Test Compilation PHY 102 14

Study Summary from Physics/Electrical Engineering Questions

This summary is compiled from the provided document containing various multiple-choice questions related to physics and electrical engineering. Where answers were marked, they are indicated. Discrepancies or corrections based on standard physics principles are noted.

1. Electrostatics & Electric Fields

Question: Force per Unit Charge (Pages 9, 15)

The force per unit charge is known as?

• a. Electric current
• b. Electric flux
• c. Electric field intensity
• d. Electric potential

Note: Electric field intensity (E) is defined as force (F) per unit charge (q), E = F/q.

Question 9: Electric Field in Hollow Metallic Sphere (Pages 2, 5, 12)

The electric field associated with a uniformly charged hollow metallic sphere is the greatest at:

• a. the center of the sphere.
• b. the sphere's outer surface.
• c. the sphere's inner surface.
• d. infinity.

Note: For a uniformly charged hollow metallic sphere, the electric field inside (including the center) is zero. The field is maximum at the surface and decreases with distance outside the sphere.

Question 11: Electric Field Direction/Magnitude (Positive Point Charge) (Page 14)

The diagram below shows a positive point charge Q. Which of the following describes the magnitude and direction of the electric field at points r and s respectively?

• a. Same and Away from Q
• b. Same and towards Q
• c. different and Away from Q
• d. different and towards Q

Note: Electric field lines originate from positive charges and point away. The magnitude of the electric field decreases with distance (E ∝ 1/r²), so it will be different at points r and s.

Question 4: Electric Field Direction/Magnitude (Positive Point Charge) (Page 19)

The diagram below shows a positive point charge Q. Which of the following describes the magnitude and direction of the electric field at points r and s respectively?

• a. Same and Away from Q
• b. Same and towards Q (Marked Answer - Incorrect)
• c. different and Away from Q
• d. different and towards Q

Correction: This marked answer is incorrect. For a positive point charge, the electric field points away from the charge, and its magnitude is different at different distances. Option 'c. different and Away from Q' is correct.

Question 12: Equipotential Surface (Pages 4, 14, 33)

An equipotential surface is adjacent points with:

• a. same electric potential
• b. zero electric potential
• c. opposite electric potential
• d. varying electric potential

Note: By definition, all points on an equipotential surface have the same electric potential.

Question 1: Electric Flux Intensity (Pages 23, 27, 30, 36)

The electric field intensity of 100 N/C is applied on circular surface area at an angle of 60° to the surface. What is the flux intensity in Nm²/C through the circular surface of radius 7m?

• a. 3850
• b. 7700/√3 (Marked Answer - Possible discrepancy)
• c. 7700√3
• d. 7700

Note: For radius 7m, Area A = π(7)² = 49π m². If the angle with the surface is 60°, the angle with the normal is 30°. Electric flux Φ = E A cos(30°) = 100 * 49π * (√3/2) ≈ 13334 Nm²/C. If the angle with the surface is 30°, the angle with the normal is 60°. Φ = E A cos(60°) = 100 * 49π * 0.5 ≈ 7700 Nm²/C (Option d). The marked answer 'b' (7700/√3 ≈ 4445) does not seem to align with standard calculations for this problem. There's also a version of this question on Pages 30 and 36 with different parameters (Area = 0.025m²) and options, where the marked answer is 'd. 0.196Nm²/C', also showing inconsistency with calculation.

Question 10: Charge on a Metal Sphere (Electrostatic Paint) (Page 20)

An electrostatic paint has a 200mm diameter metal sphere at a potential of 25KV that repels paint droplets onto a grounded object. What charge is on the sphere?

• a. 2.78μC
• b. 27.8μC
• c. 0.278μC
• d. 0.0278μC

Calculation: V = kQ/R. Given V = 25 KV = 25000 V, R = (200 mm)/2 = 0.1 m, k = 9 x 10⁹ Nm²/C². Q = VR/k = (25000 * 0.1) / (9 x 10⁹) = 2.777 x 10⁻⁷ C = 0.2777 µC. This matches option c.

