CUET UG Physics Booster Test 2-Motional EMF and Lorentz Force -

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CUET UG Physics Booster Test 2--Motional EMF and Lorentz Force

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Physics Booster Test 2--Motional EMF and Lorentz Force

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QUESTION 1 OF 20

Identify the incorrect statement about a straight conductor moving in a field:

Answer Review

1 Identify the incorrect statement about a straight conductor moving in a field:

A The conductor must form a closed loop to have an induced emf.

B The magnetic field must be uniform and time-independent to derive simple motional emf.

C The arm PQ is moved with a constant velocity.

D Friction is assumed to be zero to prevent energy loss.

💡 Short Explanation

�� Motional emf can exist in an open conductor. �� Closed circuit is needed for current, not emf. �� Magnetic force separates charges in the rod.

✅ Correct Answer Explanation

A moving conductor in a magnetic field experiences charge separation due to the Lorentz force. This creates a potential difference (emf) across its ends even if the conductor is not part of a closed circuit. → Option A is incorrect because a closed loop is required for induced current, not for induced emf. → Option B is used in the standard derivation of motional emf. → Option C is a common assumption in deriving ε = Blv. → Option D is generally assumed to simplify energy considerations.

❓ Why Other Options Are Incorrect

�� Option B → Correct assumption used in deriving motional emf.
�� Option C → Constant velocity is assumed in standard derivations.
�� Option D → Frictionless motion simplifies analysis and avoids energy dissipation.

🎯 Strategy

Used

�� Elimination

Application:

�� Check which statement contradicts the basic concept of motional emf.

Final Logic:

�� Emf can exist without a closed loop; only current requires a closed path.

🧠 Memory Trick

"EMF first, current later."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Motional EMF

2 Match List I with List II regarding velocity and field interaction.

List I	List II
(1) Constant velocity v	(a) Blx
(2) Magnetic flux ΦB	(b) Forms a closed circuit
(3) Decreasing x	(c) Increases the rate of area change negatively
(4) Loop PQRS	(d) −dx/dt

A 1-d, 2-a, 3-c, 4-b

B 1-a, 2-d, 3-b, 4-c

C 1-d, 2-b, 3-a, 4-c

D 1-b, 2-c, 3-d, 4-a

💡 Short Explanation

Velocity is related to the decreasing position coordinate by v = −dx/dt. Magnetic flux through the loop is ΦB = Blx. A decreasing x reduces the enclosed area. PQRS represents the closed conducting loop.

✅ Correct Answer Explanation

(1) Constant velocity v As the loop moves and x decreases with time, v = −dx/dt → 1 → d → (2) Magnetic flux ΦB Magnetic flux through the loop is ΦB = BA = B(lx) = Blx → 2 → a → (3) Decreasing x A decrease in x reduces the area enclosed within the magnetic field, producing a negative rate of area change. → 3 → c → (4) Loop PQRS PQRS is the conducting path through which induced current flows, forming a closed circuit. → 4 → b Therefore: 1 → d 2 → a 3 → c 4 → b Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B: Incorrectly interchanges flux and velocity relations.
Option C: Incorrectly matches magnetic flux with the closed loop.
Option D: Assigns incorrect meanings to all major quantities.

🎯 Strategy

Used

Option Grouping

Application:

Match each physical quantity with its mathematical expression or physical description.

Final Logic:

Velocity → −dx/dt
Flux → Blx
Decreasing x → Negative area change
PQRS → Closed conducting loop

🧠 Memory Trick

"Flux = Blx, Velocity = −dx/dt"

📚 Topic: Flux Change in Moving Conductors and Motional EMF.

3 If the length of the conductor is l, its velocity is v, and the magnetic field is B, the rate of change of area is given by the product of l and the velocity v. Thus, if x is the distance that changes with time, the derivative -dx/dt represents

A the acceleration of the rod

B the speed v of the conductor PQ

C the rate of change of magnetic field

D the force on the rod

💡 Short Explanation

�� Velocity is rate of change of position. �� x decreases with time. �� v = -dx/dt.

✅ Correct Answer Explanation

In motional emf derivation, area changes because the rod moves. → If x decreases with time, velocity magnitude is v = -dx/dt. → Therefore the derivative represents speed of the conductor.

