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 16 on this concept.

1. KCET 2026Set C21 markMCQ
   Q.A proton, an electron and an $\alpha$-particle enter at right angles to a uniform magnetic field with the same velocity. If $R_p$, $R_e$ and $R_\alpha$ are the radii of circular paths of these particles, then (A) $R_\alpha = R_p = R_e$ (B) $R_\alpha > R_p > R_e$ (C) $R_\alpha < R_p < R_e$ (D) $R_\alpha > R_p = R_e$

   ›Reveal solutionSolution

   For a charged particle moving perpendicular to a magnetic field, the radius of its circular path is $R = \dfrac{mv}{qB}$. Compute $R$ for each particle using the same $v$ and $B$, and compare.

   Step 1 — Radius for the proton

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

   Step 2 — Radius for the electron

   The electron has a much smaller mass than the proton ($m_e \approx m_p/1836$) but the same magnitude of charge $e$:

   $R_e = \frac{m_e v}{eB} \ll R_p$

   Step 3 — Radius for the $\alpha$-particle …

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2. KCET 2025Set D-41 markMCQ
   Q.A metallic sphere of radius R carrying a charge q is kept at certain distance from another metallic sphere of radius R/4 carrying a charge Q. What is the electric flux at any point inside the metallic sphere of radius R due to the sphere of radius R/4 ? (A) $\dfrac{Q}{\varepsilon_0} - \dfrac{q}{\varepsilon_0}$ (B) Zero (C) $\dfrac{q}{\varepsilon_0} - \dfrac{Q}{\varepsilon_0}$ (D) $\dfrac{Q}{\varepsilon_0}$

   ›Reveal solutionSolution

   Gauss's law: only charge enclosed by a surface contributes net flux, and $Q$ lies outside the sphere of radius $R$ — so its net flux through/inside that sphere is zero.

   Step 1 — The geometry.

   The figure shows two separate, non-touching metallic spheres: one of radius $R$ carrying charge $q$ on its surface, and one of radius $R/4$ carrying charge $Q$ on its surface. Neither encloses the other. We are asked for the flux inside the sphere of radius $R$, due only to the sphere of radius $R/4$.

   Step 2 — Apply Gauss's law.

   Gauss's law states that for any closed surface $S$,

   $\oint_S \vec{E}\cdot d\vec{A} = \frac{q_{\text{enclosed}}}{\varepsilon_0}$

   The crucial word is enclosed. Take any closed (Gaussian) surface lying inside the sphere of radius $R$. The charge $Q$ is on the other, distant sphere — it is not inside that Gaussian surface. Therefore the charge of the small sphere enclosed by it is zero:

   $q_{\text{enclosed, due to }Q} = 0 \qquad \Rightarrow \qquad \phi_Q = \frac{0}{\varepsilon_0} = 0$

   Step 3 — Why an external charge always gives zero net flux. …

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3. KCET 2025Set D-41 markMCQ
   Q.If the radius of first Bohr orbit is $r$, then the radius of the second Bohr orbit will be (A) $8r$ (B) $4r$ (C) $2\sqrt{2}r$ (D) $2r$

   ›Reveal solutionSolution

   Bohr's quantisation gives $r_n \propto n^2$, so going from $n=1$ to $n=2$ multiplies the radius by $2^2 = 4$.

   Step 1 — Derive the radius from Bohr's postulates.

   Bohr assumed that (i) the Coulomb attraction supplies the centripetal force, and (ii) angular momentum is quantised in units of $\hbar$.

   Force balance:

   $\frac{1}{4\pi\varepsilon_0}\frac{e^2}{r_n^2} = \frac{m v_n^2}{r_n} \quad\Longrightarrow\quad m v_n^2 r_n = \frac{e^2}{4\pi\varepsilon_0}$

   Quantisation of angular momentum:

   $m v_n r_n = \frac{nh}{2\pi}$

   Step 2 — Eliminate $v_n$.