Question 15: Potential Difference for Helium Nucleus (Pages 21, 22)

Find the potential difference required to give a helium nucleus (q = 3.2 x 10⁻¹⁹ C) whose kinetic energy is 4.8 × 10³ eV. (where 1eV = 1.6 x 10⁻¹⁹ J).

• a. 4.8 x 10³ V
• b. 1.2 x 10³ V
• c. 3.6 x 10³ V
• d. 2.4 x 10³ V

Calculation: KE = qV. Convert KE to Joules: KE = 4.8 × 10³ eV × (1.6 × 10⁻¹⁹ J/eV) = 7.68 × 10⁻¹⁶ J. Then V = KE / q = (7.68 × 10⁻¹⁶ J) / (3.2 × 10⁻¹⁹ C) = 2400 V = 2.4 × 10³ V. This matches option d.

Question 15: Electric Field from Point Charge & Force (Page 34)

Find the electric field at a point 0.2 m from a charge of 20 mC, what force will the field exert on a charge of 10 mC, placed at that point?

• a. 22.00N
• b. 47.00N
• c. 45.00N
• d. 55.00N

Calculation: Assuming charges are in microcoulombs (µC) instead of millicoulombs (mC) for the options to be reasonable. Q₁ = 20 µC = 20 × 10⁻⁶ C, Q₂ = 10 µC = 10 × 10⁻⁶ C, r = 0.2 m. E = kQ₁/r² = (9 × 10⁹) × (20 × 10⁻⁶) / (0.2)² = 4.5 × 10⁶ N/C. Force F = E × Q₂ = (4.5 × 10⁶ N/C) × (10 × 10⁻⁶ C) = 45 N. This matches option c.

Question 14: Electric Field for Electron's Weight (Page 35)

What is the magnitude of an electric field in which the force on an electron is equal in magnitude to the weight of an electron? (mass of an electron = 9.11 x 10⁻³¹ kg, charge on an electron = 1.6 x 10⁻¹⁹ C)

• a. 2.00 x 10⁻¹¹ N/C
• b. 4.65 x 10⁻¹¹ N/C
• c. 5.00 x 10⁻¹¹ N/C
• d. 5.58 x 10⁻¹¹ N/C

Calculation: qE = mg => E = mg/q = (9.11 × 10⁻³¹ kg × 9.8 m/s²) / (1.6 × 10⁻¹⁹ C) = 5.5799 × 10⁻¹¹ N/C. This matches option d.

Question 11: Electric Field Intensity (Page 42)

Find the electric field intensity at a distance of of 20cm from a charge of 2nC.

• a. 1,800 N/C
• b. 0.045 N/C
• c. N/C (Truncated)

Calculation: E = kQ/r². Q = 2 nC = 2 × 10⁻⁹ C, r = 20 cm = 0.2 m. E = (9 × 10⁹ Nm²/C²) × (2 × 10⁻⁹ C) / (0.2 m)² = 18 / 0.04 = 450 N/C. None of the visible options match.

Question 1: Relationship between Charges (Electric Field Lines) (Page 47)

Determine the most possible relationship between q1 and q2 in figure B (Diagram shows field lines originating from q1 and terminating on q2, with more lines from q1 than to q2).

• a. q1 = q2
• b. q1 = 3q2
• c. q1 < q2
• d. q1 > q2

Note: Field lines indicate q1 is positive and q2 is negative. The number of lines indicates |q1| ≈ 3|q2|. While "q1 = 3q2" is problematic for signs, it indicates the magnitude ratio.

2. Current, Resistance & Circuits

Question 6: Capacitance Calculation (Parallel Plate Capacitor) (Pages 3, 9, 10)

In a parallel-plate capacitor, the distance d between the plates is 3cm, the area of each plate is 15cm², and the voltage across them is 5V. Calculate the capacitance. Take permittivity of free space to be ε₀ = 8.854 × 10⁻¹² C/Nm².

