Physics T Gravity and Electromagnetism Unit 3 Practice Exam Study Guide

Exam Overview and Assessment Conditions

• Course Title: Physics T Integrating Australian Curriculum

• Course Code: 2331

• Weighting: 25%

• Semester and Year: Semester 1, 2024

• Teacher: Scott Carter

• Due Date: 11 June 2024

• Units Covered:

  • 1.0 Unit: Gravity and Electromagnetism

  • 0.5 Unit: Gravity and Electromagnetism Units 3a and 3b

• Unit Codes: 21979, 21992, 21993

• Assessment Item Type: In-class examination

• Assessment Conditions: individual completion, 10 minutes reading time, 90 minutes writing time. Calculators are allowed. Work must be legible and completed in blue or black pen.

Part A: True or False Statements

1. Magnetic fields can be created by current: True.

2. Doubling the distance between two positive charges will decrease the electrostatic force between the charges by a factor of two: False. (The inverse square law implies the force decreases by a factor of four).

3. When a steady DC current flowing through a solenoid is stopped, an emf is induced in the solenoid: True.

4. Magnet monopoles do not exist in nature: True.

5. A neutron will follow a curved path in a strong magnetic field: False. (Neutrons have no net charge and thus experience no magnetic force).

6. If a charged particle is moving in a straight line through some region of space, one can conclude that the magnetic field in that region is zero: False. (The particle could be moving parallel to the magnetic field, or there could be a balancing electric force).

7. Magnetic Field lines always form closed loops: True.

8. When a steady DC current is flowing through a solenoid, the magnetic field inside the solenoid is uniform: True.

9. The units of $\text{emf}$ (Electromotive Force) are Newtons: False. (The unit of $\text{emf}$ is Volts).

10. Magnetic flux is the rate of flow of magnetic field through an area per second: False. (Flux is the total magnetic field passing through a surface, not a rate of flow per second).

Part B: Multiple Choice Questions

1. Right-Hand Rule for Force ($\mathbf{F} = I\mathbf{L} \times \mathbf{B}$):

   • Correct Answer: B. If $\mathbf{L}$ and $\mathbf{B}$ are parallel, there will be zero force.

   • Reasoning: The cross product includes $\sin(\theta)$; if parallel, $\theta = 0^{\circ}$, and $\sin(0^{\circ}) = 0$.

2. Electric Field Calculation:

   • Given: $q = 0.1\,C$, $F = 10\,kN = 10,000\,N$.

   • Formula: $E = \frac{F}{q} = \frac{10,000\,N}{0.1\,C} = 100,000\,NC^{-1}$.

   • Correct Answer: B. $10^5\,NC^{-1}$.

3. Magnetic Flux through a Loop:

   • Scenario: Square loop facing directly into a uniform field.

   • Correct Answer: B. Magnetic flux is maximum because $\mathbf{B}$ and $\mathbf{A}$ (the area vector) are parallel.

4. Magnetic Field in a Toroid:

   • Following the right-hand rule for the direction of the field inside the loops of the doughnut-shape.

   • Correct Answer: A. Clockwise (based on the diagram provided in the paper).

5. Solenoid Field Comparison:

   • Solenoid 1: $10\,cm$, $250$ loops, $1\,A$ current ($n = 2500\text{ turns/m}$).

   • Solenoid 2: $20\,cm$, $500$ loops, $0.5\,A$ current ($n = 2500\text{ turns/m}$).

   • Formula: $B = \mu_0 n I$.

   • Correct Answer: C. The magnetic field inside the solenoid is halved (because current dropped by half while turns-per-unit-length remained constant).

6. Magnetic Field Direction Between Wires:

   • Wire Left: Current out of page. Wire Right: Current into page.

   • Point P is between them. Using Right Hand Grip Rule, both wires produce a downward field at point P.

   • Correct Answer: D. Downwards.

7. Electrostatic Force Vectors:

   • Particle X ($-2C$), Particle Y ($+1C$). Opposite charges attract.

   • Newton’s Third Law implies the forces are equal in magnitude and opposite in direction.

   • Correct Answer: A (Arrows pointing toward each other).

8. Rotation of a Rectangular Loop:

   • Current clockwise, field North to South (left to right).

   • Using $\mathbf{F} = I\mathbf{L} \times \mathbf{B}$: Segment ab moves into the page; segment cd moves out of the page.

   • Correct Answer: A.

9. Rotating Wire Loop in Magnetic Field:

   • Correct Answer: A. There is maximum flux when the area vector is parallel to the magnetic field.

10. Proton Location Relative to Electron Motion:

    • Electron curves upward toward a positive charge.

    • Correct Answer: A (but could also be B depending on specific pathing).

Part C: Short and Long Response Worked Solutions

Question 1: Force per Unit Length

• Given: $I = 10\,A$, $B = 1\,T$, $\theta = 45^{\circ}$.

• Formula: $F = ILB\sin(\theta)$.