❓ Why Other Options Are Incorrect

�� Option A → Acceleration is d²x/dt².
�� Option C → No magnetic field variation involved.
�� Option D → Derivative of position does not represent force.

🎯 Strategy

Used

�� Substitution

Application:

�� Replace v directly with -dx/dt.

Final Logic:

�� Velocity is defined as rate of change of position.

🧠 Memory Trick

"Position → Velocity → One derivative."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Area Change Method

4 A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a uniform magnetic field of 0.3 T normal to the loop. What is the motional emf developed across the cut if the velocity of the loop is 1 cm s⁻¹ normal to the longer side?

A 2.4 × 10⁻⁴ V

B 0.6 × 10⁻⁴ V

C 2.4 × 10⁻³ V

D 0.6 × 10⁻³ V

💡 Short Explanation

�� ε = Blv. �� l = 2 cm = 0.02 m. �� ε = 0.3 × 0.02 × 0.01.

✅ Correct Answer Explanation

Velocity is normal to the longer side (8 cm), therefore the effective length cutting flux is 2 cm. → ε = Blv = 0.3 × 0.02 × 0.01 = 6 × 10⁻⁵ V = 0.6 × 10⁻⁴ V → The provided answer (A) is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Uses wrong effective length.
�� Option C → Ten times larger than actual value.
�� Option D → Incorrect power of ten.

🎯 Strategy

Used

�� Substitution

Application:

�� Substitute values directly into ε = Blv.

Final Logic:

�� Correct effective length is 2 cm, giving 0.6 × 10⁻⁴ V.

🧠 Memory Trick

"Length cutting flux matters."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Motional EMF Calculation

5 Identify the Correct Statements regarding force on Charge Carriers
(1) The Lorentz force acts on free charge carriers in the moving conductor.
(2) The charge moves with speed v in the magnetic field B.
(3) The Lorentz force magnitude is qvB.
(4) Only positive charges experience the Lorentz force in the rod.

A (1), (2), (3)

B (1), (3), (4)

C (2), (3), (4)

D (1), (2), (4)

💡 Short Explanation

Lorentz force acts on all moving charge carriers. The magnitude of magnetic force is qvB when v ⊥ B. Both positive and negative charges experience magnetic force. Therefore, statement (4) is incorrect.

✅ Correct Answer Explanation

Statement (1) is correct. Free charge carriers in a moving conductor experience the Lorentz force when moving through a magnetic field. → Statement (2) is correct. The charge carriers move with the conductor at speed v through the magnetic field B. → Statement (3) is correct. For motion perpendicular to the magnetic field, F = qvB Hence, the Lorentz force magnitude is qvB. → Statement (4) is incorrect. The Lorentz force acts on all moving charges, whether positive or negative. The force direction differs for opposite charges, but both experience the force. Therefore, the correct statements are: (1), (2), and (3) Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B: Includes statement (4), which is false.
Option C: Includes statement (4), which is false.
Option D: Includes statement (4), which is false.

🎯 Strategy

Used

Elimination

Application:

Apply the Lorentz force law to moving charges in a magnetic field.

Final Logic:

All moving charges experience magnetic force.
The force magnitude is qvB for perpendicular motion.
Negative charges are not exempt from magnetic force.

🧠 Memory Trick

"Positive or Negative, Both Feel the Force"

📚 Topic: Lorentz Force and Motion of Charge Carriers.

6 In a moving conductor PQ, the magnetic force directs the free charge carriers, and the work is done to move them across length l. The relationship between force position-dependence and the resultant work is:

A Dependent, qE

B Independent, qvBl

C Independent, qvB/l

D Dependent, qEl

💡 Short Explanation

�� Magnetic force magnitude = qvB. �� Work = Force × distance. �� W = qvBl.

✅ Correct Answer Explanation

The magnetic force on charge q is qvB. → Over distance l: W = qvBl → Force remains constant for uniform B and constant v.

❓ Why Other Options Are Incorrect

�� Option A → Missing displacement factor l.
�� Option C → Incorrect dimensional form.
�� Option D → Uses electric field expression.

🎯 Strategy

Used

�� Dimensional/Unit Analysis

Application:

�� Verify which expression has dimensions of work.

Final Logic:

�� Only qvBl gives correct work expression.

