Q.A horizontal straight wire $10\ \text{m}$ long extending from east to west is falling with a speed of $5.0\ \text{m s}^{-1}$, at right angles to the horizontal component of the earth's magnetic field, $0.30 \times 10^{-4}\ \text{Wb m}^{-2}$.

Showing the 12 most recent of 15 on this concept.

1. KCET 2026Set C21 markMCQ
   Q.In the figure shown, the conductor PQ of length $l$ is moved from $x = 0$ to $x = b$ and then up to $x = 2b$ with a constant velocity $\vec{v}$. A uniform magnetic field $\vec{B}$ is perpendicular to the plane of the paper and extends from $x = 0$ to $x = b$ and it is zero from $x > b$. The magnitude of emf induced in the conductor is (A) $Blx; 0 \le x < b$ (B) Zero; $0 \le x < b$ (C) $Blv; 0 \le x \le b$ (D) $Blv; b \le x < 2b$

   ›Reveal solutionSolution

   The motional emf induced in a conductor of length $l$ moving with velocity $v$ perpendicular to a uniform magnetic field $B$ is $\varepsilon = Bvl$, and it exists only while the conductor is actually within the field region.

   Step 1 — Emf while PQ is inside the field region ($0 \le x \le b$)

   While PQ moves from $x=0$ to $x=b$, it is continuously within the uniform field $B$, moving with constant velocity $v$ perpendicular to $B$. The motional emf is

   $\varepsilon = Bvl$

   This value does not depend on the position $x$ within the field region — it is the same throughout, since $B$, $v$ and $l$ are all constant there.

   Step 2 — Emf beyond the field region ($x>b$) …

   🔒 Start your 14-day free trial to see the full solution →
2. KCET 2024Set D-21 markMCQ
   Q.A moving electron produces (A) only electric field (B) both electric and magnetic field (C) only magnetic field (D) neither electric nor magnetic field

   ›Reveal solutionSolution

   A charge always carries an electric field; motion of that charge is a current, which additionally creates a magnetic field.

   Step 1 — The electric field is always there

   An electron carries charge $-e$. By Coulomb's law, any charge sets up an electric field in the surrounding space:

   $E = \frac{1}{4\pi\epsilon_0}\frac{|q|}{r^2}$

   Nothing in this expression depends on whether the charge is moving. Setting the electron in motion does not remove its charge, so the electric field does not vanish — it merely becomes time-varying at a fixed observation point (and, for a uniformly moving charge, gets distorted relative to the static Coulomb field). So options (C) and (D) are ruled out immediately.

   Step 2 — Motion adds a magnetic field

   A charge $q$ moving with velocity $\vec v$ is, by definition, an element of current. The Biot–Savart law for a point charge gives the magnetic field it produces:

   $\vec{B} = \frac{\mu_0}{4\pi}\,\frac{q\,\vec{v}\times\hat{r}}{r^2}$ …

   🔒 Start your 14-day free trial to see the full solution →
3. COMEDK 2024Set 2024-A1 markMCQ
   Q.A metallic rod of length '$a$' is rotated with an angular frequency of $0.2 \mathrm{~rads}^{-1}$ about an axis normal to the rod passing through its one end. A constant and uniform magnetic field of '$\mathrm{B}$' T parallel to the axis exists everywhere. The emf developed across the ends of the rod is (A) $\frac{\mathrm{Ba}^2}{10}$ (B) $\frac{\mathrm{Ba}^2}{5}$ (C) $\frac{\mathrm{Ba}^2}{50}$ (D) $\frac{\mathrm{Ba}^2}{2}$

   ›Reveal solutionSolution

   A rod of length $a$ spinning at $\omega=0.2\ \text{rad s}^{-1}$ about one end in field $B$ generates $\varepsilon=\tfrac12 B\omega a^2=\dfrac{Ba^2}{10}$.