   From the second relation, $v_n = \dfrac{nh}{2\pi m r_n}$. Substituting into the first:

   $m\left(\frac{nh}{2\pi m r_n}\right)^2 r_n = \frac{e^2}{4\pi\varepsilon_0}$

   $\frac{n^2h^2}{4\pi^2 m r_n} = \frac{e^2}{4\pi\varepsilon_0}$

   $r_n = \frac{\varepsilon_0 n^2 h^2}{\pi m e^2}$

   Step 3 — Read off the dependence on $n$.

   Everything on the right except $n^2$ is a collection of fundamental constants, so

   $\boxed{\ r_n \propto n^2\ }$

   Numerically, for hydrogen $r_n = 0.529\,n^2\ \text{Å}$, where $0.529\ \text{Å}$ is the Bohr radius $a_0$.

   Step 4 — Apply to the two orbits. …

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4. KCET 2024Set D-21 markMCQ
   Q.Dimensional formula for activity of a radioactive substance is (A) M$^0$L$^1$T$^{-1}$ (B) M$^0$L$^{-1}$T$^0$ (C) M$^0$L$^0$T$^{-1}$ (D) M$^{-1}$L$^0$T$^0$

   ›Reveal solutionSolution

   Activity = (dimensionless count of nuclei) ÷ (time), so its dimensional formula is simply $\mathrm{M^0L^0T^{-1}}$.

   1. Definition

   The activity of a radioactive sample is the rate at which its nuclei disintegrate:

   $A = -\frac{dN}{dt} = \lambda N$

   where $N$ is the number of undecayed nuclei and $\lambda$ the decay constant.

   2. Dimensions of each factor

   • $N$ is a pure number (a count of nuclei) $\Rightarrow [N] = \mathrm{M^0L^0T^0}$.
   • $dt$ is a time $\Rightarrow [t] = \mathrm{T}$.

   3. Put them together

   $[A] = \frac{[N]}{[t]} = \frac{\mathrm{M^0L^0T^0}}{\mathrm{T}} = \mathrm{M^0L^0T^{-1}}$

   4. Consistency check with the SI unit …

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5. KCET 2023Set A-31 markMCQ
   Q.In the situation shown in the diagram, magnitude of $q \ll |Q|$ and $r \gg a$. The net force on the free charge $-q$ and net torque on it about O at the instant shown are respectively [$p = 2aQ$ is the dipole moment] (A) $\dfrac{1}{4\pi\varepsilon_0}\dfrac{pq}{r^2}\hat{k},\ \dfrac{1}{4\pi\varepsilon_0}\dfrac{pq}{r^3}\hat{i}$ (B) $-\dfrac{1}{4\pi\varepsilon_0}\dfrac{pq}{r^2}\hat{k},\ -\dfrac{1}{4\pi\varepsilon_0}\dfrac{pq}{r^3}\hat{i}$ (C) $\dfrac{1}{4\pi\varepsilon_0}\dfrac{pq}{r^3}\hat{i},\ +\dfrac{1}{4\pi\varepsilon_0}\dfrac{pq}{r^2}\hat{k}$ (D) $\dfrac{1}{4\pi\varepsilon_0}\dfrac{pq}{r^3}\hat{i},\ -\dfrac{1}{4\pi\varepsilon_0}\dfrac{pq}{r^2}\hat{k}$

   ›Reveal solutionSolution

   Use the equatorial (broadside) dipole field $E = \dfrac{p}{4\pi\varepsilon_0 r^3}$ directed anti-parallel to $\vec p$, get $\vec F = (-q)\vec E$, then take the cross product $\vec\tau = \vec r\times\vec F$ about $O$.

   1. Set up axes and the dipole moment

   Let $\hat i$ point along the dipole axis from $A(-Q)$ towards $B(+Q)$, $\hat j$ point vertically up from $O$ towards $P$, and $\hat k = \hat i \times \hat j$ out of the page.