• a. 3.33 x 10⁻¹³ F
• b. 9.31 x 10⁻¹¹ F
• c. 4.43 x 10⁻¹³ F
• d. 1.29 x 10⁻¹² F

Calculation: C = ε₀A/d. A = 15 cm² = 15 × 10⁻⁴ m². d = 3 cm = 3 × 10⁻² m. C = (8.854 × 10⁻¹² × 15 × 10⁻⁴) / (3 × 10⁻²) = (8.854 × 15 / 3) × 10⁻¹⁴ = 8.854 × 5 × 10⁻¹⁴ = 44.27 × 10⁻¹⁴ F = 4.427 × 10⁻¹³ F. This matches option c. The voltage (5V) is extraneous information for calculating capacitance.

Question 7: Circuit Element Storing Energy in Electromagnetic Field (Pages 3, 5, 11)

Which of following circuit element stores energy in electromagnetic field?

• a. Inductor
• b. Condenser
• c. Resistor
• d. Capacitor

Note: Inductors store energy in a magnetic field. Capacitors store energy in an electric field. Resistors dissipate energy as heat.

Question 15: Calculate Total Resistance (Circuit Diagram) (Page 8)

Calculate the total resistance between the points A and B. (Diagram shows a 4Ω resistor in series with a parallel combination of (1Ω+2Ω) and 3Ω).

• a. 7 ohm
• b. 0 ohm
• c. 7.67 ohm
• d. 0.48 ohm

Calculation: The series combination in the top branch is 1Ω + 2Ω = 3Ω. This 3Ω is in parallel with the 3Ω resistor in the bottom branch: R_parallel = (3Ω × 3Ω) / (3Ω + 3Ω) = 9Ω²/6Ω = 1.5Ω. This parallel combination is in series with the 4Ω resistor. Total Resistance = 4Ω + 1.5Ω = 5.5Ω. The calculated answer (5.5Ω) does not appear in the given options. No answer was marked.

Question 2: Area under Current-Time Graph (Page 16)

The area under the current-time graph is equivalent to

• a. Dielectric
• b. Potential difference
• c. Magnitude of charge
• d. Capacitance

Note: Current I = dQ/dt. Integrating current over time (∫I dt) gives the total charge (Q) that flows.

Question 13: Peak vs RMS Value (AC Signal) (Pages 4, 7)

Which of the following is the correct relationship between PEAK value and R.M.S. value of an alternating signal.

• a. peak = r.m.s. x √2
• b. r.m.s. = peak x √2
• c. peak = r.m.s./√2
• d. peak = √2/r.m.s.

Note: For a sinusoidal AC signal, V_rms = V_peak / √2, or V_peak = V_rms × √2.

Question 5: RMS Value of Sinusoidal Current (Page 18)

The equation of a sinusoidal waveform is given by I = 15sin ωt. Calculate the r.m.s value of current at t = 5s.

• a. 21.7A
• b. 15.8A
• c. 10.6A
• d. 7.5A

Calculation: For a sinusoidal current I = I_peak sin(ωt), the RMS value is I_rms = I_peak / √2. Here, I_peak = 15A. So, I_rms = 15 / √2 ≈ 10.606 A. This matches option c. The time (t=5s) is irrelevant for RMS value.

Question 6: AC Circuit Properties (Page 28)

Which of the following is true for a.c. circuit?

• a. All elements in the circuit dissipate power.
• b. Reactive elements only are associated with apparent power.
• c. power is dependent on impedance.
• d. power factor of purely resistive element is zero.

Note: In an AC circuit, real power (P) is related to impedance (Z) by P = I²R = I²Z cos(φ). Reactive elements are associated with reactive power, not apparent power exclusively. The power factor for a purely resistive element is 1, not zero.

Question: Why Transmit Electrical Energy at High Voltage? (Pages 29, 31)

Why is electrical energy usually transmitted at high voltage?

