Q.Verify that the cyclotron frequency $\omega = eB/m$ has the correct dimensions of $[T]^{-1}$.

Showing the 12 most recent of 13 on this concept.

1. COMEDK 2026Set 2026-M1 markMCQ
   Q.A proton is moving along negative $X$ axis. If a uniform magnetic field is applied parallel to the positive $Z$ axis, then choose the correct answer from the following options; The proton will experience magnetic force along positive Y direction and will move along a circular path in the XY plane The proton will experience magnetic force along negative Y direction and will move along a circular path in the XY plane The proton will experience magnetic force along positive Y direction and will move along a circular path in the XZ plane The proton will experience magnetic force along negative Y direction and will move along a circular path in the XZ plane (A) 4 (B) 2 (C) 1 (D) 3

   ›Reveal solutionSolution

   Using the right-hand rule for the Lorentz force, a proton moving along the negative X-axis in a magnetic field along the positive Z-axis experiences a force along the positive Y-axis, causing circular motion in the XY-plane. The correct option is (C).

   Concept & Intuition

   The magnetic force on a moving charge is given by $\vec{F} = q (\vec{v} \times \vec{B})$. This force is always perpendicular to both velocity and magnetic field, so it does no work but changes the direction of motion. For a uniform field, if the velocity is perpendicular to the field, the particle moves in a circle in the plane perpendicular to the field. Here, the velocity is along the negative X-axis and the field is along the positive Z-axis, so the force will lie in the XY-plane, and the circular path will also be in that plane.

   Step-by-step reasoning

   1. Identify the directions

      • Proton’s velocity: along negative X-axis → $\vec{v} = -v\,\hat{i}$
      • Magnetic field: along positive Z-axis → $\vec{B} = B\,\hat{k}$
      • Charge of proton: $q = +e$

   2. Compute the magnetic force using the cross product

      $\vec{F} = q (\vec{v} \times \vec{B}) = e [(-v\,\hat{i}) \times (B\,\hat{k})]$

      Recall $\hat{i} \times \hat{k} = -\hat{j}$, so $(-v\,\hat{i}) \times (B\,\hat{k}) = -vB (\hat{i} \times \hat{k}) = -vB (-\hat{j}) = vB\,\hat{j}$

      Thus $\vec{F} = e v B\,\hat{j}$ → force is along positive Y direction.

   3. Determine the plane of motion

      The force is perpendicular to velocity (since $\vec{F} \perp \vec{v}$) and lies in the XY-plane. The magnetic field is along Z, so the plane perpendicular to $\vec{B}$ is the XY-plane. The proton will therefore move in a circular path in the XY-plane.

   4. Match with the options …

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2. COMEDK 2025Set 2025-A1 markMCQ
   Q.In to a uniform transverse magnetic field, two charged particles having the same mass and charge enter and move in two different circular paths. The ratio of the radii of curvatures of the circular paths is $1: 4$. What would be the ratio of their respective velocities? (A) $\frac{r_1}{r_2}=4: 1$ (B) $\frac{r_1}{r_2}=2: 1$ (C) $\frac{r_1}{r_2}=1: 2$ (D) $\frac{r_1}{r_2}=1: 4$

   ›Reveal solutionSolution

   In a uniform magnetic field, the radius of a charged particle’s circular path is proportional to its velocity (for fixed mass and charge). Given a radius ratio of 1:4, the velocity ratio is also 1:4, so the correct option is (D).

   The key concept here is the magnetic force as a centripetal force. When a charged particle moves perpendicular to a uniform magnetic field, the magnetic force provides the necessary centripetal force to keep it in a circular path. Since the mass and charge are the same for both particles, the only variable affecting the radius is the velocity.

   1. Write the force balance. For a particle of mass $m$, charge $q$, moving with speed $v$ perpendicular to a uniform magnetic field $B$, the magnetic force is $F_B = qvB$. This force equals the centripetal force required for circular motion:

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

   2. Solve for the radius. Cancel one factor of $v$ (assuming $v \neq 0$):

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

   This shows that for fixed $m$, $q$, and $B$, the radius is directly proportional to the velocity: $r \propto v$.

   3. Apply the given ratio. The problem states the ratio of radii is $r_1 : r_2 = 1 : 4$. Since $r \propto v$, the velocities must be in the same proportion:

   $v_1 : v_2 = 1 : 4$

   Therefore, $\frac{v_1}{v_2} = \frac{1}{4}$.

