Q.A cylindrical bar magnet is rotated about its axis. A wire is connected from the axis and is made to touch the cylindrical surface through a contact. Then

Showing the 12 most recent of 15 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.A loop moves towards a stationary magnet at constant speed V resulting an induced $emf$ E within the loop. If the magnet also moves away from the loop at the same speed V, the new induced $emf$ in the loop is (A) E (B) $\frac{E}{2}$ (C) 2E (D) 0

   ›Reveal solutionSolution

   With loop and magnet moving in the same direction at the same speed, their separation never changes, the flux is constant, and the induced emf is zero — option (D).

   Concept and Intuition

   Faraday's law says $\varepsilon = -\dfrac{d\phi}{dt}$: an emf appears only when the flux through the loop changes. The flux from a bar magnet through a loop depends on how far the magnet is from the loop — that is, on their separation, not on either object's velocity in the laboratory.

   This is why electromagnetic induction is a purely relative phenomenon: moving the magnet towards a stationary loop, or the loop towards a stationary magnet, at the same speed produces the same emf. Conversely, if both move so that the gap between them stays fixed, nothing changes for the loop — no flux change, no emf, no induced current.

   Step-by-Step Solution

   1. Initially: loop approaches the stationary magnet at $V$, so the separation shrinks at rate $V$, and $\left|\dfrac{d\phi}{dt}\right|$ is whatever gives the emf $E$.
   2. Now: the loop still moves towards the magnet at $V$, while the magnet retreats from the loop at $V$ (same direction of motion).
   3. Relative velocity of loop with respect to magnet:

   $V_{\text{rel}} = V - V = 0 .$

   4. The separation is therefore constant, so the flux linked with the loop is constant in time:

   $\frac{d\phi}{dt} = 0 .$

   5. By Faraday's law, $\varepsilon = -\dfrac{d\phi}{dt} = 0$. There is no induced emf (and no induced current, and no opposing force — consistent with Lenz's law, which only acts when there is a change to oppose).

   Common Mistakes

   • Adding the speeds to get $2V$ and answering $2E$ — that would be the case if the magnet moved towards the loop as well (approaching each other).
   • Thinking motion alone produces an emf: it is change of flux, i.e. relative motion, that does.
   • Assuming the answer must be $E$ because "emf depends only on the loop's speed" — the magnet's motion matters just as much.

   ✓Final answer

   The correct option is (D) — $0$.

   ANSWER: D

2. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.A conducting square loop of side L moves with a uniform speed V in a region of uniform magnetic field acting perpendicular to the plane of the loop and directed into it as shown in figure. The emf induced in the loop is [FIGURE] (a square conducting loop of side L moves with velocity V to the right, fully inside a region of uniform magnetic field B directed into the page, shown by rows of x symbols) (A) $BLV$ (B) $\dfrac{BLV}{2}$ (C) zero (D) $2BLV$

   ›Reveal solutionSolution

   The key idea is that motional emf requires a change in magnetic flux through the loop. Since the entire loop moves within a uniform field, the flux remains constant, so the induced emf is zero. The correct option is (C).

   Concept and Intuition: Motional Emf

   When a conductor moves through a magnetic field, charges inside experience a magnetic force $q(\vec{v} \times \vec{B})$, which can drive a current if there is a complete circuit. This is the origin of motional emf. However, the emf induced in a closed loop is given by Faraday’s law:

   $\mathcal{E} = -\frac{d\Phi_B}{dt}$

   where $\Phi_B$ is the magnetic flux through the loop. The crucial point: only a change in flux produces an emf. If the loop moves entirely inside a uniform field, the number of field lines passing through it stays the same — no change, no emf. Many students mistakenly think that motion alone guarantees an emf, but that’s only true if the loop is entering or leaving the field, or if the field itself varies.

   Let’s work through the reasoning step by step.

