CUET UG Physics Booster Test 3-Motional EMF and Lorentz Force -

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CUET UG Physics Booster Test 3-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 3-Motional EMF and Lorentz Force

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QUESTION 1 OF 20

Identify the correct statements regarding the straight conductor in field:
1. The area enclosed by loop PQRS changes strictly linearly if the velocity v is constant.
2. The induced current direction is determined by the decreasing area opposing the magnetic flux change.
3. Φ_B = Blx is valid even if the magnetic field is rapidly varying with time.
4. The induced emf Blv requires the field to be perpendicular to the plane of the system.

Answer Review

1 Identify the correct statements regarding the straight conductor in field:
1. The area enclosed by loop PQRS changes strictly linearly if the velocity v is constant.
2. The induced current direction is determined by the decreasing area opposing the magnetic flux change.
3. Φ_B = Blx is valid even if the magnetic field is rapidly varying with time.
4. The induced emf Blv requires the field to be perpendicular to the plane of the system.

A 1, 2, 3

B 1, 2, 4

C 2, 3, 4

D 1, 3, 4

💡 Short Explanation

�� Constant velocity gives linear area change. �� Lenz's law determines current direction. �� Standard motional emf assumes uniform time-independent magnetic field.

✅ Correct Answer Explanation

Statement 1 is correct because area A = lx changes linearly when x changes uniformly with time. → Statement 2 is correct because the induced current opposes the change in magnetic flux according to Lenz's law. → Statement 3 is incorrect because Φ = Blx in this simple form assumes B is constant. For a time-varying magnetic field, flux becomes time dependent through B(t). → Statement 4 is correct because ε = Blv is derived for the case where the magnetic field is perpendicular to the plane of motion.

❓ Why Other Options Are Incorrect

�� Option A → Contains incorrect Statement 3.
�� Option C → Contains incorrect Statement 3.
�� Option D → Contains incorrect Statement 3.

🎯 Strategy

Used

�� Elimination

Application:

�� Identify the statement inconsistent with the assumptions used in motional emf derivation.

Final Logic:

�� Statement 3 is false; Statements 1, 2 and 4 are correct.

🧠 Memory Trick

"Constant B → Φ = Blx."

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

2 Analyzing the velocity and field interaction in motional emf will find that:
Statements:
1. The velocity vector must be parallel to the magnetic field vector to induce a current.
2. Replacing the constant velocity with acceleration would make the rate of change of flux independent of time.
3. Moving the arm PQ to the left decreases the enclosed area, thus reducing the magnetic flux and inducing a current to oppose this change.
4. The relative motion between the stationary rails and the magnetic field is the only cause of emf.

A 1 only

B 2 only

C 3 only

D 4 only

💡 Short Explanation

�� Area decreases. �� Flux decreases. �� Current opposes the decrease.

✅ Correct Answer Explanation

Statement 3 is correct because moving PQ left decreases area and magnetic flux. → According to Lenz's law, induced current opposes the decrease in flux. → Statement 1 is false because maximum motional emf occurs when v is perpendicular to B. → Statement 2 is false because acceleration generally makes flux change time dependent. → Statement 4 is false because emf arises from conductor-field interaction and changing flux.

❓ Why Other Options Are Incorrect

�� Option A → Statement 1 is incorrect.
�� Option B → Statement 2 is incorrect.
�� Option D → Statement 4 is incorrect.

🎯 Strategy

Used

�� Elimination

Application:

�� Test each statement using Faraday's law and Lorentz force concepts.

Final Logic:

�� Only Statement 3 is correct.

🧠 Memory Trick

"Decrease Flux → Oppose Loss."

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

3 If the magnetic flux enclosed by the loop PQRS is Φ_B = Blx, and the arm PQ moves causing x to decrease, the rate of change of flux dΦ_B/dt corresponds to the induced emf. Thus, if the direction of motion is reversed (x increases), the induced current will

A remain in the same direction

B flow in the opposite direction

C drop to zero immediately

D increase exponentially

💡 Short Explanation

�� Flux change reverses sign. �� Emf reverses sign. �� Current reverses direction.

