Q.In a chamber, a uniform magnetic field of $6.5\ \text{G}$ ($1\ \text{G} = 10^{-4}\ \text{T}$) is maintained. An electron is shot into the field with a speed of $4.8 \times 10^{6}\ \text{m s}^{-1}$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. ($e = 1.5 \times 10^{-19}\ \text{C}$, $m_e = 9.1 \times 10^{-31}\ \text{kg}$).

Showing the 12 most recent of 25 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.A proton with energy 2 MeV is moving perpendicular to a uniform magnetic field of 2.5 tesla. The force on the proton is (mass of proton is $1.66\times10^{-27}\ kg$) (A) $2.5\times10^{-10}\ N$ (B) $7.8\times10^{-11}\ N$ (C) $2.5\times10^{-11}\ N$ (D) $7.8\times10^{-12}\ N$

   ›Reveal solutionSolution

   This tests the magnetic Lorentz force on a moving charge, after first extracting speed from kinetic energy; the force is $\approx7.8\times10^{-12}\ N$.

   Concept and Intuition

   The magnetic force on a charge moving perpendicular to a field is $F = qvB$ (maximum, since $\sin90^\circ=1$). The only extra step here is that the speed isn't given directly — it must be extracted from the kinetic energy using $KE = \frac{1}{2}mv^2$, converting MeV to joules first.

   Step-by-Step Solution

   1. Convert energy: $KE = 2\ MeV = 2\times1.6\times10^{-13}\ J = 3.2\times10^{-13}\ J$.
   2. Solve for speed: $v = \sqrt{\dfrac{2KE}{m}} = \sqrt{\dfrac{2\times3.2\times10^{-13}}{1.66\times10^{-27}}} = \sqrt{3.855\times10^{14}} \approx 1.9635\times10^7\ m/s$.
   3. Since the proton moves perpendicular to $B$, the force is maximum: $F = qvB$. …

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2. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.A charged particle moves in a region where there is a uniform electric field $E = 2\times10^3\hat{i}\ N\,C^{-1}$ and a uniform magnetic field $B = 5\times10^{-2}\hat{j}\ T$. The velocity of the particle (in $ms^{-1}$), if the particle moves without any acceleration. (A) $5\times10^4\hat{k}$ (B) $2\times10^3\hat{k}$ (C) $4\times10^4\hat{k}$ (D) $4\times10^4(-\hat{k})$

   ›Reveal solutionSolution

   This is the classic velocity-selector condition — zero net force means the electric and magnetic forces exactly cancel, fixing both the speed ($v=E/B$) and the direction of $\vec v$.

   Concept and Intuition

   A charged particle experiences zero acceleration only when the total force is zero. Here the forces are electric ($q\vec E$) and magnetic ($q\vec v\times\vec B$). For these to cancel exactly (for any nonzero charge), their magnitudes must match and their directions must be opposite. This is exactly the working principle of a velocity selector — only particles with a specific speed $v=E/B$ pass through undeflected, regardless of charge magnitude.

   Step-by-Step Solution

   1. Balance condition: $q\vec E + q(\vec v\times \vec B) = 0 \Rightarrow \vec v \times \vec B = -\vec E$.
   2. Magnitude: $vB = E \Rightarrow v = \dfrac{E}{B} = \dfrac{2\times10^3}{5\times10^{-2}} = 4\times10^4\,\text{m/s}$. …

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3. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.A charged particle is moving parallel to a uniform magnetic field. Then the force on the charged particle (A) $qvB$ (B) $\dfrac{B}{qv}$ (C) $\dfrac{v^2B}{q}$ (D) zero

   ›Reveal solutionSolution

   The magnetic force on a moving charge is $F=qvB\sin\theta$; when velocity is parallel to $\vec B$, the angle between them is zero, so the force vanishes. Answer: zero.

   Concept and Intuition

   The magnetic Lorentz force is $\vec F = q\vec v \times \vec B$, whose magnitude is $qvB\sin\theta$, where $\theta$ is the angle between velocity and field. This is a cross product, and cross products of parallel (or anti-parallel) vectors are always zero — there is no component of velocity perpendicular to $\vec B$ to generate a deflecting force.