   ---

   1. Define the situation

      The loop is a square of side $L$, moving with constant velocity $V$ to the right. The magnetic field $\vec{B}$ is uniform, directed into the page (shown by ‘x’ symbols), and extends over the entire region shown — there is no boundary where the field ends within the diagram. The loop is completely inside this uniform field at all times.

   2. Calculate the magnetic flux through the loop

      Magnetic flux is given by

   $\Phi_B = \int \vec{B} \cdot d\vec{A}$

   Since $\vec{B}$ is uniform and perpendicular to the plane of the loop (into the page), and the loop’s area is $L^2$, the flux simplifies to

   $\Phi_B = B \cdot (\text{area}) = B L^2$

   The direction (sign) is constant, so we can treat it as a positive number.

   3. Does the flux change as the loop moves? The loop moves to the right, but the field is uniform everywhere. The area $L^2$ does not change, and $B$ does not change. Therefore,

   $\frac{d\Phi_B}{dt} = 0$

   No matter how fast or how far the loop moves, as long as it stays entirely within the uniform field, the flux remains constant.

   4. Apply Faraday’s law

   $\mathcal{E} = -\frac{d\Phi_B}{dt} = 0$

   Hence, the induced emf in the loop is zero.

   5. Address a common misconception

      Watch out

      A student might think: “But each side of the loop is moving through the field, so each side should have a motional emf $BLV$!” This is true for an open conductor, but in a closed loop, the emfs on opposite sides cancel. For example, the right side moving to the right has an upward induced emf (by the right-hand rule), while the left side moving to the right has a downward induced emf — they are equal and opposite, so the net emf around the loop is zero. The flux argument is simpler and more general.

   6. Confirm with the given options

      The choices are:

      (A) $BLV$

      (B) $\frac{BLV}{2}$

      (C) zero

      (D) $2BLV$

      Only option (C) matches our result.

   Tip

   A neat shortcut: If a closed loop moves entirely within a uniform magnetic field, the flux is constant → emf = 0. This holds for any shape or orientation, as long as the field is uniform and the loop doesn’t rotate.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

3. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.A horizontal telegraph wire of length 30 m spread east to west fell down freely from a height of 20 m. If the resistance of the wire is $40\ \Omega$ and the horizontal component of the earth's magnetic field at the place is $2\times10^{-5}\ \mathrm{T}$, then the induced current when the wire reaches the ground is (Acceleration due to gravity $=10\ \mathrm{m\,s^{-2}}$) (A) 0.3 mA (B) 3 mA (C) 3 A (D) 0.03 A

   ›Reveal solutionSolution

   Motional EMF is induced in the falling east-west wire by the earth's horizontal magnetic field component; combined with Ohm's law this gives a current of 0.3 mA.

   Concept and Intuition

   As the horizontal wire (oriented east-west) falls freely, it moves vertically through the earth's horizontal magnetic field component $B_H$, which is perpendicular to both the wire's length and its velocity. This generates a motional EMF $\varepsilon = B_H L v$, where $v$ is the wire's instantaneous speed. This EMF drives a current through the wire's own resistance (treating it as a simple closed-loop equivalent circuit as is conventional for this classic problem).

   Step-by-Step Solution

   1. Find the speed at the ground using free fall: $v = \sqrt{2gh} = \sqrt{2\times10\times20} = \sqrt{400} = 20\ \text{m/s}$.
   2. Compute EMF: $\varepsilon = B_H L v = (2\times10^{-5})(30)(20) = 2\times10^{-5}\times 600 = 1.2\times10^{-2}\ \text{V}$.
   3. Apply Ohm's law: $I = \varepsilon/R = \dfrac{1.2\times10^{-2}}{40} = 3\times10^{-4}\ \text{A}$.
   4. Convert: $3\times10^{-4}\ \text{A} = 0.3\ \text{mA}$.

   Common Mistakes

   • Forgetting to convert to $v$ via kinematics and instead using $h$ directly in the EMF formula.
   • Mis-converting amps to milliamps (off by factor of 10 or 1000).