✅ Correct Answer Explanation

Reversing the motion reverses the sign of dΦ/dt. → Since ε = −dΦ/dt, the induced emf changes sign. → Therefore, the induced current flows in the opposite direction.

❓ Why Other Options Are Incorrect

�� Option A → Current direction depends on the sign of flux change.
�� Option C → Flux is still changing, so emf exists.
�� Option D → No exponential dependence occurs.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Follow the sign change in flux variation.

Final Logic:

�� Reverse motion → Reverse current.

🧠 Memory Trick

"Reverse Motion = Reverse Current."

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

4 Choose the incorrect statement about the motional EMF equation:

A It provides the basis for Faraday's law when the conductor is moving in a uniform, time-independent magnetic field.

B It shows that the induced emf is directly proportional to the magnetic field B, length l, and velocity v.

C It explicitly incorporates the varying electric field E generated by static charges.

D It can be derived using the concept of rate of change of enclosed magnetic flux.

💡 Short Explanation

�� Motional emf is magnetic in origin. �� Uses Lorentz force and flux change. �� Static-charge electric fields are not involved.

✅ Correct Answer Explanation

The motional emf equation ε = Blv is derived using magnetic force on moving charges or via Faraday's law. → It does not explicitly involve electric fields generated by static charges. → Therefore Statement C is incorrect.

❓ Why Other Options Are Incorrect

�� Option A → Correct statement.
�� Option B → Direct consequence of ε = Blv.
�� Option D → Valid derivation method.

🎯 Strategy

Used

�� Odd One Out

Application:

�� Identify the statement unrelated to motional emf derivation.

Final Logic:

�� Static-charge electric fields are not part of the derivation.

🧠 Memory Trick

"Motional EMF → Magnetic Origin."

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

5 Match List I with List II for the theoretical basis of Lorentz Force Origin in motional EMF.

List I	List II
(1) Magnetic force on an arbitrary charge q	(a) Towards Q along the rod
(2) Moving charge speed	(b) Identical to rod speed v
(3) Force direction	(c) Requires moving charges in a magnetic field
(4) Basis for the derivation	(d) qvB

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

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

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

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

💡 Short Explanation

�� Magnetic force = qvB. �� Charges move with rod speed. �� Lorentz force requires moving charges.

✅ Correct Answer Explanation

1 → d because magnetic force magnitude is qvB. → 2 → b because charge carriers move with the rod at speed v. → 3 → a because force pushes charges toward Q. → 4 → c because Lorentz force requires moving charges in a magnetic field.

❓ Why Other Options Are Incorrect

�� Option B → Force magnitude and speed mismatched.
�� Option C → Speed and direction mismatched.
�� Option D → Incorrect assignment of Lorentz force expression.

🎯 Strategy

Used

�� Option Grouping

Application:

�� Match each physical quantity with its standard definition.

Final Logic:

�� Direct Lorentz-force matching gives 1-d, 2-b, 3-a, 4-c.

🧠 Memory Trick

"qvB → v → Direction."

📚 Class XII – Physics Part I – Chapter: Moving Charges and Magnetism – Topic: Lorentz Force

6 In a metallic rod moving perpendicular to a magnetic field, the uniform nature of the magnetic force on all charge carriers and the relative magnitude of the force depend on:

A Varying position, Velocity squared

B Independent of position, Speed v

C Conductor length, Constant velocity

D Electric field strength, Variable field B

💡 Short Explanation

�� F = qvB. �� Uniform B gives uniform force. �� Depends on speed.

✅ Correct Answer Explanation

The Lorentz force is F = qvB. → In a uniform magnetic field, force is independent of position. → The magnitude depends directly on speed v.

❓ Why Other Options Are Incorrect

�� Option A → Force is not proportional to v².
�� Option C → Force on a charge does not depend on rod length.
�� Option D → Electric field is not the primary cause.

🎯 Strategy

Used

�� Elimination

Application:

�� Apply Lorentz force equation.

Final Logic:

�� Uniform field + speed determines force.

🧠 Memory Trick

"qvB, not qv²B."