   For a rod rotating about an axis through one end, perpendicular to the rod, with $B$ parallel to the axis, the motional emf is

   $\varepsilon=\frac{1}{2}B\omega a^{2}.$ …

   🔒 Start your 14-day free trial to see the full solution →
4. COMEDK 2024Set 2024-M1 markMCQ
   Q.A metallic rod of $2 \mathrm{~m}$ length is rotated with a frequency $100 \mathrm{~Hz}$ about an axis passing through the centre of the circular ring of radius $2 \mathrm{~m}$. A constant magnetic field $2 \mathrm{~T}$ is applied parallel to the axis and perpendicular to the length of the rod. The emf developed across the ends of the rod is : (A) 800 $\pi$ volt (B) 1600 $\pi$ volt (C) 1600 volt (D) 400 $\pi$ volt

   ›Reveal solutionSolution

   The rod spans centre-to-rim ($L=2\ \mathrm{m}$) and rotates about one end, so $\varepsilon=\tfrac12 B\omega L^{2}=800\pi\ \mathrm{V}$ — option (A).

   Concept

   When a rod of length $L$ rotates with angular frequency $\omega$ about an axis through one end, in a magnetic field $B$ parallel to that axis, each element $\mathrm{d}r$ at radius $r$ moves at speed $v=\omega r$ and contributes $\mathrm{d}\varepsilon=B\,v\,\mathrm{d}r=B\omega r\,\mathrm{d}r$. Integrating from $0$ to $L$:

   $\varepsilon=\int_0^L B\omega r\,\mathrm{d}r=\tfrac12 B\omega L^{2}.$

   Here the axis passes through the centre of the ring and the rod length equals the ring radius, so the rod extends from the centre (on the axis) to the rim — it is rotating about one end with $L=2\ \mathrm{m}$.

   Solution

   1. Angular frequency:

   $\omega=2\pi f=2\pi(100)=200\pi\ \mathrm{rad\,s^{-1}}.$

   2. Motional emf about one end: …

   🔒 Start your 14-day free trial to see the full solution →
5. KCET 2023Set A-31 markMCQ
   Q.A positively charged particle of mass $m$ is passed through a velocity selector. It moves horizontally rightward without deviation along the line $y = \dfrac{2mv}{qB}$ with a speed $v$. The electric field is vertically downwards and magnetic field is into the plane of the paper. Now, the electric field is switched off at $t = 0$. The angular momentum of the charged particle about origin $O$ at $t = \dfrac{\pi m}{qB}$ is (A) $\dfrac{2mE^2}{qB^3}$ (B) zero (C) $\dfrac{mE^3}{qB^2}$ (D) $\dfrac{mE^2}{qB^3}$

   ›Reveal solutionSolution

   $t=\dfrac{\pi m}{qB}$ is exactly half the cyclotron period, so the particle completes a semicircle and recrosses the horizontal axis through $O$ moving along it; there $\vec r$ and $\vec v$ are collinear, so $\vec L=m(\vec r\times\vec v)=0$ - option (B).

   Reasoning

   With the electric field switched off, only the magnetic field acts and the particle moves on a circle of radius $r=\dfrac{mv}{qB}$ with cyclotron period

   $T=\frac{2\pi m}{qB}.$

   The given instant

   $t=\frac{\pi m}{qB}=\frac{T}{2}$

   is half a period, so the particle traverses a semicircle to the diametrically opposite point of its path. …

   🔒 Start your 14-day free trial to see the full solution →
6. KCET 2023Set A-31 markMCQ
   Q.A metallic rod of length 1 m held along east-west direction is allowed to fall down freely. Given horizontal component of earth’s magnetic field $B_H = 3 \times 10^{-5}$ T. The emf induced in the rod at an instant $t = 2$ s after it is released is (Take $g = 10$ ms$^{-2}$) (A) $3 \times 10^{-3}$ V (B) $3 \times 10^{-4}$ V (C) $6 \times 10^{-3}$ V (D) $6 \times 10^{-4}$ V

   ›Reveal solutionSolution

   A falling rod cuts the horizontal component of Earth’s magnetic field, generating motional emf. The induced emf at $t = 2$ s is $6 \times 10^{-4}$ V.

   The key idea is motional emf: when a conductor moves perpendicular to a magnetic field, the free charges inside experience a magnetic force that pushes them to one end, creating a potential difference. For a rod of length $l$ moving with velocity $v$ perpendicular to a uniform field $B$, the induced emf is $\mathcal{E} = B l v$.