   ✓Final answer

   The correct option is (A) — 0.3 mA.

   ANSWER: A

4. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.A metallic disc of radius 0.3 m is rotating with a constant angular speed of $60\ \mathrm{rad\,s^{-1}}$ in a plane perpendicular to a uniform magnetic field of $5\times10^{-2}$ T. The emf induced between a point on the rim and centre of the disc is (A) 0.06 V (B) 0.612 V (C) 1.35 V (D) 0.135 V

   ›Reveal solutionSolution

   This tests the motional-emf formula for a rotating conducting disc in a magnetic field. Using $\varepsilon=\tfrac12 B\omega R^2$ gives $\varepsilon=0.135$ V.

   Concept and Intuition

   Every radial element of the spinning disc is itself a tiny "rod" moving through the magnetic field with a speed that increases linearly from zero at the centre to $\omega R$ at the rim. Each element contributes a motional emf $d\varepsilon=B v\,dr=B(\omega r)\,dr$, and integrating these contributions from the centre to the rim gives the total emf between those two points — analogous to a rotating conducting rod (the Faraday disc / homopolar generator).

   Step-by-Step Solution

   1. Consider a thin element at radius $r$, thickness $dr$, moving with speed $v=\omega r$.
   2. Motional emf of this element: $d\varepsilon=Bv\,dr=B\omega r\,dr$.
   3. Integrate from $r=0$ to $r=R$:

   $\varepsilon=\int_0^R B\omega r\,dr=\frac12 B\omega R^2$

   4. Substitute $B=5\times10^{-2}$ T, $\omega=60\ \mathrm{rad\,s^{-1}}$, $R=0.3$ m:

   $\varepsilon=\frac12(5\times10^{-2})(60)(0.3)^2=\frac12(0.05)(60)(0.09)$

   $=\frac12(0.27)=0.135\ \text{V}$

   Common Mistakes

   • Forgetting the factor of $\tfrac12$ that comes from integrating $r$ over the disc (using $\varepsilon=B\omega R^2$ instead).
   • Using diameter instead of radius for $R$.

   ✓Final answer

   The correct option is (D) — 0.135 V.

   ANSWER: D

5. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.If a wheel with 24 metallic spokes each 40 cm long is rotated with a speed of 180 rev/min in a plane normal to the horizontal component of earth's magnetic field, the emf induced between the axle and the rim of the wheel is E. If the number of spokes is made 12 and the wheel is rotated with a speed of 90 rev/min in the same field, the induced emf is (A) E (B) 2E (C) 4E (D) 0.5E

   ›Reveal solutionSolution

   All spokes of the wheel are in parallel between the same two terminals (axle and rim), so the induced emf depends only on $B$, $\omega$, and spoke length $L$ — not on the number of spokes. Halving $\omega$ (spoke length unchanged) halves the emf.

   Concept and Intuition

   Each metallic spoke is a conducting rod rotating about one end (the axle) in a magnetic field perpendicular to the plane of rotation. A rotating rod of length $L$ with angular velocity $\omega$ generates an emf $\varepsilon = \tfrac12 B\omega L^2$ between its two ends, exactly like a rod sweeping out area. Since every spoke reaches from the same axle to the same rim, all the spokes are connected between the same pair of nodes — they are in parallel, not in series. A parallel combination of identical emf sources (each with the same emf and internal resistance) still delivers that same single-spoke emf between the two terminals; adding more spokes in parallel does not add up the emfs.