📚 Class XII – Physics Part I – Chapter: Moving Charges and Magnetism – Topic: Magnetic Force

7 A charge q of 2.0 C is moved from P to Q in a rod of length 0.5 m, moving at 4.0 m/s in a 0.1 T magnetic field. What is the work done on this charge by the magnetic force?

A 0.4 J

B 0.2 J

C 0.8 J

D 1.6 J

💡 Short Explanation

�� W = qvBl. �� Substitute values. �� Obtain 0.4 J.

✅ Correct Answer Explanation

W = qvBl → W = (2)(4)(0.1)(0.5) → W = 0.4 J

❓ Why Other Options Are Incorrect

�� Option B → Half the correct value.
�� Option C → Double the correct value.
�� Option D → Four times the correct value.

🎯 Strategy

Used

�� Substitution

Application:

�� Direct substitution into W = qvBl.

Final Logic:

�� W = 0.4 J.

🧠 Memory Trick

"Work = qvBl."

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

8 Choose the correct statements about EMF as Work per Charge:
Statements:
1. ε = W/q implies that work done in moving a unit charge determines the emf.
2. This definition mathematically connects the Lorentz force derivation directly to Faraday's law.
3. It requires the magnetic field to be non-uniform to perform work.
4. The work done is purely due to the q(v × B) component of the Lorentz force.

A 1, 2, 3

B 1, 2, 4

C 2, 3, 4

D 1, 3, 4

💡 Short Explanation

�� EMF = Work/Charge. �� Links Lorentz force to Faraday's law. �� Uniform magnetic field is sufficient.

✅ Correct Answer Explanation

Statement 1 is the definition of emf. → Statement 2 correctly connects microscopic and macroscopic descriptions of induction. → Statement 3 is incorrect because a uniform magnetic field can also produce motional emf. → Statement 4 is correct in the standard motional emf derivation.

❓ Why Other Options Are Incorrect

�� Option A → Contains incorrect Statement 3.
�� Option C → Contains incorrect Statement 3.
�� Option D → Contains incorrect Statement 3.

🎯 Strategy

Used

�� Elimination

Application:

�� Identify the false statement.

Final Logic:

�� Statement 3 is incorrect.

🧠 Memory Trick

"EMF = Work ÷ Charge."

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

9 While moving charges experience a magnetic force q(v × B), an emf is nevertheless induced in a completely stationary conductor when

A the charges within it move at the drift velocity

B it is placed in a static magnetic field

C it is exposed to a time-varying magnetic field

D it is disconnected from a closed circuit

💡 Short Explanation

�� Changing magnetic field creates electric field. �� Stationary conductor can still have emf. �� Faraday's law applies.

✅ Correct Answer Explanation

A changing magnetic field generates an induced electric field. → This electric field produces emf even in a stationary conductor.

❓ Why Other Options Are Incorrect

�� Option A → Drift velocity is not the cause.
�� Option B → Static magnetic field alone does not induce emf.
�� Option D → Circuit connection does not determine emf generation.

🎯 Strategy

Used

�� Elimination

Application:

�� Apply Faraday's law.

Final Logic:

�� Changing B induces emf.

🧠 Memory Trick

"Changing B → Induced E."

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

10 Identify the correct statements regarding induced electric fields:
Statements:
1. They are generated by time-varying magnetic fields.
2. They explain the existence of induced emf when the conductor is stationary.
3. They are identical in property to electric fields produced by static charges.
4. The force on a charge in a stationary conductor is given solely by qE.

A 1, 2, 4

B 1, 3, 4

C 2, 3, 4

D 1, 2, 3

💡 Short Explanation

�� Induced fields arise from changing B. �� They explain stationary-conductor emf. �� They are non-conservative.

✅ Correct Answer Explanation

Statement 1 is correct by Faraday's law. → Statement 2 is correct because induced electric fields explain emf in stationary conductors. → Statement 3 is incorrect because induced electric fields are non-conservative, unlike electrostatic fields. → Statement 4 is correct because for stationary charges magnetic force is zero and electric force qE acts.