🧠 Memory Trick

"Force × length = qvBl."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Work Done on Charge Carriers

7 Choose the correct Statements about work done on charges in motional EMF
(1) The work done is calculated as W = qvBl.
(2) Work is done by the magnetic force to move the charge from P to Q.
(3) The expression for work is independent of the magnetic field strength.
(4) The work done gives rise to the induced emf across the rod.

A (1), (2), (3)

B (1), (2), (4)

C (2), (3), (4)

D (1), (3), (4)

💡 Short Explanation

The work associated with charge separation is W = qvBl. This work results in a potential difference (emf) across the rod. The work depends directly on the magnetic field strength B. Therefore, statement (3) is incorrect.

✅ Correct Answer Explanation

Statement (1) is correct. The work done in moving a charge q across a rod of length l moving with speed v in a magnetic field B is: W = qvBl → Statement (2) is correct. The magnetic force causes charge separation between the ends of the conductor, effectively moving charges from one end to the other. → Statement (3) is incorrect. Since W = qvBl the work done depends directly on B. Hence, it is not independent of magnetic field strength. → Statement (4) is correct. The work done per unit charge produces the induced emf across the rod: ε = W/q = Blv Therefore, the correct statements are: (1), (2), and (4) Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A: Includes statement (3), which is false.
Option C: Includes statement (3), which is false.
Option D: Includes statement (3), which is false.

🎯 Strategy

Used

Elimination

Application:

Use the relations:
W = qvBl
and
ε = W/q = Blv
to test each statement.

Final Logic:

Work depends on B, v, l, and q.
Therefore, statement (3) is incorrect.

🧠 Memory Trick

"More B → More Work → More EMF"

📚 Topic: Motional EMF and Work Done on Charge Carriers.

8 The principle of EMF as Work per Charge implies:
1. implies that ε = q / W
2. states that emf is the work done per unit charge, yielding ε = Blv
3. is only valid for stationary charges
4. contradicts Faraday's law of induction
Choose the correct option:

A 1

B 2

C 3

D 4

💡 Short Explanation

�� EMF = Work/Charge. �� ε = W/q. �� Motional emf becomes Blv.

✅ Correct Answer Explanation

By definition, ε = W/q → Using W = qvBl, ε = qvBl/q = Blv → This agrees with Faraday's law.

❓ Why Other Options Are Incorrect

�� Option A → Formula reversed.
�� Option C → Valid for moving charges also.
�� Option D → Completely consistent with Faraday's law.

🎯 Strategy

Used

�� Substitution

Application:

�� Substitute W = qvBl into ε = W/q.

Final Logic:

�� ε = Blv.

🧠 Memory Trick

"EMF = Work ÷ Charge."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Definition of EMF

9 When explaining the existence of induced emf in a stationary conductor, we must assume that

A a time-varying magnetic field generates an electric field

B provides the required push

C the velocity of the conductor suddenly becomes non-zero

D the Lorentz force q(v ×

💡 Short Explanation

�� Stationary charges have v = 0. �� Magnetic force becomes zero. �� Induced electric field causes emf.

✅ Correct Answer Explanation

Faraday's law states that a changing magnetic field creates a non-conservative electric field. → This electric field drives charges even when the conductor is stationary.

❓ Why Other Options Are Incorrect

�� Option B → Magnetic field cannot push stationary charges.
�� Option C → Conductor remains stationary.
�� Option D → Lorentz force requires motion.

🎯 Strategy

Used

�� Elimination

Application:

�� Remove options requiring moving charges.

Final Logic:

�� Changing magnetic field creates electric field.

🧠 Memory Trick

"Changing B creates E."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Faraday's Law

10 Choose the incorrect statement about induced electric fields:

A They are produced by a time-varying magnetic field.

B They exert a force qE on a stationary charge.

C They have properties perfectly identical to electric fields produced by static charges.

D They account for the induced current in stationary coils according to Faraday's discovery.

💡 Short Explanation

�� Induced fields are non-conservative. �� Electrostatic fields are conservative. �� They are not identical.

✅ Correct Answer Explanation

Induced electric fields arise from changing magnetic fields. → Unlike electrostatic fields, induced electric fields form closed loops and are non-conservative. → Therefore statement C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct consequence of Faraday's law.
�� Option B → Electric force qE acts on stationary charges.
�� Option D → Explains induced current in stationary coils.