   Step-by-Step Solution

   1. EMF of a single rotating spoke: $\varepsilon = \tfrac12 B\omega L^2$.
   2. Since all spokes connect the same axle to the same rim, they act as parallel identical sources; the net emf between axle and rim equals that of one spoke: $E = \tfrac12 B\omega_1 L^2$, independent of the spoke count (24 originally).
   3. In the new situation, the number of spokes changes to 12 — irrelevant to the emf, since spokes are in parallel regardless of count.
   4. Spoke length $L$ is unchanged (still 40 cm) and the magnetic field is unchanged.
   5. Angular speed changes from $\omega_1$ (180 rev/min) to $\omega_2$ (90 rev/min) = $\omega_1/2$.
   6. New emf: $E' = \tfrac12 B\omega_2 L^2 = \tfrac12 B\left(\tfrac{\omega_1}{2}\right)L^2 = \tfrac{1}{2}E$.

   Common Mistakes

   • Assuming the emf scales with the number of spokes (treating them as in series), which would wrongly give $E' = \tfrac{12}{24}\cdot\tfrac{90}{180}E$ or similar — spokes sharing both terminals are in parallel, and parallel identical sources don't add.
   • Forgetting that the number of spokes is a distractor entirely irrelevant to the answer.

   ✓Final answer

   The correct option is (D) — $0.5E$.

   ANSWER: D

6. AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
   Q.A helicopter has metallic blades with length 4 m extending outward from the central point and rotating at $3\ \mathrm{rev\,s^{-1}}$. If the vertical component of earth's magnetic field is $40\ \mu T$, then the emf induced between the blade tip and the central point is (A) 3.14 mV (B) 2.83 mV (C) 16 mV (D) 6 mV

   ›Reveal solutionSolution

   A rotating rod in a perpendicular magnetic field generates an EMF $\varepsilon=\tfrac12B\omega L^2$ between its centre and tip; substituting the given values gives about $6$ mV. Answer: (D) 6 mV.

   Concept and Intuition

   Each helicopter blade is a straight conductor rotating about one end (the hub) in the (locally uniform) vertical component of Earth's magnetic field. Every small element of the rotating rod experiences a motional EMF $d\varepsilon=B\,v\,dr=B(\omega r)dr$, and integrating from the hub ($r=0$) to the tip ($r=L$) gives the classic result $\varepsilon=\tfrac12B\omega L^2$ — the same formula used for a conducting rod/disc rotating in a magnetic field (e.g. the Faraday disc).

   Step-by-Step Solution

   1. Angular speed: $\omega=2\pi f=2\pi(3)=6\pi\approx18.85$ rad/s.
   2. Blade length $L=4$ m, field $B=40\ \mu\text{T}=40\times10^{-6}$ T.
   3. EMF for a rod rotating about one end: $\varepsilon=\dfrac{1}{2}B\omega L^2$.
   4. Substitute: $\varepsilon=\dfrac{1}{2}(40\times10^{-6})(18.85)(16)$.
   5. Compute: $40\times10^{-6}\times18.85=7.54\times10^{-4}$; times $16=1.206\times10^{-2}$; times $\dfrac12=6.03\times10^{-3}$ V $=6.03$ mV $\approx6$ mV.

   Common Mistakes

   • Using $L$ instead of $L^2$ (or forgetting the factor of $\tfrac12$ that comes from integrating over the rod's length).
   • Using frequency $f$ directly in the formula instead of converting to angular frequency $\omega=2\pi f$.

   ✓Final answer

   The correct option is (D) — 6 mV.

   ANSWER: D

7. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.A metallic wire loop of side ($l$) 0.1 m and resistance of $1\Omega$ is moved with a constant velocity in a uniform magnetic field of $2\ \mathrm{Wm^{-2}}$ as shown in the figure. The magnetic field is perpendicular to the plane of the loop. The loop is connected to a network of resistors. The velocity of loop so as to have a steady current of 1 mA in loop is [FIGURE] (a square wire loop of side $l$ is shown moving with velocity v to the right through a region of magnetic field directed into the page (marked with x symbols); the loop's two terminals P (top) and Q (bottom) connect via wires to a network of five $3\Omega$ resistors arranged in a diamond/bridge pattern -- two $3\Omega$ resistors forming the upper branches, two $3\Omega$ resistors forming the lower branches, and one $3\Omega$ resistor as the central bridging element) (A) $0.67\ \mathrm{cm\,s^{-1}}$ (B) $2\ \mathrm{cm\,s^{-1}}$ (C) $3\ \mathrm{cm\,s^{-1}}$ (D) $4\ \mathrm{cm\,s^{-1}}$