❓ Why Other Options Are Incorrect

�� Option B → Contains incorrect Statement 3.
�� Option C → Contains incorrect Statement 3.
�� Option D → Contains incorrect Statement 3.

🎯 Strategy

Used

�� Elimination

Application:

�� Identify the statement inconsistent with induced electric fields.

Final Logic:

�� Statement 3 is false; 1, 2 and 4 are correct.

🧠 Memory Trick

"Induced E forms loops."

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

11 Which of the following statements is incorrect regarding the behavior of a metallic rod rotating about a central pivot in a uniform magnetic field?

A The free electrons move towards the outer end due to the Lorentz force.

B An emf is generated across the length of the rod.

C The Lorentz force acts equally on charge carriers at the pivot and at the rim.

D A steady state is reached where no more electrons flow outwardly.

💡 Short Explanation

�� Velocity depends on radius. �� Lorentz force depends on velocity. �� Force is not the same everywhere.

✅ Correct Answer Explanation

For a rotating rod, linear speed is v = ωr. → At the pivot, r = 0 and hence v = 0, so magnetic force is zero. → At the rim, velocity is maximum and Lorentz force is maximum. → Therefore, the force is not equal at the pivot and rim.

❓ Why Other Options Are Incorrect

�� Option A → Correct; electrons are displaced due to Lorentz force.
�� Option B → Correct; charge separation produces emf.
�� Option D → Correct; equilibrium is eventually reached.

🎯 Strategy

Used

�� Elimination

Application:

�� Use v = ωr to check force variation.

Final Logic:

�� Since F = qvB and v varies with r, force cannot be equal everywhere.

🧠 Memory Trick

"Rim fastest, force greatest."

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

12 If free electrons move towards the outer end of a rotating rod, a resulting separation of charges produces an emf. Thus, if the rod rotates at a steady state, the potential difference across the ends of the rod will be

A zero

B a constant non-zero value

C constantly increasing without bound

D alternating smoothly between positive and negative

💡 Short Explanation

�� Charge separation occurs. �� Equilibrium develops. �� Constant emf is produced.

✅ Correct Answer Explanation

Rotation continuously produces charge separation. → A balancing electric field develops. → At steady state, the potential difference becomes constant and non-zero.

❓ Why Other Options Are Incorrect

�� Option A → Charge separation creates emf.
�� Option C → Potential does not increase indefinitely.
�� Option D → Uniform rotation in constant B gives steady emf.

🎯 Strategy

Used

�� Contextual/Tonal Matching

Application:

�� Interpret the meaning of "steady state".

Final Logic:

�� Steady state implies constant potential difference.

🧠 Memory Trick

"Steady rotation → Steady emf."

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

13 When integrating the segment EMF for a rotating rod, the dependence of velocity on radius r and the resulting EMF expression are:

A Linear, Proportional to R²

B Quadratic, Proportional to R

C Constant, Proportional to R²

D Linear, Proportional to R

💡 Short Explanation

�� v = ωr. �� Velocity varies linearly with r. �� Integration produces R² dependence.

✅ Correct Answer Explanation

Linear velocity: v = ωr → Elemental emf: dε = Bωrdr → Integrating from 0 to R: ε = ½BωR² → Therefore velocity varies linearly with r and emf is proportional to R².

❓ Why Other Options Are Incorrect

�� Option B → Velocity is not quadratic in r.
�� Option C → Velocity is not constant.
�� Option D → Final emf depends on R², not R.

🎯 Strategy

Used

�� Substitution

Application:

�� Use v = ωr in emf integration.

Final Logic:

�� Linear velocity dependence yields quadratic radius dependence.

🧠 Memory Trick

"v ∝ r, emf ∝ R²."

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

14 Match List I with List II for variables in the Rotational EMF equation.

List I	List II
(1) Angular frequency ω	(a) R²/2
(2) Linear velocity v	(b) 2πν
(3) Radius R	(c) Upper limit of integration
(4) Integral of r dr from 0 to R	(d) rω

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

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

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

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

💡 Short Explanation

�� ω = 2πν. �� v = rω. �� R is upper integration limit. �� ∫rdr = R²/2.