🎯 Strategy

Used

�� Odd One Out

Application:

�� Identify the statement conflicting with the nature of induced electric fields.

Final Logic:

�� Induced and electrostatic electric fields are fundamentally different.

🧠 Memory Trick

"Induced E loops; electrostatic E ends."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Induced Electric Field

11 If a metallic rod is rotated in a uniform magnetic field, the free electrons get distributed over the ring due to the Lorentz force. Thus, if a steady state is reached at a certain value of emf, the net flow of electrons along the rod will be

A constantly increasing

B alternating periodically

C zero

D maximum

💡 Short Explanation

�� Electrons redistribute initially. �� Internal electric field develops. �� Steady state implies no net electron flow.

✅ Correct Answer Explanation

During rotation, free electrons experience Lorentz force and accumulate at one end. → This charge separation creates an electric field opposing further movement of charges. → At steady state, electric force balances magnetic force. → Therefore, no further net flow of electrons occurs along the rod.

❓ Why Other Options Are Incorrect

�� Option A → Electron flow does not continuously increase.
�� Option B → EMF is steady, not alternating.
�� Option D → Maximum flow occurs only during transient redistribution.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� The phrase "steady state" implies equilibrium and no net charge transport.

Final Logic:

�� Force balance at steady state results in zero net electron flow.

🧠 Memory Trick

"Steady state = No net drift."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Rotating Conductors

12 In a rotating metallic rod, the accumulation of free electrons and the resulting potential behavior are:

A Center to rim, Alternating EMF

B Distributed over ring, Steady state EMF

C Concentrated at center, Zero EMF

D Lost to surroundings, Transient EMF

💡 Short Explanation

�� Charges redistribute. �� Potential difference develops. �� Steady motional emf exists.

✅ Correct Answer Explanation

Rotation in a magnetic field causes charge separation. → Electrons redistribute over the conductor until equilibrium is reached. → A steady potential difference exists between the ends of the rod. → Hence, a steady-state emf is produced.

❓ Why Other Options Are Incorrect

�� Option A → EMF is not alternating under uniform rotation.
�� Option C → Charge concentration at center does not eliminate emf.
�� Option D → Charges are not lost to surroundings.

🎯 Strategy

Used

�� Elimination

Application:

�� Remove options contradicting steady rotational induction.

Final Logic:

�� Charge redistribution leads to a stable emf.

🧠 Memory Trick

"Redistribution → Stable EMF."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Rotating Rod in Magnetic Field

13 Match List I with List II regarding the integration of segment EMF in a rotating rod.

List I	List II
(1) Velocity of segment dr	(a) Bvdr
(2) Elemental emf dε	(b) rω
(3) Total emf limits	(c) ½BωR²
(4) Total emf expression	(d) 0 to R

A 1-b, 2-a, 3-d, 4-c

B 1-a, 2-b, 3-c, 4-d

C 1-b, 2-d, 3-a, 4-c

D 1-d, 2-c, 3-b, 4-a

💡 Short Explanation

Velocity of an element at radius r is rω. Elemental emf is dε = Bvdr. Integration is performed from 0 to R. The resulting emf is ½BωR².

✅ Correct Answer Explanation

(1) Velocity of segment dr For a rod rotating with angular velocity ω, the linear velocity of an element at distance r from the axis is: v = rω → 1 → b → (2) Elemental emf dε The small emf induced across an elemental length dr is: dε = Bvdr → 2 → a → (3) Total emf limits To find the total emf across the rod, integrate from the axis (r = 0) to the outer end (r = R). → 3 → d → (4) Total emf expression ε = ∫₀ᴿ Bωr dr = ½BωR² → 4 → c Therefore: 1 → b 2 → a 3 → d 4 → c Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B: Interchanges velocity and elemental emf expressions.
Option C: Incorrectly matches integration limits with elemental emf.
Option D: Contains multiple incorrect correspondences.

🎯 Strategy

Used

Option Grouping

Application:

Follow the standard derivation of emf induced in a rotating rod.