   ›Reveal solutionSolution

   The motional emf of a sliding loop drives current through a balanced Wheatstone-bridge network of five equal resistors; recognizing the balance simplifies the network to $3\,\Omega$, and setting the resulting current to 1 mA gives the required velocity.

   Concept and Intuition

   A conducting loop moving with velocity $v$ through a uniform field $B$ perpendicular to its plane acts as a source of motional emf $\varepsilon = Bvl$ (only the leading/trailing side cutting field lines matters for a simple translating loop of width $l$). This emf drives current through the loop's own resistance in series with whatever external network is connected across its terminals P and Q. Here that external network is a Wheatstone bridge of five identical $3\,\Omega$ resistors — since all four arms are equal ($3:3::3:3$), the bridge is balanced, meaning the two "middle" nodes sit at the same potential and the fifth (bridging) resistor carries no current at all. It can therefore be deleted without changing the network's behaviour.

   Step-by-Step Solution

   1. Balanced-bridge simplification: with the bridge resistor removed, the network becomes two series pairs of $3\,\Omega+3\,\Omega = 6\,\Omega$ each, connected in parallel between P and Q: $R_{ext} = 6\parallel 6 = 3\,\Omega$.
   2. Total resistance in the circuit (loop's own resistance in series with the external network, as the loop is the emf source): $R_{total} = R_{loop}+R_{ext} = 1+3 = 4\,\Omega$.
   3. Motional emf: $\varepsilon = Bvl = 2\times v\times0.1 = 0.2v$.
   4. Current: $I = \dfrac{\varepsilon}{R_{total}} = \dfrac{0.2v}{4} = 0.05v$.
   5. Set $I = 1\ \text{mA} = 1\times10^{-3}$ A: $0.05v = 10^{-3} \Rightarrow v = 0.02\ \text{m/s} = 2\ \text{cm/s}$.

   Common Mistakes

   • Trying to reduce the bridge network by naive series-parallel combination without first checking for balance (which would give the wrong, more complicated, equivalent resistance).
   • Forgetting to add the loop's own $1\,\Omega$ resistance to the external $3\,\Omega$ before computing the current.

   ✓Final answer

   The correct option is (B) — $2\ \mathrm{cm\,s^{-1}}$.

   ANSWER: B

8. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.In an ac generator, if coil of N turns and area A is rotated at $\upsilon$ revolutions per second in a uniform magnetic field B, then the motional emf produced is equal to (At $t = 0$ s, the coil is perpendicular to the field) (A) $NBA(2\pi\upsilon)\sin(2\pi\upsilon t)$ (B) $NBA^2(2\pi\upsilon)\sin(2\pi\upsilon t)$ (C) $N^2B^2A^2(2\pi\upsilon)\sin(2\pi\upsilon t)$ (D) $NBA(4\pi\upsilon)\sin(2\pi\upsilon t)$

   ›Reveal solutionSolution

   Faraday's law applied to a rotating coil gives the standard AC-generator EMF $e=NBA\omega\sin(\omega t)$ with $\omega=2\pi\upsilon$, matching option (A).

   Concept and Intuition

   An AC generator works by rotating a coil in a uniform magnetic field, continuously changing the flux linkage through it and thereby inducing an EMF (Faraday's law of electromagnetic induction). If the coil starts perpendicular to $\vec B$ at $t=0$ (flux maximum at $t=0$), the flux varies as a cosine and the EMF — being $-d\Phi/dt$ — comes out as a sine function, peaking a quarter-cycle later.