• a. the resistance of the transmission cables is as small as possible
• b. the transmission cables are safer to handle
• c. as little energy as possible is wasted in the transmission cables
• d. the transmission system does not require transformers e. the current in the transmission cables is as large as possible

Note: Power loss in transmission lines is given by P_loss = I²R. By transmitting at high voltage, the current (I) for a given amount of power (P = VI) is reduced, significantly reducing energy loss.

Question 8: Electroscope with Negatively Charged Rod (Page 31)

What are you likely to observe when a negatively charge rod is brought near the knob of a charged electroscope, with a positive residual charge?

• a. The leaf will collapse immediately
• b. The leaf diverges permanently
• c. The leaf remains unchanged
• d. The leaf collapses and then diverges

Note: A positively charged electroscope has diverged leaves. When a negatively charged rod is brought near, electrons from the rod repel electrons in the knob down to the leaves. These electrons neutralize some of the positive charge, causing the leaves to collapse. If enough electrons are transferred, the leaves can become negatively charged, causing them to diverge again.

Question 3: Coulomb's Law - Force between Charges (Page 33)

Two point charges are 6 cm apart. They are moved to a new separation of 3 cm. By what factor does the resulting mutual force between them change?

• a. 1/2
• b. 2
• c. 4
• d. 1/4

Calculation: Coulomb's force F ∝ 1/r². If the distance r is halved (from 6 cm to 3 cm), the force increases by a factor of (r_old/r_new)² = (6/3)² = 2² = 4.

Question 12: Force between Parallel Wires (Page 36)

Two long parallel wire are 40mm apart. If current of 20A and 30A are applied to the wire in the same direction. Find the force of attraction of wire per unit length. (Question text states 'A. 0.003N/m' as part of the question)

• a. 0.3N/m
• b. 3N/m
• c. 30N/m

Calculation: Force per unit length F/L = (μ₀ * I₁ * I₂) / (2πd). μ₀ = 4π × 10⁻⁷ T m/A. I₁ = 20A, I₂ = 30A, d = 40 mm = 0.04 m. F/L = (4π × 10⁻⁷ × 20 × 30) / (2π × 0.04) = (2 × 10⁻⁷ × 600) / 0.04 = 0.03 N/m. Neither the value given in the question (0.003 N/m) nor the options match the calculation for d = 40mm. If d were 4mm, then F/L would be 0.3N/m (option a).

Question 7: Peak Voltage of AC Signal (Pages 37, 38)

If a 50µF capacitor is connected across an alternating electromotive force describe by v = 30 sin (2000πt) volts. What is the peak voltage?

• a. 1500 V
• b. 60000 V
• c. 20002000 V
• d. 3030 V (Marked Answer - Incorrect)

Correction: For an alternating electromotive force given by v = V_peak sin(ωt), the peak voltage (V_peak) is the amplitude of the sine function. From the equation v = 30 sin (2000πt) volts, the peak voltage is directly 30 V. The marked answer 3030 V is incorrect. The capacitance value is extraneous for finding the peak voltage.

Question 13: Resistance of a Stretched Wire (Page 39)

If the resistance of a wire is r, and the wire is stretched to double its length, then its resistance is?

• a. r/2
• b. 2r
• c. 4r
• d. r/4

Calculation: When a wire is stretched, its volume (V) remains constant. Resistance R = ρL/A. Since V = AL, A = V/L. So, R = ρL/(V/L) = ρL²/V. As ρ and V are constant, R ∝ L². If length L is doubled (L_new = 2L_old), then R_new = (2L_old)² * (ρ/V) = 4 * (ρL_old²/V) = 4 * R_old. The new resistance is 4r.