✅ Correct Answer Explanation

1 → b because ω = 2πν. → 2 → d because linear velocity is v = rω. → 3 → c because R is the upper limit of integration. → 4 → a because: ∫₀ᴿ r dr = R²/2

❓ Why Other Options Are Incorrect

�� Option B → Angular frequency and velocity mismatched.
�� Option C → Integral and velocity mismatched.
�� Option D → Multiple incorrect pairings.

🎯 Strategy

Used

�� Option Grouping

Application:

�� Match standard rotational formulas.

Final Logic:

�� Direct formula matching gives 1-b, 2-d, 3-c, 4-a.

🧠 Memory Trick

"2πν → rω → R → R²/2."

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

15 The method of finding emf using the rate of change of area evaluates the flux through a closed loop OPQ, demonstrating that

A the induced emf equals B multiplied by the rate at which the sector area is swept

B a resistor must be absent for the area rule to apply

C the potential difference only exists if the magnetic field changes over time

D the angle θ is independent of the angular speed ω

💡 Short Explanation

�� ε = B(dA/dt). �� Area sweep changes flux. �� Faraday's law applies.

✅ Correct Answer Explanation

Magnetic flux: Φ = BA → Therefore: ε = B(dA/dt) → The induced emf equals magnetic field multiplied by the rate of area swept.

❓ Why Other Options Are Incorrect

�� Option B → Resistor presence is irrelevant.
�� Option C → Area change alone can induce emf.
�� Option D → θ changes with ω.

🎯 Strategy

Used

�� Substitution

Application:

�� Use Faraday's law with changing area.

Final Logic:

�� Flux change comes from area sweep.

🧠 Memory Trick

"Area Sweep = EMF."

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

16 The area of a sector swept by a rod of length 1 m rotating at 50 rev/s is increasing. If the uniform magnetic field is 1 T, what is the value of d(Area)/dt, and consequently the induced emf?

A 50 m²/s and 50 V

B 157 m²/s and 157 V

C 314 m²/s and 314 V

D 25 m²/s and 25 V

💡 Short Explanation

�� ω = 2πν. �� dA/dt = ½R²ω. �� ε = B(dA/dt).

✅ Correct Answer Explanation

ν = 50 rev/s → ω = 2π(50) = 100π rad/s → dA/dt = ½R²ω = ½(1²)(100π) = 50π ≈ 157 m²/s → ε = B(dA/dt) = 1 × 157 = 157 V

❓ Why Other Options Are Incorrect

�� Option A → Uses ν instead of ω.
�� Option C → Double the correct value.
�� Option D → Incorrect calculation.

🎯 Strategy

Used

�� Substitution

Application:

�� Convert frequency to angular speed before calculation.

Final Logic:

�� dA/dt = 50π ≈ 157.

🧠 Memory Trick

"50 rev/s → 100π rad/s."

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

17 When determining the potential between the axle and rim of a metallic wheel with spokes rotating in a magnetic field,
1. the total emf is the sum of the emfs of individual spokes
2. the presence of the rim shorts out the spokes, resulting in zero emf
3. the emf is equivalent to that of a single solid rotating rod of the same radius
4. the emf depends on the thickness of the individual spokes
Choose the correct options:

A 1

B 2

C 3

D 4

💡 Short Explanation

�� Each spoke behaves like a rotating rod. �� Spokes are in parallel. �� Same emf as one rod.

✅ Correct Answer Explanation

Each spoke generates the same emf. → Since all spokes connect the same axle and rim, they are effectively in parallel. → The wheel behaves like a single rotating rod of radius R.

❓ Why Other Options Are Incorrect

�� Option A → Parallel emfs do not add.
�� Option B → Rim does not eliminate emf.
�� Option D → Thickness does not determine induced emf.

🎯 Strategy

Used

�� Elimination

Application:

�� Analyze spoke connection configuration.

Final Logic:

�� Wheel acts as one rotating rod.

🧠 Memory Trick

"Wheel = One Rod."

