicircular/parabolic paths of different radii inside the field, emerging at the top separated by a horizontal distance d) (A) $\dfrac{|(m_2 - m_1)V|}{|qB|}$ (B) $\dfrac{|(m_2 - m_1)V|}{|2qB|}$ (C) $\dfrac{|2(m_2 - m_1)V|}{|qB|}$ (D) $\dfrac{|(m_2 - m_1)V|}{|4qB|}$

   ›Reveal solutionSolution

   The maximum separation between the two particles is twice the difference in their cyclotron radii, because each follows a semicircular path in the magnetic field. The correct expression is $\dfrac{2|(m_2 - m_1)V|}{|qB|}$, which corresponds to option (C).

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   Concept and Intuition

   When a charged particle moves perpendicular to a uniform magnetic field, it experiences a magnetic force that is always perpendicular to its velocity. This force provides the centripetal force for circular motion. The radius of that circle — the cyclotron radius — is given by:

   $r = \frac{mv}{|q|B}$

   Here, both particles have the same charge $q$ and the same initial speed $V$, but different masses $m_1$ and $m_2$. So they will follow circular paths of different radii inside the field. Since they enter the field moving straight upward (velocity perpendicular to $\vec{B}$), each will trace out exactly a semicircle before exiting the field region. The maximum separation between them occurs at the moment they are farthest apart — which is when they are on opposite sides of their circular paths, i.e., when one is at the far left of its semicircle and the other at the far right. But here, because both start from the same point and curve in the same direction (both bend to the right, as shown in the figure), the maximum separation is simply the difference in the diameters of their semicircular paths.

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   Step-by-Step Solution

   1. Determine the radius of each particle's path For a particle of mass $m$ and charge $q$ moving with speed $V$ perpendicular to a uniform magnetic field $B$, the magnetic force provides the centripetal force:

   $qVB = \frac{mV^2}{r} \quad \Rightarrow \quad r = \frac{mV}{qB}$

   (We take absolute values; the sign of $q$ determines direction of bending, but both particles have the same sign, so they bend the same way.)

   Thus:

   $r_1 = \frac{m_1 V}{|q|B}, \qquad r_2 = \frac{m_2 V}{|q|B}$

   2. Path inside the field is a semicircle

      The particles enter from the bottom, moving upward. The magnetic force is always perpendicular to velocity, so they will follow a circular arc. Since they enter at the bottom of the field region and exit at the top, and the field is uniform, each completes exactly half a circle (a semicircle) before leaving. The horizontal displacement from entry point to exit point is equal to the diameter of that circle: $2r$.

   3. Find the horizontal positions of the exit points …

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8. AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
   Q.The radius of the path of a charged particle of mass 'm' and charge 'q' moving with a speed v in a magnetic field B is given by (A) $\dfrac{mv^2}{qB}$ (B) $\dfrac{mv}{qB}$ (C) $\dfrac{2m\pi v}{qB}$ (D) $\dfrac{mv}{qB^2}$

   ›Reveal solutionSolution

   Equating the magnetic (Lorentz) force to the required centripetal force directly gives the standard cyclotron-radius formula.

   Concept and Intuition

   A charged particle moving perpendicular to a uniform magnetic field experiences a force $qvB$ that is always perpendicular to its velocity — this makes it move in a circle, with the magnetic force itself supplying the centripetal force needed for that circular path.

   Step-by-Step Solution

   1. Magnetic force: $F=qvB$.
   2. Centripetal force required: $F=\dfrac{mv^2}{r}$. …

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9. AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
   Q.To generate the high magnetic fields, the synchrotron uses (A) solenoid only (B) toroid only (C) combination of solenoid and toroid (D) metal dee

   ›Reveal solutionSolution

   A synchrotron needs very high magnetic fields to bend energetic particles around its ring; this is achieved using a combination of solenoid and toroid winding configurations, not either alone.