   Step-by-Step Solution

   1. Force law: $F = qvB\sin\theta$.
   2. "Moving parallel to the field" means the angle between $\vec v$ and $\vec B$ is $\theta = 0^\circ$. …

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4. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.A velocity selector is to be constructed to select ions with a velocity of $6\ \mathrm{km\,s^{-1}}$. If the electric field used is $400\ \mathrm{V\,m^{-1}}$, then the magnetic field to be used is (A) $\dfrac{11}{20}$ T (B) $\dfrac{2}{3}$ T (C) $\dfrac{1}{15}$ T (D) $\dfrac{2}{15}$ T

   ›Reveal solutionSolution

   This tests the velocity selector principle: crossed electric and magnetic fields let through only one speed. Answer: $B=\dfrac{1}{15}$ T.

   Concept and Intuition

   In a velocity selector, an electric field and a magnetic field are arranged perpendicular to each other and to the ion's velocity, so the electric force $qE$ and magnetic force $qvB$ act in opposite directions. Only ions moving at the exact speed where these forces balance pass through undeflected: $qE=qvB \Rightarrow v=\dfrac{E}{B}$.

   Step-by-Step Solution

   1. Balance condition: $v=\dfrac{E}{B} \Rightarrow B=\dfrac{E}{v}$.
   2. Convert velocity: $v=6\ \text{km/s}=6000\ \text{m/s}$. …

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5. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
   Q.When an electron accelerated from rest through a potential difference $V_1$ enters a uniform magnetic field, the maximum force on it is F. If the potential difference is changed to $V_2$, then the maximum force on the electron in the same magnetic field is 4F, then $\dfrac{V_1}{V_2} =$ (A) 1 : 4 (B) 4 : 1 (C) 2 : 1 (D) 1 : 16

   ›Reveal solutionSolution

   The maximum magnetic force on a charge scales with its speed, and speed scales with the square root of the accelerating voltage; a 4-fold force increase means a 16-fold voltage increase, so $V_1:V_2=1:16$.

   Concept and Intuition

   When a charge is accelerated from rest through a potential difference $V$, all the electrical work goes into kinetic energy: $qV=\tfrac12 mv^2$, so $v=\sqrt{2qV/m}$ — speed grows only as the square root of the voltage. Once in a uniform magnetic field, the maximum magnetic force (when velocity is perpendicular to $B$) is $F_{max}=qvB$, which is directly proportional to $v$, and hence to $\sqrt{V}$.

   Step-by-Step Solution

   1. Speed after acceleration through $V_1$: $v_1=\sqrt{2qV_1/m}$, giving force $F=qv_1B$.
   2. Speed after acceleration through $V_2$: $v_2=\sqrt{2qV_2/m}$, giving force $4F=qv_2B$. …

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6. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.If a proton of kinetic energy 8.35 MeV enters a uniform magnetic field of 10 T at right angles to the direction of the field, then the force acting on the proton is (Mass of proton $= 1.67 \times 10^{-27}$ kg and charge of proton $= 1.6 \times 10^{-19}$ C) (A) $48 \times 10^{-12}$ N (B) $16 \times 10^{-12}$ N (C) $64 \times 10^{-12}$ N (D) $32 \times 10^{-12}$ N

   ›Reveal solutionSolution

   This tests converting kinetic energy to speed and then applying the magnetic Lorentz force $F=qvB$ for a charge moving perpendicular to $B$; the answer is $64\times10^{-12}$ N.

   Concept and Intuition

   A charged particle moving at right angles to a uniform magnetic field feels a force $F=qvB\sin\theta=qvB$ (since $\theta=90^\circ$). To use this we first need the proton's speed, which we get from its kinetic energy.

   Step-by-Step Solution

   1. Convert kinetic energy to joules using the proton charge as the eV-to-J factor: $KE=8.35\times1.6\times10^{-13}\,\text{J}=1.336\times10^{-12}$ J.
   2. From $KE=\tfrac12 mv^2$: $v=\sqrt{\dfrac{2KE}{m}}=\sqrt{\dfrac{2\times1.336\times10^{-12}}{1.67\times10^{-27}}}=\sqrt{1.60\times10^{15}}\approx4.0\times10^{7}\ \text{m/s}$. …