   The dipole moment always points from the negative to the positive charge, so

   $\vec p = 2aQ\,\hat i = p\,\hat i$

   The free charge $-q$ sits at $P$, on the perpendicular bisector: $\vec r_{P} = r\,\hat j$, with $r \gg a$ so the point-dipole (far-field) formulae apply. Also $q \ll |Q|$, so $-q$ does not disturb the dipole.

   2. Field at an equatorial point

   For a short dipole, at a point on the equatorial line,

   $\vec E = -\frac{1}{4\pi\varepsilon_0}\frac{\vec p}{r^{3}} = -\frac{1}{4\pi\varepsilon_0}\frac{p}{r^{3}}\,\hat i$

   The minus sign is the physics: on the broadside position the field is anti-parallel to $\vec p$ (the components along the axis from the two charges cancel, and the components anti-parallel to $\vec p$ add). Note the $1/r^3$ dependence — this alone kills options (A) and (B), which quote a $1/r^2$ force.

   3. Force on the free charge

   $\vec F = (-q)\,\vec E = (-q)\left(-\frac{1}{4\pi\varepsilon_0}\frac{p}{r^{3}}\hat i\right) = +\frac{1}{4\pi\varepsilon_0}\frac{pq}{r^{3}}\,\hat i$

   So the force on $-q$ is along $+\hat i$, of magnitude $\dfrac{pq}{4\pi\varepsilon_0 r^{3}}$. (Sensible: the negative charge is pulled towards the $+Q$ end.)

   4. Torque about $O$ …

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6. KCET 2022Set B-31 markMCQ
   Q.A charged particle of mass ‘$m$’ and charge ‘$q$’ is released from rest in an uniform electric field $\vec{E}$. Neglecting the effect of gravity, the kinetic energy of the charged particle after ‘$t$’ seconds is (A) $\frac{Eqm}{t}$ (B) $\frac{E^2 q^2 t^2}{2m}$ (C) $\frac{2E^2 t^2}{mq}$ (D) $\frac{E^2 q^2 m}{2t^2}$

   ›Reveal solutionSolution

   The electric force is constant, so the motion is uniformly accelerated: find $a = qE/m$, then $v = at$, then $KE = \tfrac12 mv^2$.

   Step 1 — Identify the force and the acceleration.

   The only force acting (gravity is to be neglected) is the electric force on the charge:

   $F = qE$

   Since $\vec{E}$ is uniform, $F$ is constant, and therefore so is the acceleration. By Newton's second law:

   $a = \frac{F}{m} = \frac{qE}{m}$

   A constant acceleration means the standard kinematic equations apply directly.

   Step 2 — Find the velocity after time $t$.

   The particle is released from rest, so $u = 0$. Using $v = u + at$:

   $v = 0 + \left(\frac{qE}{m}\right)t = \frac{qEt}{m}$

   Step 3 — Compute the kinetic energy.

   $KE = \frac{1}{2}mv^2 = \frac{1}{2}\,m\left(\frac{qEt}{m}\right)^{2}$

   $KE = \frac{1}{2}\,m \cdot \frac{q^2E^2t^2}{m^2}$

   One power of $m$ cancels:

   $\boxed{KE = \frac{E^2q^2t^2}{2m}}$

   Step 4 — Cross-check by the work–energy theorem.

   An independent route must give the same result. The work done by the constant force over the distance travelled $s = \tfrac12 at^2$ is:

   $W = F \cdot s = (qE)\left(\frac{1}{2}\cdot\frac{qE}{m}\,t^2\right) = \frac{q^2E^2t^2}{2m}$

   By the work–energy theorem, $KE = W$ (since it started from rest) — identical ✓.