   4. Match with the options. …

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3. COMEDK 2025Set 2025-M1 markMCQ
   Q.Two very long, straight, parallel wires carry steady currents I and 21 respectively. The distance between the wires is d . At a certain instant of time, a point charge $q$ is at a point equidistant from the two wires, in the plane of the wires. Its instantaneous velocity is v perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is (A) $\frac{\mu_o I q v}{\pi d}$ (B) zero (C) $\frac{2 \mu_o I q v}{\pi d}$ (D) $\frac{\mu_o I q v}{2 \pi d}$

   ›Reveal solutionSolution

   At the midpoint the magnetic field from each wire is directed perpendicular to the plane of the wires — the same direction as the charge's velocity — so $\vec v\parallel\vec B$ and $\vec F=q\,\vec v\times\vec B=0$. The correct option is (B).

   The force on a moving charge is $\vec F=q\,\vec v\times\vec B$, whose magnitude is $qvB\sin\theta$. It vanishes when $\vec v$ is parallel (or antiparallel) to $\vec B$, whatever the field's magnitude.

   1. Set up geometry. Let the wires run along the $z$-axis at $x=\pm\tfrac{d}{2}$, so the plane of the wires is the $xz$-plane. The charge sits at the midpoint (origin) and moves with velocity $\vec v$ perpendicular to this plane, i.e. along $\hat y$. …

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4. COMEDK 2025Set 2025-M1 markMCQ
   Q.A charged particle is released from rest in a region of space in which steady and uniform electric and magnetic fields are parallel to each other. The particle will move in a (A) Helix (B) Straight line (C) Cycloid (D) Circle

   ›Reveal solutionSolution

   When a charged particle is released from rest in parallel, uniform electric and magnetic fields, the initial velocity is zero, so the magnetic force is zero, and the particle accelerates along the field direction — it moves in a straight line.

   The key insight here is that the magnetic force on a charged particle depends on its velocity relative to the magnetic field. If the particle starts from rest, there is no magnetic force at the very beginning. The electric field, however, immediately exerts a force along its own direction. Since the fields are parallel, the particle will simply accelerate along that common direction, never acquiring a velocity component perpendicular to the fields — so no helical or cycloidal motion arises.

   Let’s walk through the reasoning step by step.

   1. Forces on the particle A particle of charge $q$ and mass $m$ in electric field $\mathbf{E}$ and magnetic field $\mathbf{B}$ experiences the Lorentz force:

   $\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})$

   Here, $\mathbf{E}$ and $\mathbf{B}$ are steady, uniform, and parallel. Let’s take the common direction as the $z$-axis: $\mathbf{E} = E\,\hat{\mathbf{z}}$, $\mathbf{B} = B\,\hat{\mathbf{z}}$.

   2. Initial condition: released from rest At $t=0$, the velocity $\mathbf{v} = 0$. Therefore, the magnetic force $q(\mathbf{v} \times \mathbf{B}) = 0$. The only force is the electric force:

   $\mathbf{F} = qE\,\hat{\mathbf{z}}$

   So the particle begins to accelerate purely along the $z$-direction.

   3. What happens next? As the particle gains velocity, it now has $\mathbf{v} = v_z\,\hat{\mathbf{z}}$. The cross product $\mathbf{v} \times \mathbf{B}$ becomes: $v_z\,\hat{\mathbf{z}} \times B\,\hat{\mathbf{z}} = 0$ …

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5. KCET 2024Set D-21 markMCQ
   Q.The magnetic field at the centre of a circular coil of radius $R$ carrying current $I$ is 64 times the magnetic field at a distance $x$ on its axis from the centre of the coil. Then the value of $x$ is (A) $\dfrac{R}{4\sqrt{15}}$ (B) $R\sqrt{3}$ (C) $\dfrac{R}{4}$ (D) $R\sqrt{15}$

   ›Reveal solutionSolution

   Take the ratio of the on-axis field to the centre field; the $64$ becomes a perfect cube, giving $\sqrt{R^2+x^2} = 4R$.

   Step 1 — The two standard results (from Biot–Savart)

   For a circular coil of radius $R$ carrying current $I$:

   At the centre:

   $B_{\text{centre}} = \frac{\mu_0 I}{2R}$

   On the axis, at distance $x$ from the centre:

   $B_{\text{axis}} = \frac{\mu_0 I R^2}{2\left(R^2 + x^2\right)^{3/2}}$

   (Note the centre result is just the $x = 0$ case of the axial formula — a useful check.)