📚 Class XII – Physics Part I – Chapter: Electromagnetic Induction – Topic: Axle-Rim Potential

18 Choose the correct statements about the parallel EMF configuration in a wheel:
1. The spokes act as identical emf sources connected in parallel.
2. The potential difference across any individual spoke equals the potential difference of the entire wheel.
3. The number of spokes alters the total internal resistance, but not the theoretical induced emf.
4. Adding more spokes increases the induced emf proportionally.

A 1, 2, 3

B 1, 2, 4

C 2, 3, 4

D 1, 3, 4

💡 Short Explanation

�� Spokes are parallel sources. �� Voltage remains same. �� Internal resistance may change.

✅ Correct Answer Explanation

Statement 1 is correct because spokes connect the same axle and rim. → Statement 2 is correct because parallel branches have the same potential difference. → Statement 3 is correct because adding spokes can alter effective resistance but not emf. → Statement 4 is incorrect because parallel emfs do not add.

❓ Why Other Options Are Incorrect

�� Option B → Contains incorrect Statement 4.
�� Option C → Contains incorrect Statement 4.
�� Option D → Contains incorrect Statement 4.

🎯 Strategy

Used

�� Elimination

Application:

�� Identify the incorrect statement about parallel sources.

Final Logic:

�� Parallel branches keep the same emf.

🧠 Memory Trick

"Parallel → Same Voltage."

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

19 Horizontal Component Effects on a rotating wheel statements:
Statements:
1. A wheel rotating in a plane normal to H_E generates an emf proportional to H_E.
2. If the wheel is rotated perfectly parallel to H_E, the induced motional emf across its spokes is maximized.
3. 1 Gauss (G) equals 10⁻⁴ T, which is crucial for converting H_E into standard SI units.
4. Earth's magnetic field is a realistic source of a uniform B field over the area of a small wheel.

A 1, 2, 3

B 1, 3, 4

C 2, 3, 4

D 1, 2, 4

💡 Short Explanation

�� Maximum emf requires field normal to rotation plane. �� Unit conversion is correct. �� Earth's field is approximately uniform locally.

✅ Correct Answer Explanation

Statement 1 is correct because emf is proportional to magnetic field strength. → Statement 2 is incorrect because emf is maximized when the wheel rotates in a plane normal to the field, not parallel to it. → Statement 3 is correct since 1 G = 10⁻⁴ T. → Statement 4 is correct for small laboratory dimensions.

❓ Why Other Options Are Incorrect

�� Option A → Contains incorrect Statement 2.
�� Option C → Contains incorrect Statement 2.
�� Option D → Contains incorrect Statement 2.

🎯 Strategy

Used

�� Elimination

Application:

�� Identify the false statement regarding field orientation.

Final Logic:

�� Statement 2 violates the condition for maximum flux cutting.

🧠 Memory Trick

"Normal Field → Maximum EMF."

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

20 If a wire of length l falls horizontally with a velocity v perpendicular to the horizontal component of the earth's magnetic field B, the motional emf is Blv. Thus, if a 10 m long wire falls at 5.0 m/s and the induced emf is 1.5 × 10⁻³ V, the horizontal component of the earth's magnetic field must be

A 0.30 × 10⁻⁴ T

B 3.0 × 10⁻⁴ T

C 0.15 × 10⁻⁴ T

D 1.5 × 10⁻⁴ T

💡 Short Explanation

�� ε = Blv. �� Rearrange for B. �� Substitute values.

✅ Correct Answer Explanation

B = ε/(lv) → B = (1.5 × 10⁻³)/(10 × 5) → B = 3 × 10⁻⁵ T → B = 0.30 × 10⁻⁴ T

❓ Why Other Options Are Incorrect

�� Option B → Ten times larger than actual value.
�� Option C → Half the correct value.
�� Option D → Five times larger than actual value.

🎯 Strategy

Used

�� Substitution

Application:

�� Rearrange ε = Blv and solve for B.

Final Logic:

�� B = 0.30 × 10⁻⁴ T.

🧠 Memory Trick

"Find B? Divide by lv."

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

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