• Calculation:   $F = 10\,A \times L \times 1\,T \times \sin(45^{\circ})$   $\frac{F}{L} = 10 \times 1 \times \sin(45^{\circ}) = 7.07\,N/m$

• Result: $7.07\,N/m$.

Question 2: Electrostatic Force on an Electron in Carbon-12

• Given: Distance $r = 5 \times 10^{-11}\,m$. Charge of Electron $q_2 = -1.6 \times 10^{-19}\,C$. Charge of Carbon nucleus (6 protons) $q_1 = 6 \times (1.6 \times 10^{-19}\,C)$.

• Calculation:   $F = \frac{k q_1 q_2}{r^2}$   $F = \frac{9 \times 10^9 \times [6 \times (1.6 \times 10^{-19})] \times [ -1.6 \times 10^{-19}]}{(5 \times 10^{-11})^2}$

• Result: $5.52 \times 10^{-7}\,N$ towards the nucleus.

Question 3: Dipoles

• Definition: An electric dipole is a pair of equal and opposite electric charges.

• Diagram Details: Field lines originate from the positive charge and terminate at the negative charge.

Question 4: Potential Energy ($U$) Changes

• a) Positive charge moving against field (toward positive plate): $\Delta U$ Increases.

• b) Positive charge moving with field (toward negative plate): $\Delta U$ Decreases.

• c) Positive charge moving perpendicular to field lines: $\Delta U$ remains the same (effectively zero work done perpendicular to the force), or decreases if it begins to curve.

• d) Negative charge moving against field lines (toward positive plate): $\Delta U$ Decreases.

Question 5: Charges on a Square

• Field Lines: Lines point away from positive charges and toward negative charges.

• Field at Centre: The magnitude is $0\,NC^{-1}$. This occurs because the electric fields from the four equal charges are equidistant and symmetrical, causing them to cancel each other out at the center point.

Question 6: Proton Acceleration between Plates

• Given: Potential Difference $\Delta V = 120\,V$, Plate distance $s = 1\,cm = 0.01\,m$.

• Energy Gain: $\Delta U = \Delta V \times q = -120\,V \times (1.6 \times 10^{-19}\,C) = -1.93 \times 10^{-17}\,J$. Kinetic Energy gained ($KE$) = $+1.93 \times 10^{-17}\,J$.

• Maximum Velocity:   $KE = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2 \times KE}{m}}$   $v = \sqrt{\frac{2 \times 1.926 \times 10^{-17}\,J}{1.6726 \times 10^{-27}\,kg}} = 1.52 \times 10^5\,ms^{-1}$

• Acceleration:   $v^2 = u^2 + 2as \implies a = \frac{v^2}{2s}$   $a = \frac{(151,519.86)^2}{2 \times 0.01} = 1.15 \times 10^{12}\,ms^{-2}$ towards the negative plate.

• Force:   $F = ma = 1.6726 \times 10^{-27}\,kg \times 1.15 \times 10^{12}\,ms^{-2} = 1.92 \times 10^{-15}\,N$ towards the negative plate.

• Constancy of Force: Yes, force remains constant because the electric field between two charged parallel plates is uniform (magnitude and direction are the same everywhere). Since $F = Eq$, and $E$ and $q$ are constant, $F$ is constant.

• Electric Field Calculation:   $E = \frac{F}{q} = \frac{1.92 \times 10^{-15}\,N}{1.6 \times 10^{-19}\,C} = 12,000\,NC^{-1}$.

Question 7: Electric Field Magnitude for Neon Nucleus

• Given: $F = 3.2 \times 10^{-20}\,N$. Neon nucleus ($Z=10$) charge $q = 10 \times (1.6 \times 10^{-19}\,C)$.

• Calculation:   $E = \frac{F}{q} = \frac{3.2 \times 10^{-20}}{1.6 \times 10^{-18}} = 0.02\,NC^{-1}$.

Question 8: Magnet Interaction

• Zero Magnetic Field Point: Located exactly between two opposing magnets of equal strength where the fields cancel.

• Electron Motion: To move an electron "out of page" ($F$) when the field ($B$) is to the left, the electron velocity ($v$) must be downwards (applying right-hand rule and reversing for negative charge).

• Stationary Proton: Experiences no force because a charge must be moving relative to a magnetic field to experience a force.

• Path along Field Line: A particle following a field line experiences no force because its velocity vector $\mathbf{v}$ is parallel to the magnetic field vector $\mathbf{B}$. Since $F = qvB\sin(\theta)$ and $\theta = 0^{\circ}$, $\sin(0^{\circ}) = 0$, resulting in $F = 0$.

Question 9: Hovering Oil Drop

• Plate Polarity: To balance gravity ($F_g$ downwards), the electric force ($F_e$) must be upwards. Since the oil drop is negative, it is repelled by a negative plate. Thus, the bottom plate is negative.