   Here the rod is falling freely under gravity, so its velocity increases linearly with time. The rod is oriented east-west, and the horizontal component of Earth’s field is given — that’s the component the rod cuts as it falls vertically. The vertical component of Earth’s field would be parallel to the rod’s motion and doesn’t contribute to the emf (since $v$ and $B$ would be parallel, giving zero cross product). So we only need $B_H$.

   1. Find the velocity after 2 seconds.

      The rod is released from rest and falls freely with $g = 10$ m/s$^2$.

      $v = g t = 10 \times 2 = 20$ m/s (downward).

   2. Identify the effective field and length.

      Rod length $l = 1$ m.

      Horizontal component $B_H = 3 \times 10^{-5}$ T. …

   🔒 Start your 14-day free trial to see the full solution →
7. COMEDK 2023Set 2023-E1 markMCQ
   Q.A metallic rod of $10 \mathrm{~cm}$ is rotated with a frequency 100 revolution per second about an axis perpendicular to its length and passing through its one end in uniform transverse magnetic field of strength $1 \mathrm{~T}$. The emf developed across its ends is: (A) 628 V (B) 3.14 V (C) 31.4 V (D) 6.28 V

   ›Reveal solutionSolution

   A rod rotating about one end in a field $B$ develops $\varepsilon=\tfrac12 B\omega L^2=\pi\approx3.14$ V.

   For a rod of length $L$ rotating with angular velocity $\omega$ about an axis through one end, perpendicular to a field $B$:

   $\varepsilon=\frac{1}{2}B\omega L^2.$ …

   🔒 Start your 14-day free trial to see the full solution →
8. KCET 2020Set A-11 markMCQ
   Q.A potentiometer has a uniform wire of length 5 m. A battery of emf 10 V and negligible internal resistance is connected between its ends. A secondary cell connected to the circuit gives balancing length at 200 cm. The emf of the secondary cell is (A) 4 V (B) 6 V (C) 2 V (D) 8 V

   ›Reveal solutionSolution

   The emf of the secondary cell is found by the ratio of the balancing length to the total length of the potentiometer wire, multiplied by the total voltage across the wire. The answer is 4 V.

   The key idea behind a potentiometer is that the potential drop across a uniform wire is directly proportional to its length. Since the wire has uniform cross-section and material, and a steady current flows through it, the voltage per unit length (the potential gradient) is constant. When you connect a secondary cell in the circuit, you slide the jockey until the galvanometer shows zero deflection — that balancing length tells you the emf of the cell equals the potential drop across that portion of the wire.

   So the problem reduces to: if the full 5 m wire carries 10 V, then each metre carries 2 V. At a balancing length of 200 cm (which is 2 m), the emf of the secondary cell is simply the voltage across those 2 m.

   Let’s go through it step by step.

   1. Find the potential gradient The total length of the potentiometer wire is $L = 5\ \text{m}$. The total potential difference across it is $V = 10\ \text{V}$ (the battery’s emf, since internal resistance is negligible). The potential gradient $k$ (voltage per unit length) is:

   $k = \frac{V}{L} = \frac{10\ \text{V}}{5\ \text{m}} = 2\ \text{V/m}$

   2. Interpret the balancing length The balancing length is given as $l = 200\ \text{cm} = 2\ \text{m}$. At this point, the potential drop across the wire from the start to the jockey exactly equals the emf $E$ of the secondary cell. So:

   $E = k \times l = (2\ \text{V/m}) \times (2\ \text{m}) = 4\ \text{V}$

   3. Check the logic …

   🔒 Start your 14-day free trial to see the full solution →
9. KCET 2020Set A-11 markMCQ
   Q.The current in a coil of inductance 0.2 H changes from 5 A to 2 A in 0.5 sec. The magnitude of the average induced emf in the coil is (A) 0.6 V (B) 1.2 V (C) 30 V (D) 0.3 V

   ›Reveal solutionSolution

   The average induced emf is found using Faraday’s law: $\mathcal{E} = -L \frac{\Delta I}{\Delta t}$. Substituting $L = 0.2\ \text{H}$, $\Delta I = -3\ \text{A}$, $\Delta t = 0.5\ \text{s}$ gives magnitude $1.2\ \text{V}$. The correct option is (B).