   Step-by-Step Solution

   1. Flux through the coil at angle $\theta=\omega t$ from the perpendicular position: $\Phi(t)=NBA\cos(\omega t)$, where $\omega=2\pi\upsilon$ ($\upsilon$ = revolutions per second).
   2. Induced EMF: $e=-\dfrac{d\Phi}{dt}=-NBA\times(-\omega\sin\omega t)=NBA\omega\sin(\omega t)$.
   3. Substitute $\omega=2\pi\upsilon$: $e=NBA(2\pi\upsilon)\sin(2\pi\upsilon t)$.
   4. This matches the form of option (A) exactly, with correct powers of $N$, $B$, $A$ (all to the first power) and the correct angular frequency factor.

   Common Mistakes

   • Squaring $N$, $B$, or $A$ (as in some distractor options) — the EMF is linear in each of these quantities, not quadratic.
   • Using $4\pi\upsilon$ instead of $2\pi\upsilon$ for the angular frequency — a common factor-of-two slip.

   ✓Final answer

   The correct option is (A) — $NBA(2\pi\upsilon)\sin(2\pi\upsilon t)$.

   ANSWER: A

9. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.An ac generator converts (A) electrical energy into mechanical energy. (B) electrical energy into magnetic energy. (C) mechanical energy into magnetic energy. (D) mechanical energy into electrical energy.

   ›Reveal solutionSolution

   An AC generator uses electromagnetic induction to turn mechanical rotation into electrical energy.

   Concept and Intuition

   A generator works on Faraday's law of electromagnetic induction: mechanically rotating a coil within a magnetic field (or a magnet within a coil) changes the magnetic flux through the coil, inducing an EMF. This is the exact reverse of a motor, which converts electrical energy into mechanical energy.

   Step-by-Step Solution

   1. In an AC generator, an external mechanical agent (e.g. a turbine, hand-crank, or engine) does work to rotate the coil/armature.
   2. As the coil rotates in the magnetic field, the flux linkage changes with time, inducing an EMF (Faraday's law).
   3. This induced EMF drives a current in the external circuit — meaning mechanical energy input has been converted into electrical energy output.
   4. This is the opposite conversion of what a motor does (electrical → mechanical).

   Common Mistakes

   • Confusing a generator with a motor (which does the reverse conversion, electrical to mechanical).

   ✓Final answer

   The correct option is (D) — mechanical energy into electrical energy.

   ANSWER: D

10. AP EAPCET 2022Set ap-2022-07-12-AN1 markMCQ
    Q.A wire of length 20 cm is moving with a velocity of 180 m min$^{-1}$ perpendicular to a magnetic field. If the induced emf in the wire is 3 V, the magnitude of the field in tesla is (A) 2 (B) 3 (C) 5 (D) 10

    ›Reveal solutionSolution

    A straight conductor moving perpendicular to a magnetic field develops a motional emf $\varepsilon=BLv$; solve for $B$ given the other three quantities.

    Concept and Intuition

    When a conducting rod of length $L$ moves with speed $v$ perpendicular to a magnetic field $B$, the free charges in the rod experience a magnetic force that separates charge until an equilibrium (motional) emf develops: $\varepsilon = BLv$.

    Step-by-Step Solution

    1. Convert velocity: $v = 180\ \text{m/min} = \dfrac{180}{60} = 3$ m/s.
    2. Length: $L = 0.2$ m.
    3. From $\varepsilon = BLv$: $B = \dfrac{\varepsilon}{Lv} = \dfrac{3}{0.2\times3} = \dfrac{3}{0.6} = 5$ T.

    Common Mistakes

    • Forgetting to convert the velocity from m/min to m/s before using it.

    ✓Final answer

    The correct option is (C) — 5.