   Concept and Intuition

   A solenoid produces a strong, nearly uniform field along its axis but the field outside is weak; a toroid confines the field entirely within its doughnut-shaped core, giving a strong field along a closed loop path. Large accelerators like synchrotrons need very high, well-confined fields over an extended, closed path to bend and focus particle beams — a purpose best served by combining aspects of both configurations rather than using either winding type in isolation.

   Step-by-Step Solution

   1. Solenoid alone: strong field but only along its own axis, does not confine a field around a full ring by itself. …

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10. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
    Q.The magnetic force q [v x B] is (A) parallel to both v and B (B) perpendicular to v (C) perpendicular to both v and B (D) parallel to B

    ›Reveal solutionSolution

    This tests the direction of the magnetic (Lorentz) force; being a cross product, it is perpendicular to both the velocity and the field.

    Concept and Intuition

    The magnetic force on a charge is defined through the vector (cross) product $\vec F = q(\vec v \times \vec B)$. By the very definition of a cross product $\vec A \times \vec B$, the resulting vector is always perpendicular to the plane containing $\vec A$ and $\vec B$ — never parallel to either one (unless the force is zero). Physically this is why a magnetic field can change a charge's direction of motion but never do work on it: the force is always sideways to the velocity.

    Step-by-Step Solution

    1. Write the force law: $\vec F = q(\vec v \times \vec B)$.
    2. By the geometric definition of a cross product, $\vec v \times \vec B$ is a vector perpendicular to the plane formed by $\vec v$ and $\vec B$. …

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11. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
    Q.If 'q' is electric charge, B is magnetic field, 'R' is the dee radius and 'm' is the mass of ions, the kinetic energy of the ions in cyclotron is given by (A) $\dfrac{qBR}{2m}$ (B) $\dfrac{qBR}{m}$ (C) $\dfrac{q^2B^2R^2}{4\pi m}$ (D) $\dfrac{q^2B^2R^2}{2m}$

    ›Reveal solutionSolution

    Balancing magnetic force with centripetal force gives the ion's speed at the dee radius; squaring and substituting into $KE=\tfrac12mv^2$ gives $\dfrac{q^2B^2R^2}{2m}$.

    Concept and Intuition

    In a cyclotron, charged ions move in circular arcs inside the dees under a uniform magnetic field $B$ perpendicular to their motion. The magnetic (Lorentz) force supplies the centripetal force needed for circular motion. As the ions gain energy from the accelerating electric field between dees, their radius of curvature grows, reaching maximum radius $R$ (the dee radius) just before ejection — that determines their maximum kinetic energy.

    Step-by-Step Solution

    1. Centripetal force condition: $qvB = \dfrac{mv^2}{R}$.
    2. Solve for speed: $v = \dfrac{qBR}{m}$.
    3. Kinetic energy: $KE = \dfrac12 mv^2$. …

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12. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.A cyclotron's oscillator frequency is 20 MHz. The operating magnetic field for accelerating protons is (charge of proton = $1.6\times10^{-19}$ C, mass of proton = $1.67\times10^{-27}$ kg) (A) 0.66 T (B) 1.1 T (C) 0.33 T (D) 1.31 T

    ›Reveal solutionSolution

    This tests the cyclotron resonance formula relating oscillator frequency, magnetic field, and the accelerated particle's charge-to-mass ratio. The required field is 1.31 T.

    Concept and Intuition

    In a cyclotron, a charged particle moves in a circular path under a magnetic field, and it must cross the gap between the dees exactly in phase with the oscillating electric field. This requires the particle's own cyclotron (revolution) frequency to match the oscillator's frequency: $f = \dfrac{qB}{2\pi m}$. Rearranging gives the magnetic field needed for a given oscillator frequency and particle type.

    Step-by-Step Solution

    1. Cyclotron frequency condition: $f = \dfrac{qB}{2\pi m} \Rightarrow B = \dfrac{2\pi f m}{q}$.
    2. Substitute values: $f=20\times10^6$ Hz, $m=1.67\times10^{-27}$ kg, $q=1.6\times10^{-19}$ C.
    3. Numerator: $2\pi\times20\times10^6\times1.67\times10^{-27} = 6.2832\times20\times10^6\times1.67\times10^{-27}$. …

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