A Level H2 Physics Electricity Magnetism Quiz

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A-Level Physics H2 Quiz - Electricity Magnetism

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Duration: 60 minutes
Total Marks: 40
Topic: Electricity & Magnetism (Syllabus 9478, Section V)

Instructions:

• Answer all 20 questions.
• Show all working clearly where calculation is required.
• Use the spaces provided.
• Marks for each question are shown in brackets.
• This quiz is syllabus-first practice content generated from LLM-inferred templates; it is not derived from any specific past-year paper.

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Section A: Electric Fields and Circuits (Questions 1–7)

1. State Coulomb’s law for the force between two point charges. [2]

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2. Two point charges $Q_1 = +4.0\ \text{nC}$ and $Q_2 = -2.0\ \text{nC}$ are separated by a distance of $0.10\ \text{m}$ in vacuum. Calculate the magnitude of the electrostatic force between them.
(Use $\frac{1}{4\pi\varepsilon_0} = 9.0\times10^9\ \text{N m}^2\text{C}^{-2}$.) [3]

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3. A battery of emf $9.0\ \text{V}$ and internal resistance $1.0\ \Omega$ is connected to a $4.0\ \Omega$ resistor. Calculate the terminal voltage of the battery. [3]

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4. Resistors of $6.0\ \Omega$ and $3.0\ \Omega$ are connected in parallel. This combination is then connected in series with a $2.0\ \Omega$ resistor across a $12\ \text{V}$ supply of negligible internal resistance. Calculate the current through the $3.0\ \Omega$ resistor. [4]

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5. Define electric field strength at a point. [2]

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6. The diagram below shows a uniform electric field between two parallel plates.

Image pending generation: diagram for Q6.

Calculate the electric field strength between the plates. [2]

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7. A capacitor of capacitance $220\ \mu\text{F}$ is charged to $12\ \text{V}$. Calculate the energy stored in the capacitor. [3]

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Section B: Magnetism and Electromagnetic Induction (Questions 8–14)

8. State the direction of the magnetic field at the centre of a circular coil carrying current in the clockwise direction when viewed from above. [1]

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9. A straight wire carries a current of $5.0\ \text{A}$ perpendicular to a uniform magnetic field of $0.40\ \text{T}$. The length of wire in the field is $0.12\ \text{m}$. Calculate the magnetic force on the wire. [3]

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10. The diagram shows a proton moving horizontally to the right entering a uniform magnetic field directed into the page.

Image pending generation: diagram for Q10.

State the direction of the magnetic force on the proton and sketch the path. [2]

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11. A coil of $200$ turns and area $0.015\ \text{m}^2$ rotates at $25\ \text{Hz}$ in a uniform magnetic field of $0.50\ \text{T}$. Calculate the maximum emf induced in the coil. [3]

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12. Explain, using Faraday’s law, why the induced emf in a coil is zero when the magnetic flux through it is maximum but constant. [3]

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13. A rectangular coil of $N = 50$ turns, area $0.02\ \text{m}^2$, is pulled out of a $0.30\ \text{T}$ magnetic field at a speed such that the flux drops to zero in $0.10\ \text{s}$. Calculate the average induced emf. [3]

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14. A step-down transformer has $1000$ primary turns and $200$ secondary turns. If the primary is connected to $240\ \text{V}$ a.c., calculate the secondary voltage assuming ideal conditions. [2]

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Section C: Combined and Applied E&M (Questions 15–20)

15. Define magnetic flux linkage. [2]

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16. A $12\ \text{V}$ battery with internal resistance $0.5\ \Omega$ supplies a current of $2.0\ \text{A}$ to a circuit. Calculate the power dissipated in the internal resistance. [2]

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17. The circuit below shows a potential divider with two resistors.

Image pending generation: circuit for Q17.

Calculate the output voltage across $R_2$. [3]

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18. A charge of $+3.0\ \mu\text{C}$ is moved from a point at $+50\ \text{V}$ to a point at $+20\ \text{V}$. Calculate the change in electric potential energy. [3]

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19. Describe one precaution when setting up a circuit with an LED to improve accuracy and safety. [2]

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20. A metal rod of length $0.40\ \text{m}$ moves perpendicular to a $0.25\ \text{T}$ magnetic field at $3.0\ \text{m s}^{-1}$. Calculate the motional emf induced across the rod. [3]

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A-Level Physics H2 Quiz - Electricity Magnetism (Answer Key)

Topic: Electricity & Magnetism
Total Marks: 40
Syllabus-first practice generated from LLM-inferred templates. Not past-year derived.