   Step 5 — Dimensional check on the options. …

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7. KCET 2022Set B-31 markMCQ
   Q.A magnetic field of flux density 1.0 Wb m$^{-2}$ acts normal to a 80 turn coil of 0.01m$^2$ area. If this coil is removed from the field in 0.2 second, the emf induced in it is (A) 0.8V (B) 5V (C) 4V (D) 8V

   ›Reveal solutionSolution

   Apply Faraday's law of electromagnetic induction: the induced emf is $N$ times the rate of change of magnetic flux through one turn.

   Step 1 — The law.

   Faraday's law says an emf appears whenever the flux linked with a coil changes:

   $\varepsilon = -N\frac{d\Phi}{dt}\qquad\Longrightarrow\qquad |\varepsilon| = N\frac{\Delta\Phi}{\Delta t}\ \ \text{(for a uniform change)}$

   The minus sign is Lenz's law — the induced current opposes the change that produced it — and it fixes only the direction, not the magnitude asked for here. The factor $N$ appears because the $N$ turns are in series, so their emfs add.

   Step 2 — Initial flux through one turn.

   The field is normal to the coil ($\theta=0$, $\cos\theta=1$), so

   $\Phi_i = B A\cos\theta = (1.0\ \mathrm{Wb\,m^{-2}})(0.01\ \mathrm{m^{2}})(1) = 0.01\ \mathrm{Wb}.$

   (Note $1\ \mathrm{Wb\,m^{-2}} = 1\ \mathrm{T}$.)

   Step 3 — Final flux.

   The coil is removed from the field, so

   $\Phi_f = 0\ \mathrm{Wb},\qquad |\Delta\Phi| = 0.01 - 0 = 0.01\ \mathrm{Wb}.$

   Step 4 — Substitute. …

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8. KCET 2022Set B-31 markMCQ
   Q.An alternating current is given by $i=i_1 \sin \omega t + i_2 \cos \omega t$. The r.m.s current is given by (A) $\sqrt{\dfrac{i_1^2 + i_2^2}{2}}$ (B) $\sqrt{\dfrac{i_1^2 + i_2^2}{2}}$ (C) $\dfrac{i_1 + i_2}{\sqrt{2}}$ (D) $\dfrac{i_1 - i_2}{\sqrt{2}}$

   ›Reveal solutionSolution

   For i = i1sin(wt) + i2cos(wt), the mean square current over a full cycle is <i^2> = <i1^2 sin^2(wt)> + 2i1i2*<sin(wt)cos(wt)> + <i2^2 cos^2(wt)> = i1^2*(1/2) + 0 + i2^2*(1/2) =…

   For i = i1sin(wt) + i2cos(wt), the mean square current over a full cycle is <i^2> = <i1^2 sin^2(wt)> + 2i1i2*<sin(wt)cos(wt)> + <i2^2 cos^2(wt)> = i1^2*(1/2) + 0 + i2^2*(1/2) = (i1^2+i2^2)/2 (using <sin^2>=<cos^2>=1/2 and <sin*cos>=0 over a full cycle). So i_rms = sqrt((i1^2+i2^2)/2). Note: options (A) and (B) a …

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9. KCET 2022Set B-31 markMCQ
   Q.The de-Broglie wavelength of a particle of kinetic energy ‘K’ is $\lambda$ ; the wavelength of the particle, if its kinetic energy is $K/4$ is (A) $2\lambda$ (B) $4\lambda$ (C) $\lambda/2$ (D) $\lambda/4$

   ›Reveal solutionSolution

   Since $\lambda \propto 1/\sqrt{K}$, quartering the kinetic energy doubles the wavelength.

   Step 1 — De Broglie relation in terms of kinetic energy.

   $\lambda = \dfrac{h}{p} = \dfrac{h}{\sqrt{2mK}}$

   Step 2 — Apply the new kinetic energy $K/4$. …

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10. KCET 2022Set B-31 markMCQ
    Q.In a photo electric experiment, if both the intensity and frequency of the incident light are doubled, then the saturation photo electric current (A) Is doubled (B) Becomes four times (C) Remains constant (D) Is halved

    ›Reveal solutionSolution

    Saturation current tracks the photon arrival rate (intensity), never the photon energy (frequency) — so doubling intensity alone doubles the current, regardless of what frequency does.