   Step 2 — Form the ratio

   The condition given is $B_{\text{centre}} = 64\,B_{\text{axis}}$:

   $\frac{B_{\text{centre}}}{B_{\text{axis}}} = \frac{\dfrac{\mu_0 I}{2R}}{\dfrac{\mu_0 I R^2}{2\left(R^2+x^2\right)^{3/2}}} = \frac{\mu_0 I}{2R}\times\frac{2\left(R^2+x^2\right)^{3/2}}{\mu_0 I R^2}$

   Everything ($\mu_0$, $I$, the 2's) cancels:

   $\frac{B_{\text{centre}}}{B_{\text{axis}}} = \frac{\left(R^2+x^2\right)^{3/2}}{R^3} = 64$

   Step 3 — Solve for $x$

   Write $R^3 = \left(R^2\right)^{3/2}$, so the whole left side is a $3/2$ power:

   $\left(\frac{R^2+x^2}{R^2}\right)^{3/2} = 64$ …

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6. COMEDK 2024Set 2024-A1 markMCQ
   Q.A charge of $+1 \mathrm{C}$ is moving with velocity $\vec{V}=(2 \mathrm{i}+2 \mathrm{j}-\mathrm{k}) \mathrm{ms}^{-1}$ through a region in which electric field $\vec{E}=(\mathrm{i}+\mathrm{j}-3 \mathrm{k}) \quad \mathrm{NC}^{-1}$ and magnetic field $\vec{B}=(\mathrm{i}-2 \mathrm{j}+3 \mathrm{k}) \mathrm{T}$ are present. The force experienced by the charge is (A) $(5 i+6 j-7 k) N$ (B) $(5 i-6 j-9 k) N$ (C) $(-5 i-6 j-7 k) N$ (D) $(-5 i+6 j+9 k) N$

   ›Reveal solutionSolution

   Lorentz force $\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$ gives $\vec{F}=(5\hat{i}-6\hat{j}-9\hat{k})\,\text{N}$ — option (B).

   The charge moves through combined electric and magnetic fields, so it feels the Lorentz force

   $\vec{F}=q\left(\vec{E}+\vec{v}\times\vec{B}\right).$

   Given: $q=+1\,\text{C}$, $\vec{v}=2\hat{i}+2\hat{j}-\hat{k}$, $\vec{E}=\hat{i}+\hat{j}-3\hat{k}$, $\vec{B}=\hat{i}-2\hat{j}+3\hat{k}$.

   Step 1 - the magnetic term $\vec{v}\times\vec{B}$:

   $\vec{v}\times\vec{B}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&2&-1\\1&-2&3\end{vmatrix}$

   • $\hat{i}:\ (2)(3)-(-1)(-2)=6-2=4$
   • $\hat{j}:\ -\big[(2)(3)-(-1)(1)\big]=-(6+1)=-7$
   • $\hat{k}:\ (2)(-2)-(2)(1)=-4-2=-6$ …

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7. COMEDK 2024Set 2024-E1 markMCQ
   Q.A negative charge particle is moving upward in a magnetic field which is towards north. The particle is deflected towards (A) North (B) South (C) East (D) West

   ›Reveal solutionSolution

   Using $\vec F = q(\vec v\times\vec B)$ with $\vec v$ upward and $\vec B$ toward north: for a positive charge the force would point west, and since the charge is negative the force reverses to point east — option (C).

   Concept & Intuition

   The Lorentz force on a moving charge is $\vec F = q(\vec v\times\vec B)$. Its direction depends on both the cross product $\vec v\times\vec B$ and the sign of the charge.

   Step-by-step reasoning

   1. Set up a right-handed coordinate system.

      Using East–North–Up (ENU): $\hat x=\text{East}$, $\hat y=\text{North}$, $\hat z=\text{Up}$, with $\hat x\times\hat y=\hat z$ (a standard right-handed system).

   2. Compute $\vec v\times\vec B$.

      $\vec v = \hat z$ (up), $\vec B=\hat y$ (north).

   $\hat z\times\hat y = -(\hat y\times\hat z) = -\hat x = \text{West}$

   3. Apply the charge's sign. …

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8. COMEDK 2023Set 2023-E1 markMCQ
   Q.A magnetic field does not interact with: (A) An electric charge at rest. (B) A moving electric charge (C) A current carrying straight conductor (D) A moving permanent magnet

   ›Reveal solutionSolution

   So the magnetic field does not interact with an electric charge at rest.

   Concept: the magnetic Lorentz force is F = q (v x B). It is proportional to the velocity of the charge.