    ANSWER: C

11. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
    Q.An electric generator is based on _________ (A) Faraday's laws of electromagnetic induction (B) Motion of charged particles in an electromagnetic field (C) Fission of Uranium by slow neutrons (D) Newton's laws of motion

    ›Reveal solutionSolution

    The electric generator is a direct application of Faraday's law: rotating a coil in a magnetic field changes the flux through it, inducing an EMF.

    Concept and Intuition

    A generator is essentially the reverse of a motor: mechanical rotation of a coil inside a magnetic field (or vice versa) continuously changes the magnetic flux linked with the coil. By Faraday's law, $\varepsilon=-\dfrac{d\Phi_B}{dt}$, this changing flux induces an EMF, which drives current through an external circuit.

    Step-by-Step Solution

    1. Identify the working principle needed: converting mechanical rotation into electrical EMF.
    2. Faraday's law of electromagnetic induction states that an EMF is induced whenever the magnetic flux through a circuit changes.
    3. In a generator, the coil's orientation relative to the field constantly changes as it rotates, so $\Phi_B(t)=BA\cos(\omega t)$ varies with time, inducing a sinusoidal EMF.
    4. None of the other listed principles (motion of charged particles in an EM field, nuclear fission, Newton's laws) describe the generator's basic working.
    5. Hence the generator is based on Faraday's laws of electromagnetic induction.

    Common Mistakes

    • Confusing the generator (uses Faraday's law) with devices based on the motion of charged particles in fields, like a cyclotron or CRT.

    ✓Final answer

    The correct option is (A) — Faraday's laws of electromagnetic induction.

    ANSWER: A

12. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.A rectangular loop circuit has a sliding wire $PQ$ as shown in the figure. The loop is placed in a magnetic field '$B$', perpendicular to its plane. The resistance of the wire $PQ$ is $R$. If the wire moves with constant velocity '$v$', then find the current flowing through the wire $PQ$? [FIGURE] (a rectangular loop with a resistor $R$ forming the left side, another resistor $R$ forming the right side, and a sliding wire $PQ$ (also of resistance $R$) vertically across the middle at distance $l$ from the left side, moving to the right with velocity $v$) (A) $\dfrac{Blv}{3R}$ (B) $\dfrac{Blv}{2R}$ (C) $\dfrac{3Blv}{2R}$ (D) $\dfrac{2Blv}{3R}$

    ›Reveal solutionSolution

    The moving wire PQ is an EMF source of its own resistance $R$; the fixed resistors on either side act in parallel as its external load, giving total resistance $3R/2$ and hence a current of $\dfrac{2Blv}{3R}$ through PQ.

    Concept and Intuition

    As PQ slides with velocity $v$ in field $B$, motional EMF $\varepsilon=Blv$ is generated across it, and PQ itself has resistance $R$. Since the rails are (implicitly) resistance-free, current from PQ can return via either the left resistor or the right resistor — both connect the same top and bottom rails, so they are electrically in parallel with each other, both acting as the external circuit for the source PQ.

    Step-by-Step Solution

    1. EMF source: PQ, with $\varepsilon=Blv$ and internal resistance $R$.
    2. External resistance seen by PQ: the left $R$ and right $R$ are both connected across the same two rails (top and bottom), so they are in parallel: $R_{ext}=\dfrac{R\cdot R}{R+R}=\dfrac R2$.
    3. Total resistance in the circuit as seen by the source: $R_{total}=R+\dfrac R2=\dfrac{3R}2$.
    4. Current delivered by the source (this is the current that flows through PQ, since all current must pass through the source itself): $I=\dfrac{\varepsilon}{R_{total}}=\dfrac{Blv}{3R/2}=\dfrac{2Blv}{3R}$.

    Common Mistakes

    • Treating the two outer resistors as being in series instead of parallel (they are in parallel because both connect the same top/bottom rail pair).
    • Forgetting that PQ's own resistance $R$ must be added in series with the external parallel combination to get total circuit resistance.

    ✓Final answer

    The correct option is (D) — $\dfrac{2Blv}{3R}$.

    ANSWER: D