   The key idea here is self-induction. When the current through a coil changes, the magnetic flux linked with the coil itself changes, inducing an emf that opposes the change (Lenz’s law). The magnitude of this induced emf is proportional to the rate of change of current, with the constant of proportionality being the inductance $L$.

   Why does this work? Faraday’s law for a coil says induced emf $\mathcal{E} = -\frac{d\Phi}{dt}$. For a single coil, the flux $\Phi$ is proportional to the current: $\Phi = L I$, so $\frac{d\Phi}{dt} = L \frac{dI}{dt}$. Hence $\mathcal{E} = -L \frac{dI}{dt}$. For a finite time interval, the average induced emf uses the average rate of change: $\mathcal{E}_{\text{avg}} = -L \frac{\Delta I}{\Delta t}$.

   The negative sign indicates direction (opposing the change), but the question asks for magnitude, so we take the absolute value.

   Let’s work it out step by step.

   1. Identify the given data

      Inductance, $L = 0.2\ \text{H}$

      Initial current, $I_i = 5\ \text{A}$

      Final current, $I_f = 2\ \text{A}$

      Time interval, $\Delta t = 0.5\ \text{s}$

   2. Find the change in current

      $\Delta I = I_f - I_i = 2 - 5 = -3\ \text{A}$

      The negative sign means current is decreasing. For magnitude, we’ll use $|\Delta I| = 3\ \text{A}$.

   3. Apply the formula for average induced emf

      $\mathcal{E}_{\text{avg}} = -L \frac{\Delta I}{\Delta t}$

      Substitute the values:

      $\mathcal{E}_{\text{avg}} = -0.2 \times \frac{(-3)}{0.5}$

      The two negatives cancel:

      $\mathcal{E}_{\text{avg}} = 0.2 \times \frac{3}{0.5}$

   4. Simplify the arithmetic

      $\frac{3}{0.5} = 6$

      So $\mathcal{E}_{\text{avg}} = 0.2 \times 6 = 1.2\ \text{V}$ …

   🔒 Start your 14-day free trial to see the full solution →
10. KCET 2020Set A-11 markMCQ
    Q.In the given circuit the peak voltages across C, L and R are 30 V, 110 V and 60 V respectively. The rms value of the applied voltage is (A) 100 V (B) 200 V (C) 70.7 V (D) 141 V

    ›Reveal solutionSolution

    In a series LCR circuit, the applied peak voltage is the phasor sum of the individual peak voltages (not the arithmetic sum). Here, the peak applied voltage is $\sqrt{60^2 + (110 - 30)^2} = 100\ \text{V}$, so its rms value is $100 / \sqrt{2} \approx 70.7\ \text{V}$. The correct option is (C).

    The trap in this problem is to simply add the three peak voltages: $30 + 110 + 60 = 200\ \text{V}$. That would be true only if all voltages were in phase — but in a series LCR circuit, the voltages across the inductor and capacitor are $180^\circ$ out of phase with each other, and both are $90^\circ$ out of phase with the resistor voltage. So we must use phasor addition.

    1. Identify the phase relationships.

       In a series LCR circuit driven by an AC source, the current is the same through all components.

       • The voltage across the resistor, $V_R$, is in phase with the current.
       • The voltage across the inductor, $V_L$, leads the current by $90^\circ$.
       • The voltage across the capacitor, $V_C$, lags the current by $90^\circ$. Therefore, $V_L$ and $V_C$ are exactly opposite in phase ($180^\circ$ apart). Their net effect is the difference $V_L - V_C$ (or $V_C - V_L$, depending on which is larger).

    2. Find the net reactive voltage.

       Given: $V_L = 110\ \text{V}$ (peak), $V_C = 30\ \text{V}$ (peak).