Question 11: Capacitance of Parallel Plate Capacitor (Pages 41, 43)

The dimensions of plates of a parallel plate capacitor are 8 cm by 8 cm and they are separated by a distance of 2 mm. Calculate the capacitance of the capacitor if air is between the plates

• a. 3.54 x 10⁻¹⁰ F
• b. 2.77 x 10⁻¹² F
• c. 2.83 x 10⁻¹¹ F
• d. 3.54 x 10⁻¹⁰ F

Calculation: C = ε₀A/d. A = (8 cm)² = (0.08 m)² = 0.0064 m². d = 2 mm = 0.002 m. ε₀ = 8.854 × 10⁻¹² F/m. C = (8.854 × 10⁻¹² × 0.0064) / 0.002 = 2.833 × 10⁻¹¹ F. This matches option c.

Question: Current Flow in Parallel DC Circuit (Page 43)

Connected in parallel across a dc source, greater current flows through the

• a. higher resistance
• b. lower resistance
• c. same current flows through each
• d. same charge and same voltage

Note: In a parallel circuit, the voltage across each branch is the same. According to Ohm's Law (I = V/R), a lower resistance will result in a greater current for a constant voltage.

Question 8: Dielectric Insertion in Capacitor (Potential Constant) (Page 44)

A dielectric is slowly inserted between the plates of a parallel plate capacitor, while the potential difference between the plates is held constant by a battery. As it is being inserted:

• a. the capacitance, the potential difference, and the charge on the positive plate all increase
• b. the capacitance and the charge on the positive plate increase but the potential difference between the plates remains the same
• c. the capacitance, the potential difference, and the charge on the positive plate all decrease
• d. the potential difference between the plates increases, the charge on the positive plate decreases, and the capacitance remains the same

Note: When a battery maintains constant potential difference (V), and a dielectric is inserted, capacitance (C) increases (C = kC₀ where k > 1). Since Q = CV and V is constant, the charge (Q) on the plates must also increase.

Question 1: Calculate Equivalent Resistance (Circuit Diagram) (Page 45)

Calculate the equivalent resistance between A and B. (Diagram shows two parallel combinations in series: (5Ω || 10Ω) in series with (15Ω || 20Ω)).

• a. 60 ohm
• b. 15 ohm
• c. 12 ohm
• d. 48 ohm

Calculation: R_p1 = (5×10)/(5+10) = 50/15 = 3.333Ω. R_p2 = (15×20)/(15+20) = 300/35 = 8.571Ω. R_total = R_p1 + R_p2 = 3.333 + 8.571 = 11.904Ω ≈ 12Ω. This matches option c.

Question: Ideal Transformer Calculation (Page 46)

An ideal transformer has 550 turns on the primary and 30 turns on the secondary. What is the maximum output pd if the maximum input voltage is 3300V? And what maximum primary current is required if the maximum current of 11A is drawn from the secondary? Assuming the transformer is 100% efficient

• a. V_s = 180V and I_p = 0.6A
• b. V_s = 108V and I_p = 0.06A
• c. V_s = 180V and I_p = 1.20A
• d. V_s = 180V and I_p = 0.06A (Marked Answer - Incorrect for I_p)

Calculation: 1. V_s / V_p = N_s / N_p => V_s = 3300V × (30 / 550) = 180V. 2. I_p / I_s = N_s / N_p => I_p = 11A × (30 / 550) = 0.6A. The correct answer is V_s = 180V and I_p = 0.6A, which corresponds to option 'a'. The marked answer 'd' has an incorrect primary current.

Question 10: RMS Value of AC Current (Page 48)

The equation of alternating current is i = 42.4sin628t. Then value of current is

• a. 2A
• b. 38 A.
• c. 22 A.
• d. 42.42 A.

Note: The equation i = I_peak sin(ωt) gives I_peak = 42.4 A. If "value of current" refers to the peak current, then option d (42.42 A) is the closest. If it refers to RMS current, I_rms = I_peak / √2 = 42.4 / √2 ≈ 30A, which is not an option.