   • A charge AT REST has v = 0, so F = q(0 x B) = 0. A magnetic field exerts no force on a stationary charge. (This is the defining difference between the electric and magnetic parts of the Lorentz force.)
   • A moving charge experiences q v B sin(theta) - there is interaction.
   • A current-carrying conductor is moving charge in bulk; F = B I L sin(theta) - there is interaction. …

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9. COMEDK 2023Set 2023-M1 markMCQ
   Q.Two very long straight parallel wires carry currents $i$ and $2 i$ in opposite directions. The distance between the wires is $r$. At a certain instant of time a point charge $q$ is at. a point equidistant from the two wires in the plane of the wires. Its instantaneous velocity $v$ is perpendicular to this plane. The magnitude of the force due to the magnetic field acting on the charge at this instant is (A) zero (B) $\frac{3 \mu_0}{2 \pi} \frac{i q v}{r}$ (C) $\frac{\mu_0}{\pi} \frac{i q v}{r}$ (D) $\frac{\mu_0}{2 \pi} \frac{i q v}{r}$

   ›Reveal solutionSolution

   Both wires lie in a plane; at any point in that plane the net magnetic field is perpendicular to the plane. The charge moves perpendicular to the plane too, so its velocity is parallel to $\vec B$ and the magnetic force $q\,\vec v\times\vec B$ vanishes.

   The two parallel wires define a plane. At the mid-point $P$ (in that plane) each wire produces a field directed perpendicular to the plane (into or out of it), because $P$ lies in the plane of the currents. Even after adding the two contributions, the resultant $\vec B$ is still perpendicular to the plane. …

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10. COMEDK 2022Set 20221 markMCQ
    Q.An electric current $I$ enters and leaves a uniform circular wire of radius $r$ through diametrically opposite points. A charged particle $q$ moves along the axis of circular wire passes through its centre with speed $v$. The magnetic force on the particle when it passes through the centre has a magnitude (A) $\frac{q v \mu_0 I}{2 \pi r}$ (B) $q v \frac{\mu_0 I}{\pi r}$ (C) $\frac{q v \mu_0 I}{r}$ (D) $0$

    ›Reveal solutionSolution

    (Also note: even if B were non-zero it would be along the axis, parallel to v, so v × B = 0 anyway.)

    Concept: Magnetic field at the centre of a circular loop fed at two diametrically opposite points.

    The current entering divides into the two semicircular halves. Both halves have the same length and hence the same resistance, so each carries I/2 — but they carry it in opposite senses around the circle (one clockwise, one anticlockwise, both going from the entry point to the exit point).

    Field at the centre from a semicircle of current i: B = μ₀i/(4r). The two contributions are equal in magnitude and opposite in direction, so they cancel exactly:

    B_centre = μ₀(I/2)/(4r) − μ₀(I/2)/(4r) = 0. …

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11. KCET 2021Set B-21 markMCQ
    Q.A tightly wound long solenoid has ‘n’ turns per unit length, a radius ‘r’ and carries a current I. A particle having charge ‘q’ and mass ‘m’ is projected from a point on the axis in a direction perpendicular to the axis. The maximum speed of the particle for which the particle does not strike the solenoid is (A) $\dfrac{\mu_0 n I q r}{m}$ (B) $\dfrac{\mu_0 n I q r}{2m}$ (C) $\dfrac{\mu_0 n I q r}{4m}$ (D) $\dfrac{\mu_0 n I q r}{8m}$

    ›Reveal solutionSolution

    The particle circles with radius $mv/(qB)$ starting on the axis, so it reaches a maximum distance of one diameter from the axis; setting that diameter equal to the solenoid radius gives $v_{\max}=\mu_0 nIqr/(2m)$.

    Step 1 — Field inside a long solenoid.

    For a tightly wound long solenoid with $n$ turns per unit length,

    $B=\mu_0 n I$

    directed along the axis, and it is uniform over the cross-section.

    Step 2 — Motion of the charge.

    The particle is launched from a point on the axis with velocity $v$ perpendicular to $\vec B$. A charge moving perpendicular to a uniform field executes a circle in the plane perpendicular to $\vec B$ (the plane containing the initial velocity), with radius

    $R_c=\frac{mv}{qB}=\frac{mv}{q\,\mu_0 n I}.$

    Step 3 — Geometry: how far from the axis does it get?

    The circular orbit passes through the launch point (which lies on the axis). The point of the orbit farthest from the launch point is the diametrically opposite point, at distance $2R_c$. Hence the maximum distance the particle ever gets from the axis is

    $d_{\max}=2R_c.$ …

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12. COMEDK 2021Set 20211 markMCQ
    Q.Two particles of masses $m_1=m,m_2=2m$ and charges $q_1=q,q_2=2q$ entered into uniform magnetic field. Find $F_1/F_2$ (force ratio). (A) $\frac{1}{2}$ (B) $1$ (C) $\frac{1}{3}$ (D) $2$

    ›Reveal solutionSolution

    The masses m and 2m only affect the radius (r = mv/qB) and period, not the instantaneous force.

    Concept: magnetic force on a moving charge, F = qvB sin(theta).

    The force does NOT depend on the mass at all - only on charge, speed and field. The two particles enter the SAME uniform field with the same speed (nothing else is stated), so …

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