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7. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.An electron is moving with a velocity $(2\bar{i} + 3\bar{j})\ ms^{-1}$ in an electric field $(3\bar{i} + 6\bar{j} + 2\bar{k})\ Vs^{-1}$ and a magnetic field of $(2\bar{j} + 3\bar{k})\ T$. Then the magnitude and direction (with x-axis) of the Lorentz force acting on the electron is (A) $9.6 \times 10^{-19}$ N, $\theta = \cos^{-1}\left(\dfrac{2}{\sqrt{5}}\right)$ (B) $9.6 \times 10^{-19}$ N, $\theta = \cos^{-1}\left(\dfrac{5}{\sqrt{2}}\right)$ (C) $2.15 \times 10^{-18}$ N, $\theta = \cos^{-1}\left(\dfrac{2}{\sqrt{5}}\right)$ (D) $2.15 \times 10^{-18}$ N, $\theta = \cos^{-1}\left(\dfrac{5}{\sqrt{2}}\right)$

   ›Reveal solutionSolution

   Computing $\vec v\times\vec B$ and adding $\vec E$ gives the net field vector $(12,0,6)$ V/m; its magnitude times $e$ gives $2.15\times10^{-18}$ N at $\cos^{-1}(2/\sqrt5)$ from the x-axis.

   Concept and Intuition

   The Lorentz force is $\vec F=q(\vec E+\vec v\times\vec B)$. We compute the cross product component-wise, add the electric field, then find magnitude and direction of the resultant.

   Step-by-Step Solution

   1. $\vec v\times\vec B$ with $\vec v=(2,3,0)$, $\vec B=(0,2,3)$:

   $\vec v\times\vec B = (3\cdot3-0\cdot2,\ 0\cdot0-2\cdot3,\ 2\cdot2-3\cdot0) = (9,-6,4)$

   2. Add $\vec E=(3,6,2)$: $\vec E+\vec v\times\vec B = (12,0,6)$.
   3. Magnitude: $|\vec F\text{-field}| = \sqrt{12^2+0^2+6^2}=\sqrt{180}=6\sqrt5$.
   4. Force: $F = e\times6\sqrt5 = 1.6\times10^{-19}\times13.42 \approx 2.15\times10^{-18}$ N. …

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8. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.A proton and an alpha particle moving with energies in the ratio 1:4 enter a uniform magnetic field of 3T at right angles to the direction of magnetic field. The ratio of the magnetic forces acting on the proton and the alpha particle is (A) 1:2 (B) 1:4 (C) 2:3 (D) 1:3

   ›Reveal solutionSolution

   Express the magnetic force in terms of kinetic energy (not just charge and mass separately), then substitute the known charge/mass/energy ratios for proton and alpha. Answer: 1:2.

   Concept and Intuition

   The magnetic force on a charged particle moving perpendicular to a field is $F = qvB$. Since we're given kinetic energies rather than speeds, it helps to write $v = \sqrt{2KE/m}$ and substitute, so that $F = qB\sqrt{2KE/m}$.

   Step-by-Step Solution

   1. Proton: charge $q_p = e$, mass $m_p$, kinetic energy $K$.
   2. Alpha particle: charge $q_\alpha = 2e$, mass $4m_p$, kinetic energy $4K$ (given ratio 1:4).
   3. $F_p = eB\sqrt{\dfrac{2K}{m_p}}$.
   4. $F_\alpha = 2eB\sqrt{\dfrac{2(4K)}{4m_p}} = 2eB\sqrt{\dfrac{2K}{m_p}}$. …

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9. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.A charged particle moving along a straight line path enters a uniform magnetic field of 4 mT at right angles to the direction of the magnetic field. If the specific charge of the charged particle is $8 \times 10^7\ \mathrm{C\,kg^{-1}}$, the angular velocity of the particle in the magnetic field is (A) $64 \times 10^4\ \mathrm{rad\,s^{-1}}$ (B) $32 \times 10^4\ \mathrm{rad\,s^{-1}}$ (C) $16 \times 10^4\ \mathrm{rad\,s^{-1}}$ (D) $48 \times 10^4\ \mathrm{rad\,s^{-1}}$

   ›Reveal solutionSolution

   Cyclotron angular frequency depends only on the specific charge (charge-to-mass ratio) and the field strength. Answer: $32\times10^4$ rad/s.

   Concept and Intuition

   When a charged particle moves perpendicular to a uniform magnetic field, it moves in a circle with angular velocity $\omega = \dfrac{qB}{m}$. This is independent of the particle's speed or the radius of its path — it depends only on the specific charge $q/m$ and the field $B$.