Question 14: AC Signal through Capacitor (Phase Relationship) (Page 50)

When an alternating signal flows through a capacitor,

• a. Voltage leads current by 90°
• b. current leads voltage by 90°
• c. current lags voltage by 90°
• d. current and voltage are in phase

Note: In a purely capacitive AC circuit, the current leads the voltage by 90 degrees (π/2 radians). A common mnemonic is "ELI the ICE man", where "ICE" stands for Current leads voltage in a Capacitor.

3. Semiconductors & Materials

Question 10: Energy Band for Free Electrons (Pages 2, 12, 13)

The energy band in which free electrons exist is

• a. first band
• b. second band
• c. conduction band
• d. valence band

Note: In materials, the conduction band is where electrons are free to move and conduct electricity. The valence band contains electrons bound to atoms.

Question 14: Doping Silicon with Boron (Page 7)

When silicon is doped with boron, it becomes

• a. neutral conductor
• b. N-type semiconductor
• c. P-type semiconductor
• d. neutral atom

Note: Boron is a trivalent impurity. When added to silicon (which is tetravalent), it creates "holes" (electron deficiencies) in the crystal lattice, making it a P-type (positive-type) semiconductor.

Question 4: Hall Probe Material (Pages 15, 17)

Hall probe is made up of

• a. metals
• b. non metals
• c. semiconductor
• d. radioactive material

Note: Hall probes are typically made from semiconductor materials (like InSb or GaAs) due to their high carrier mobility and sensitivity to the Hall effect.

Question 13: Biasing of P-N Junction (Page 24)

Which of the following is correct about biasing of a p-n junction?

• a. it is reversed biased when p is connected to the positive terminal of a battery
• b. it is reversed biased when p is connected to the negative terminal of a battery
• c. it is forward biased when n is connected to the positive terminal of a battery (Marked Answer - Incorrect)
• d. it is forward biased when p is connected to the negative terminal of a battery

Correction: For reverse bias, the P-type side should be connected to the negative terminal of the battery, and the N-type side to the positive terminal. Option 'b' correctly describes reverse bias. Option 'c' describes reverse bias, not forward bias.

Question 13: Biasing of P-N Junction (Detailed) (Page 26)

Which of the following is/are correct?

1. reverse biasing a diode causes the holes in the N-type material to be withdrawn from the junction;
2. reverse biasing a diode causes the holes in the P-type material to be withdrawn from the junction;
3. forward biasing a diode causes the electron in the N-type material to travel towards the junction;
4. Forward biasing a diode causes the holes in the P-type material to be withdrawn from the junction.

 

• a. II, III
• b. III, IV
• c. I, II
• d. II, III, IV (Marked Answer - Incorrect)

Analysis and Correction:

1. (False) In reverse bias, minority holes in N-type are attracted to the negative terminal, but holes from the N-type are not withdrawn from the junction. Majority carriers (electrons) are pulled away.
2. (True) In reverse bias, majority holes in P-type are pulled away from the junction by the negative terminal.
3. (True) In forward bias, majority electrons in N-type are pushed towards the junction by the negative terminal.
4. (False) In forward bias, majority holes in P-type are pushed towards the junction by the positive terminal, not withdrawn.

Therefore, only statements II and III are correct. The correct option should be 'a'. The marked option 'd' is incorrect because statement IV is false.

Question 10: P-N Junction Diode Truth Statements (Page 40)

Which of the following is true about p-n junction diode?

1. Forward bias current rises linearly.
2. Reverse bias current is absolutely zero.
3. Width of depletion region is unchanged when forward biased.

 

• a. i, iii. (Marked Answer - Partially incorrect)
• b. ii, iii.
• c. 1, ii
• d. none

Analysis and Correction:

1. (Partially True/Approximation) Forward bias current rises exponentially (Shockley equation) after the knee voltage, but can be approximated as linear in some contexts.
2. (False) Reverse bias current is not absolutely zero; there's a small reverse saturation current.
3. (False) The width of the depletion region narrows in forward bias, it is not unchanged.

Given the analysis, neither 'i' nor 'iii' are strictly true. This suggests an error in the question or marked answer. If only 'i' is considered approximately true, then none of the options fit.