Final Logic:

v = rω
��
dε = Bvdr
��
Integrate from 0 to R
��
ε = ½BωR²

🧠 Memory Trick

"rω → Bvdr → 0 to R → ½BωR²"

📚 Topic: EMF Induced in a Rotating Rod (Derivation).

14 Choose the correct statements regarding angular speed relations in a rotating rod:
(1) The linear velocity v is related to angular speed ω by v = ωr.
(2) The angular speed ω equals 2πν, where ν is frequency.
(3) The linear velocity v is uniform across the entire rod length.
(4) Substituting v = ωr into the integration yields ε = 1/2 BωR².

A (1), (2), (3)

B (1), (2), (4)

C (2), (3), (4)

D (1), (3), (4)

💡 Short Explanation

�� v = ωr. �� ω = 2πν. �� Velocity varies with r.

✅ Correct Answer Explanation

Statements A and B are standard rotational motion relations. → Using v = ωr in the emf integration gives: ε = ∫₀ᴿ Bωrdr = ½BωR² → Statement 3 is incorrect because different points on the rod have different values of r and hence different speeds.

❓ Why Other Options Are Incorrect

�� Option A → Includes false statement 3.
�� Option C → Includes false statement 3.
�� Option D → Includes false statement 3.

🎯 Strategy

Used

�� Elimination

Application:

�� Identify the incorrect rotational motion statement.

Final Logic:

�� Linear speed increases with radius.

🧠 Memory Trick

"Farther out, faster."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Rotational Motion Relations

15 The rate of change of area method:
Choose correct:

A 1

B 2

C 3

D 4

💡 Short Explanation

�� Area swept changes with time. �� Flux changes due to area change. �� Same emf as integration method.

✅ Correct Answer Explanation

The area method uses Faraday's law: ε = B(dA/dt) → A closed loop is considered to calculate changing magnetic flux. → It gives the same result as direct integration.

❓ Why Other Options Are Incorrect

�� Option B → 2 may remain constant.
�� Option C → Both methods give identical results.
�� Option D → θ changes continuously during rotation.

🎯 Strategy

Used

�� Elimination

Application:

�� Identify the statement consistent with flux-change derivation.

Final Logic:

�� Changing area, not changing B, produces emf.

🧠 Memory Trick

"Changing area = Changing flux."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Area Sweep Method

16 If the radius R of the circle swept by the rod is 2.0 m, the uniform magnetic field B is 0.5 T, and the rod rotates such that the rate of change of angle dθ/dt is 10 rad/s, what is the induced emf using the area of sector equation?

A 5 V

B 10 V

C 20 V

D 15 V

💡 Short Explanation

�� ε = ½BR²ω. �� R = 2 m. �� ω = 10 rad/s.

✅ Correct Answer Explanation

Using: ε = ½BR²ω = ½ × 0.5 × (2)² × 10 = 0.25 × 4 × 10 = 10 V

❓ Why Other Options Are Incorrect

�� Option A → Half the correct value.
�� Option C → Double the correct value.
�� Option D → Arithmetic error.

🎯 Strategy

Used

�� Substitution

Application:

�� Substitute values into the standard formula.

Final Logic:

�� ε = 10 V.

🧠 Memory Trick

"Half B R-square omega."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Rotating Rod Numerical

17 Choose the correct Statements Regarding the Axle and Rim Potential of a Rotating Wheel in a Magnetic Field
(1) An induced emf develops between the axle and the rim.
(2) The wheel acts like a rotating metallic rod with length equal to its radius.
(3) The induced emf is calculated using ε = ½ωBR².
(4) Adding more identical spokes in parallel increases the total output voltage.

A (1), (2), (3)

B (1), (2), (4)

C (2), (3), (4)

D (1), (3), (4)

💡 Short Explanation

A rotating wheel develops an emf between its axle and rim. Each spoke behaves like a rotating conducting rod. The induced emf is given by ε = ½ωBR². Adding more spokes does not increase the emf.

✅ Correct Answer Explanation

Statement (1) is correct. When a conducting wheel rotates in a magnetic field, charge separation occurs, producing a potential difference between the axle and the rim. → Statement (2) is correct. Each spoke can be treated as a conducting rod rotating about one end, with effective length equal to the wheel radius R. → Statement (3) is correct. The emf induced between the axle and rim is: ε = ½ωBR² where: ω = angular velocity B = magnetic field strength R = wheel radius → Statement (4) is incorrect. Adding identical spokes in parallel does not increase the emf. The potential difference between axle and rim remains the same. Additional spokes only reduce internal resistance and increase current-carrying capability. Therefore, the correct statements are: (1), (2), and (3) Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B: Includes statement (4), which is false.
Option C: Includes statement (4), which is false.
Option D: Includes statement (4), which is false.