    Why frequency doesn't matter for saturation current. Above the threshold frequency, every absorbed photon ejects one electron. The RATE of electron ejection is set by how many photons arrive per second — that's what intensity measures. Frequency only sets each photoelectron's kinetic energy (via Einstein's equation), not how many …

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11. KCET 2021Set B-21 markMCQ
    Q.The maximum range of a gun on horizontal plane is $16~\mathrm{km}$. If $g = 10~\mathrm{m~s^{-2}}$, then muzzle velocity of a shell is (A) $160~\mathrm{m~s^{-1}}$ (B) $200\sqrt{2}~\mathrm{m~s^{-1}}$ (C) $400~\mathrm{m~s^{-1}}$ (D) $800~\mathrm{m~s^{-1}}$

    ›Reveal solutionSolution

    Use $R_{\max} = u^2/g$ (the range at the optimum launch angle of $45^\circ$) and solve for the muzzle speed $u$.

    Step 1 — The range formula and why $45^\circ$ maximises it.

    For projectile launched at speed $u$ and angle $\theta$ over level ground,

    $R(\theta) = \frac{u^2 \sin 2\theta}{g}.$

    Since $\sin 2\theta \le 1$ with equality at $2\theta = 90^\circ$, i.e. $\theta = 45^\circ$, the maximum range is

    $R_{\max} = \frac{u^2}{g}.$

    The gun's "maximum range" is exactly this quantity — it is the reach at the best possible elevation.

    Step 2 — Convert units.

    $R_{\max} = 16\ \mathrm{km} = 16000\ \mathrm{m}.$

    Step 3 — Solve for $u$. …

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12. KCET 2019Set A-11 markMCQ
    Q.An electron is moving with an initial velocity $\vec{v} = V_0\hat{i}$ and is in a uniform magnetic field $\vec{B} = B_0\hat{j}$. Then its de Broglie wavelength (A) remains constant (B) increases with time (C) decreases with time (D) increase and decreases periodically

    ›Reveal solutionSolution

    The electron moves in a circular path perpendicular to the magnetic field, so its speed stays constant. Since de Broglie wavelength depends only on speed, it remains constant.

    The key idea here is that de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$, where $p$ is the magnitude of the momentum. For an electron, $p = mv$ (non-relativistic), so $\lambda = \frac{h}{mv}$. The question reduces to: does the speed $v$ of the electron change in a uniform magnetic field?

    A magnetic field exerts a force $\vec{F} = q(\vec{v} \times \vec{B})$ on a moving charge. This force is always perpendicular to the velocity. A force perpendicular to velocity does no work — it changes the direction of motion but not the speed. So $v$ remains constant in magnitude.

    Let’s trace the motion step by step.

    1. Initial conditions: The electron has velocity $\vec{v} = V_0 \hat{i}$ (along the x-axis) and enters a uniform magnetic field $\vec{B} = B_0 \hat{j}$ (along the y-axis). The charge of an electron is $q = -e$.

    2. Force on the electron:

       $\vec{F} = q(\vec{v} \times \vec{B}) = -e (V_0 \hat{i} \times B_0 \hat{j}) = -e V_0 B_0 (\hat{i} \times \hat{j}) = -e V_0 B_0 \hat{k}$

       So the force is along the negative z-direction (into the page, if we set axes conventionally). This force is perpendicular to the velocity (which is along $\hat{i}$).

    3. Nature of motion: Since the force is always perpendicular to velocity and constant in magnitude, the electron undergoes uniform circular motion in the x-z plane. The magnetic field direction (y-axis) is perpendicular to the plane of motion. The speed $v = V_0$ never changes — only the direction of $\vec{v}$ changes. …

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