• Field Strength Calculation:   $F_{down} = mg = (1 \times 10^{-9}\,kg) \times 9.8\,ms^{-2} = 9.8 \times 10^{-9}\,N$   $E = \frac{F}{q} = \frac{9.8 \times 10^{-9}\,N}{1.6 \times 10^{-19}\,C} = 6.125 \times 10^{10}\,NC^{-1}$.

Question 10: Comparison of Electric and Magnetic Forces

• Similarities:

  • Both can be experienced by a moving charge.

  • Like entities (charges/poles) repel, while opposite entities attract.

• Differences:

  • Stationary charges experience electric force but not magnetic force.

  • Magnetic force is always perpendicular to field lines; electric force is parallel to field lines.

Question 11: Transformer Design

• Goal: Step-down from $11,000\,V$ to $240\,V$.

• Formula: $\frac{V_p}{V_s} = \frac{N_p}{N_s}$.

• Ratio: $\frac{11,000}{240} = \frac{275}{6}$.

• Example solution: $N_p = 275$ turns, $N_s = 6$ turns (or any multiples like $550:12$).

Question 12: Magnetic Field Between Wires

• Given: Wire 1 ($1\,A$), Wire 2 ($3\,A$). Point P is halfway (distance $r$ from each).

• Calculations:

  • $B_{left} = \frac{\mu_0 I}{2\pi r} = \frac{2 \times 10^{-7} \times 1}{r}$ (into page).

  • $B_{right} = \frac{\mu_0 I}{2\pi r} = \frac{2 \times 10^{-7} \times 3}{r} = \frac{6 \times 10^{-7}}{r}$ (out of page).

• Net Field: $\frac{6 \times 10^{-7}}{r} - \frac{2 \times 10^{-7}}{r} = \frac{4 \times 10^{-7}}{r}\,T$ (out of page).

Question 13: Superposition of Electric Fields from Three Electrons

• Physical Setup: Three electrons in a row, Point P is $1\,cm$ above the middle electron.

• Middle Electron Field:   $E_m = \frac{9 \times 10^9 \times 1.6 \times 10^{-19}}{(0.01)^2} = 1.44 \times 10^{-5}\,NC^{-1}$ (downward).

• Side Electron Fields (Left and Right):   Distance $r = \sqrt{(1\,cm)^2 + (1\,cm)^2} = 1.41\,cm = 0.0141\,m$.   $E = \frac{9 \times 10^9 \times 1.6 \times 10^{-19}}{(0.0141)^2} = 7.2 \times 10^{-6}\,NC^{-1}$.

• Vector Components (Left/Right):

  • E_{vert} = .2 \times 10^{-6} \times \sin(45^{\circ}) = 5.09 \times 10^{-6}\,NC^{-1} (downwards).

  • $E_{horiz}$ components cancel ($5.09 \times 10^{-6}$ left vs $5.09 \times 10^{-6}$ right).

• Total Field at P:   $E_{total} = E_{m} + E_{vert, left} + E_{vert, right}$   $E_{total} = (1.44 \times 10^{-5}) + (5.09 \times 10^{-6}) + (5.09 \times 10^{-6}) = 2.46 \times 10^{-5}\,NC^{-1}$ (downwards).

Reference Constants and Formulas

Constants

• $q_e = -1.6 \times 10^{-19}\,C$

• $q_p = +1.6 \times 10^{-19}\,C$

• $m_e = 9.11 \times 10^{-31}\,kg$

• $m_p = 1.6726 \times 10^{-27}\,kg$

• $\varepsilon_o = 8.854 \times 10^{-12}\,C^2/Nm^2$

• $\mu_0 = 4\pi \times 10^{-7}\,Tm/A$

• $k = \frac{1}{4π\varepsilon_o} \approx 9 \times 10^9\,N\,m^2/C^2$

• $g = 9.8\,ms^{-2}$

Unit Conversions

• Giga ($G$): $10^9$

• Mega ($M$): $10^6$

• Kilo ($k$): $10^3$

• Centi ($c$): $10^{-2}$

• Milli ($m$): $10^{-3}$

• Micro ($\mu$): $10^{-6}$

• Nano ($n$): $10^{-9}$

• Pico ($p$): $10^{-12}$

Core Formulas

• Electric Force/Field: $E = \frac{F}{q}$, $E = \frac{kQ}{r^2}$, $F = \frac{kQq}{r^2}$

• Potential: $V = \frac{U}{q}$, $\Delta V = \frac{\Delta U}{q}$

• Magnetism: $B = \frac{\mu_o I}{2\pi r}$, $F = ILB\sin(\theta)$, $F = qvB\sin(\theta)$

• Flux & Induction: $\Phi_B = BA\cos\theta$, $emf = -\frac{\Delta\Phi}{\Delta t}$

• Transformers: $\frac{V_s}{V_p} = \frac{I_p}{I_s} = \frac{N_s}{N_p}$

• Energy: $KE = \frac{1}{2}mv^2$, $|\Delta U| = |\Delta KE|$