🎯 Strategy

Used

Elimination

Application:

Apply the rotating rod emf formula and parallel-source concept.

Final Logic:

Rotating wheel → Axle-rim emf.
Spoke behaves like a rotating rod.
More spokes → More current capacity, not more voltage.

🧠 Memory Trick

"Wheel = Many Rotating Rods"

📚 Topic: Rotating Wheel Generator and Motional EMF.

18 In a wheel with multiple metallic spokes rotating in a magnetic field, the number of spokes is immaterial to the total induced emf because

A they all cancel each other out

B the emfs across the spokes are in parallel

C they are connected in series

D the magnetic field does not pass through the spokes

💡 Short Explanation

�� Same axle and rim terminals. �� Parallel connection. �� Voltage remains unchanged.

✅ Correct Answer Explanation

Every spoke develops the same emf. → Since all spokes connect between the same axle and rim, they are effectively connected in parallel. → Parallel sources of equal emf do not increase output voltage.

❓ Why Other Options Are Incorrect

�� Option A → Emfs reinforce rather than cancel.
�� Option C → Spokes are not connected in series.
�� Option D → Magnetic field interacts with each spoke.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Identify the physical arrangement of the spokes.

Final Logic:

�� Same terminals imply parallel connection.

🧠 Memory Trick

"Same ends = Parallel."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Rotating Wheel EMF

19 Incorrect statement about the effects of the horizontal component of earth's magnetic field (HE):

A A wheel rotated in a plane normal to HE will develop an induced emf.

B HE acts uniformly over the area of the rotating wheel.

C A stationary coil placed normal to HE will develop a continuous motional emf.

D The value of HE at a given place can be used to calculate motional emf.

💡 Short Explanation

�� Motion is necessary. �� Stationary conductor has no motional emf. �� Earth's field can induce emf only with motion.

✅ Correct Answer Explanation

Motional emf requires motion through a magnetic field. → A stationary coil experiences no motional emf in a constant magnetic field. → Therefore statement C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct application of motional induction.
�� Option B → Reasonable approximation over wheel dimensions.
�� Option D → HE can be used in emf calculations.

🎯 Strategy

Used

�� Elimination

Application:

�� Remove the option violating the condition for motional emf.

Final Logic:

�� No motion means no motional emf.

🧠 Memory Trick

"No motion, no motional emf."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Earth's Magnetic Field Effects

20 If a horizontal straight wire extending east to west falls with a speed v at right angles to the horizontal component of the earth's magnetic field B, the motional emf induced depends on its length l, speed v, and B. Thus, if the wire's length is 10 m, falling at 5.0 m/s in a field of 0.30 × 10⁻⁴ T, the instantaneous emf will be

A 1.5 × 10⁻³ V

B 1.5 × 10⁻⁴ V

C 0.15 × 10⁻³ V

D 15 × 10⁻³ V

💡 Short Explanation

�� ε = Blv. �� B = 0.30 × 10⁻⁴ T. �� l = 10 m, v = 5 m/s.

✅ Correct Answer Explanation

Using: ε = Blv = (0.30 × 10⁻⁴)(10)(5) = 15 × 10⁻⁴ = 1.5 × 10⁻³ V → Therefore the provided answer is correct.

❓ Why Other Options Are Incorrect

�� Option B → One power of ten smaller.
�� Option C → Numerical error.
�� Option D → Ten times larger than correct value.

🎯 Strategy

Used

�� Substitution

Application:

�� Direct substitution into ε = Blv.

Final Logic:

�� Multiplying B, l and v gives 1.5 × 10⁻³ V.

🧠 Memory Trick

"EMF = B × l × v."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Motional EMF in Earth's Magnetic Field

CUET UG Physics Booster Test 2–Motional EMF and Lorentz Force

CUET UG Physics Booster Test 2--Motional EMF and Lorentz Force

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