       Since they oppose each other, the net reactive peak voltage is:

    $V_{\text{reactive}} = |V_L - V_C| = |110 - 30| = 80\ \text{V}$

    3. Combine with the resistive voltage using phasor addition. The resistor voltage $V_R = 60\ \text{V}$ (peak) is at $90^\circ$ to the net reactive voltage. So the total applied peak voltage $V_{\text{peak}}$ is the hypotenuse of a right triangle: $V_{\text{peak}} = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{60^2 + 80^2} = \sqrt{3600 + 6400} = \sqrt{10000} = 100\ \text{V}$ …

    🔒 Start your 14-day free trial to see the full solution →
11. KCET 2019Set A-11 markMCQ
    Q.A circular current loop of magnetic moment M is in an arbitrary orientation in an external uniform magnetic field B. The work done to rotate the loop by $30°$ about an axis perpendicular to its plane is (A) MB (B) $\sqrt{3}\frac{MB}{2}$ (C) $\frac{MB}{2}$ (D) Zero

    ›Reveal solutionSolution

    The axis of rotation is the direction of $\vec{M}$ itself, so $\vec{M}$ never turns relative to $\vec{B}$; the potential energy doesn't change, so no work is done.

    Step 1 — Potential energy of a magnetic dipole in a uniform field.

    $U(\theta) = -\vec{M}\cdot\vec{B} = -MB\cos\theta$

    where $\theta$ is the angle between the magnetic moment $\vec{M}$ and the field $\vec{B}$. The work done by an external agent in reorienting the loop is

    $W = \Delta U = U(\theta_2)-U(\theta_1) = -MB\left(\cos\theta_2-\cos\theta_1\right)$

    So work depends only on how $\theta$ changes — nothing else.

    Step 2 — Where does $\vec{M}$ point?

    For a planar current loop of area $A$ carrying current $I$,

    $\vec{M} = I A\,\hat{n}$

    where $\hat{n}$ is the unit normal to the plane of the loop (right-hand rule). $\vec{M}$ is perpendicular to the plane, i.e. along the axis of the loop.

    Step 3 — Read the rotation axis carefully (this is the whole trick).

    The loop is rotated about an axis perpendicular to its plane — that axis is precisely the direction of $\hat{n}$, i.e. of $\vec{M}$.

    Rotating a vector about its own direction does nothing to it:

    $\vec{M}_{\text{final}} = \vec{M}_{\text{initial}}$

    The circular loop simply spins in its own plane; the current distribution is symmetric about that axis, so the loop's magnetic state is completely unaltered. …

    🔒 Start your 14-day free trial to see the full solution →
12. KCET 2019Set A-11 markMCQ
    Q.Consider the situation given in figure. The wire AB is slid on the fixed rails with a constant velocity. If the wire AB is replaced by a semicircular wire, the magnitude of the induced current will (A) increase (B) remain same (C) decrease (D) increase or decrease depending on whether the semicircle bulges towards the resistance or away from it

    ›Reveal solutionSolution

    Motional emf uses the effective (straight-line, end-to-end) length between the rails, not the arc length of the wire — so bending the slider into a semicircle changes nothing.

    Step 1 — The emf as a line integral.

    For a wire moving with uniform velocity $\vec v$ in a uniform field $\vec B$, the emf is

    $\varepsilon=\int_{A}^{B}\big(\vec v\times\vec B\big)\cdot d\vec l$

    Because $\vec v$ and $\vec B$ are both constant, the vector $\vec v\times\vec B$ is a constant vector and can come outside the integral:

    $\varepsilon=\big(\vec v\times\vec B\big)\cdot\int_{A}^{B} d\vec l=\big(\vec v\times\vec B\big)\cdot\vec L_{AB}$

    where $\vec L_{AB}$ is simply the displacement vector from A to B — the straight line joining the two rail contact points. The path taken between A and B is irrelevant.

    Step 2 — Apply it to the semicircular wire.

    The wire still touches the same two rails, so A and B (and hence $\vec L_{AB}$, of magnitude $\ell$ = rail separation) are unchanged. Therefore

    $\varepsilon=B\ell v\quad\text{(unchanged)}$

    Step 3 — Cross-check with Faraday's law. …

    🔒 Start your 14-day free trial to see the full solution →