   Step-by-Step Solution

   1. Specific charge: $q/m = 8\times10^7\ \mathrm{C\,kg^{-1}}$.
   2. Field: $B = 4\ \mathrm{mT} = 4\times10^{-3}\ \mathrm{T}$. …

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10. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.When an electron placed in a uniform magnetic field is accelerated from rest through a potential difference $V_1$, it experiences a force F. If the potential difference is changed to $V_2$, the force experienced by the electron in same magnetic field is 2F, then the ratio of potential differences $V_2/V_1$ is (A) 2:1 (B) 1:4 (C) 4:1 (D) 1:2

    ›Reveal solutionSolution

    Speed after acceleration scales as $\sqrt V$, and magnetic force scales linearly with speed, so force scales as $\sqrt V$; doubling the force means quadrupling the potential difference.

    Concept and Intuition

    Accelerating the electron through a potential difference converts electrical PE into KE: $qV = \tfrac12 mv^2$, so $v=\sqrt{2qV/m}$ — speed grows only as the square root of the accelerating voltage. Once moving in the magnetic field, the magnetic (Lorentz) force is $F=qvB$, which is directly proportional to speed $v$ (with $q$, $B$ fixed). Chaining these two proportionalities, $F\propto\sqrt V$.

    Step-by-Step Solution

    1. From energy: $v = \sqrt{\dfrac{2qV}{m}} \Rightarrow v\propto\sqrt V$.
    2. Magnetic force: $F = qvB \propto v \propto \sqrt V$.
    3. Given $F_2 = 2F_1$: $\dfrac{F_2}{F_1}=\sqrt{\dfrac{V_2}{V_1}}=2$. …

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11. AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
    Q.An electron moving along positive x-direction with velocity $2 \times 10^5\ \mathrm{ms^{-1}}$, enters a magnetic field of $\vec{B}=(\hat{i}+4\hat{j}+3\hat{k})$ T. The magnitude of the force acting on the electron is (A) $1.6 \times 10^{13}$ N (B) $1.6 \times 10^{-13}$ N (C) $1.6 \times 10^{-14}$ N (D) $1.6 \times 10^{-3}$ N

    ›Reveal solutionSolution

    Computing $\vec v \times \vec B$ for the given vectors gives a magnitude of $5v$; multiplying by the electron charge yields a magnetic force of $1.6\times10^{-13}$ N.

    Concept and Intuition

    A charged particle moving through a magnetic field experiences the Lorentz magnetic force $\vec F = q\vec v \times \vec B$, whose magnitude depends only on the component of velocity perpendicular to $\vec B$. Direct vector cross-product computation is the cleanest way to handle a $\vec B$ with multiple non-zero components.

    Step-by-Step Solution

    1. Write $\vec v = 2\times10^5\,\hat i$ m/s (only x-component) and $\vec B = \hat i + 4\hat j + 3\hat k$ T.
    2. Compute $\vec v\times\vec B = \begin{vmatrix}\hat i&\hat j&\hat k\\ v&0&0\\1&4&3\end{vmatrix} = \hat i(0\cdot3-0\cdot4)-\hat j(v\cdot3-0\cdot1)+\hat k(v\cdot4-0\cdot1) = (0,-3v,4v)$.
    3. Magnitude: $|\vec v\times\vec B| = v\sqrt{3^2+4^2} = v\times5 = 2\times10^5\times5 = 1\times10^6$ (in SI units, this is $\mathrm{m/s\cdot T}$).
    4. Force magnitude: $F = |q|\,|\vec v\times\vec B| = 1.6\times10^{-19}\times10^6 = 1.6\times10^{-13}$ N.
    5. This matches option (B).

    Common Mistakes …

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12. AP EAPCET 2023Set ap-2023-05-23-AN1 markMCQ
    Q.If the magnetic field is along the positive y-axis and the electron is moving along the positive x-axis, the direction of force on the electron is (A) along -y axis (B) along +y axis (C) along +z axis (D) along -z axis

    ›Reveal solutionSolution

    Using $\vec F = q\vec v\times\vec B$ with $v$ along $+x$, $B$ along $+y$, and the electron's negative charge, the force points along $-z$.

    Concept and Intuition

    The magnetic (Lorentz) force on a moving charge is $\vec F = q\vec v \times \vec B$. The cross product direction is found from the right-hand rule, and then the sign of the charge flips the direction if the charge is negative (as for an electron).

    Step-by-Step Solution

    1. $\vec v = v\hat x$, $\vec B = B\hat y$.
    2. $\vec v \times \vec B = vB(\hat x \times \hat y) = vB\hat z$. …

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