4. Magnetism & Electromagnetic Induction

Question 8: Electromagnetic Induction Definition (Pages 5, 11)

The process by which an emf and hence current is generated or induced in a conductor when there is a change in the magnetic flux linking the conductor is called

• a. electromagnetic induction
• b. mutual induction
• c. Faraday's law
• d. Electromagnetic interference

Note: This is the definition of electromagnetic induction, discovered by Faraday. Faraday's law is the quantitative relationship for this phenomenon.

Question 2: Magnetic Flux Calculation (Page 6)

Calculate the flux through a coil loop of area 0.45m² when its plane is inclined at an angle of 60° to a magnetic field of 1.15T

• a. 0.17wb/m²
• b. 0.45wb/m²
• c. 4.06wb/m² (Marked Answer - Unit discrepancy)
• d. 5.73wb/m²

Note: Magnetic flux (Φ) is measured in Weber (Wb), not Wb/m². The given options use units of magnetic flux density (Wb/m² or Tesla). If Flux = B A cos(θ), where θ is the angle between the magnetic field and the normal to the plane: If plane is at 60° to field, then normal is at 30° to field. Φ = 1.15T * 0.45m² * cos(30°) ≈ 0.448 Wb. If plane's normal is at 60° to field, Φ = 1.15T * 0.45m² * cos(60°) ≈ 0.259 Wb. None of these calculations match the numerical value of the marked answer (4.06) with correct units. This question has issues with units and values.

Question 3: Induced EMF Phenomenon (Page 17)

The phenomenon of producing an induced emf with the aid of a magnetic field is

• a. called electromotive production.
• b. almost never observed.
• c. a scientific curiosity with no practical application.
• d. called electromagnetic induction.

Note: This is a rephrasing of the definition of electromagnetic induction.

Question 5: Magnetic Force on a Positive Charge (Page 32)

A positive charge moving with a constant velocity v enters a region of a uniform magnetic field pointing to the top of the page. What is the direction of the magnetic force on the charge?

• a. There is no magnetic force on the charge.
• b. Right.
• c. To the bottom of the page.
• d. To the top of the page.

Note: The direction of velocity (v) is not specified. The magnetic force F = q(v x B). If v is perpendicular to B and out of the page, the force is to the right. If v is parallel or anti-parallel to B, the force is zero. Without the direction of v, a definitive answer cannot be given. No answer was marked.

Question 11: Induced EMF (Faraday's Law) (Page 49)

The flux linking a coil changes from 5.0 x 10⁻⁴ Wb to 2.2 x 10⁻⁴ Wb in 0.025 seconds. Calculate the average e.m.f. induced in the coil if it has 500turns

• a. -5.6V
• b. -4.3V
• c. -3.8V
• d. -2.7V

Calculation: ε = -N * (ΔΦ/Δt). ΔΦ = (2.2 - 5.0) × 10⁻⁴ Wb = -2.8 × 10⁻⁴ Wb. Δt = 0.025 s. N = 500. ε = -500 × (-2.8 × 10⁻⁴ Wb / 0.025 s) = 500 × (0.0112 V) = 5.6 V. The calculated induced EMF is +5.6V. None of the options are positive 5.6V. Option 'a' is -5.6V. This indicates a potential issue with the expected sign or the options provided.

5. Other Concepts

Question 11: Properties of Electric Fields (NOT True) (Page 25)

Which of the following statements is NOT true of electric fields?

• a. The magnitude of the field is dependent on the test charge placed there
• b. It is a vector field that can be associated each point in space
• c. It is the force per unit charge exerted on a test positive charge at rest at that point
• d. It can be generated by electric charges or by time-varying magnetic fields

Note: An electric field is a property of the source charges and the space itself; its magnitude is independent of the test charge used to measure it. The test charge is assumed to be infinitesimally small so as not to affect the source field. Thus, statement 'a' is NOT true, making it the correct answer to the question.