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Section A: Electric Fields and Circuits

Q1 [2 marks]
Coulomb’s law: The force $F$ between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance $r$ between them:
$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q_1 Q_2}{r^2}$.
Marking: 1 mark for proportionality statement, 1 mark for inverse-square / formula.

Q2 [3 marks]
$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{|Q_1 Q_2|}{r^2}$
$= (9.0\times10^9) \dfrac{(4.0\times10^{-9})(2.0\times10^{-9})}{(0.10)^2}$
$= (9.0\times10^9) \dfrac{8.0\times10^{-18}}{0.010}$
$= 7.2\times10^{-6}\ \text{N}$.
Answer: $7.2\ \mu\text{N}$ attractive.
Marks: 1 formula, 1 substitution, 1 final value + unit.

Q3 [3 marks]
Total resistance $R = 1.0 + 4.0 = 5.0\ \Omega$.
Current $I = \dfrac{9.0}{5.0} = 1.8\ \text{A}$.
Terminal voltage $V = \mathcal{E} - Ir = 9.0 - (1.8)(1.0) = 7.2\ \text{V}$.
Marks: 1 circuit eq, 1 current, 1 terminal V.

Q4 [4 marks]
Parallel combo: $R_p = \dfrac{6\times3}{6+3} = 2.0\ \Omega$.
Total $R = 2.0 + 2.0 = 4.0\ \Omega$.
Total current $I = 12/4 = 3.0\ \text{A}$.
Voltage across parallel = $I \times R_p = 3.0\times2.0 = 6.0\ \text{V}$.
Current through $3.0\ \Omega$: $I_3 = 6.0/3.0 = 2.0\ \text{A}$.
Marks: 1 parallel, 1 total, 1 V across, 1 current.

Q5 [2 marks]
Electric field strength $E$ is the force per unit positive charge at a point: $E = F/q$.
Marks: 1 definition, 1 formula.

Q6 [2 marks]
$E = V/d = 200 / 0.05 = 4000\ \text{V m}^{-1}$ (downward).
Marks: 1 formula, 1 value + direction.

Q7 [3 marks]
$E = \frac{1}{2}CV^2 = \frac{1}{2}(220\times10^{-6})(12)^2$
$= 0.5 \times 220\times10^{-6} \times 144 = 1.58\times10^{-2}\ \text{J}$.
Answer: $15.8\ \text{mJ}$.
Marks: 1 formula, 1 sub, 1 answer.

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Section B: Magnetism and Electromagnetic Induction

Q8 [1 mark]
Downward (into the plane) when viewed from above. (Right-hand grip rule.)

Q9 [3 marks]
$F = BIL = (0.40)(5.0)(0.12) = 0.24\ \text{N}$.
Marks: 1 formula, 1 calc, 1 unit.

Q10 [2 marks]
Force direction: downward (by Fleming’s left-hand rule: current = velocity of + charge right, field into page → force down). Path: semicircular arc curving downward.
Marks: 1 direction, 1 sketch description.

Q11 [3 marks]
$\omega = 2\pi f = 2\pi(25) = 157\ \text{rad s}^{-1}$.
$\varepsilon_{\max} = NBA\omega = 200\times0.015\times0.50\times157 = 235.5\ \text{V}$.
Answer: $236\ \text{V}$ (3 s.f.).
Marks: 1 omega, 1 formula, 1 value.

Q12 [3 marks]
Faraday’s law: $\varepsilon = -N\dfrac{d\Phi}{dt}$. When flux is maximum but constant, $d\Phi/dt = 0$, so $\varepsilon = 0$.
Marks: 1 law, 1 derivative zero, 1 conclusion.

Q13 [3 marks]
Initial flux $\Phi = NBA = 50\times0.30\times0.02 = 0.30\ \text{Wb}$.
Change $\Delta\Phi = 0.30\ \text{Wb}$ in $0.10\ \text{s}$.
$\varepsilon_{\text{avg}} = N\dfrac{\Delta\Phi}{\Delta t} = 50\times\dfrac{0.30}{0.10} = 150\ \text{V}$.
Marks: 1 flux, 1 rate, 1 emf.

Q14 [2 marks]
$\dfrac{V_s}{V_p} = \dfrac{N_s}{N_p} \Rightarrow V_s = 240\times\dfrac{200}{1000} = 48\ \text{V}$.
Marks: 1 ratio, 1 answer.

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Section C: Combined and Applied E&M

Q15 [2 marks]
Magnetic flux linkage = $N\Phi$, product of number of turns and magnetic flux through coil. Unit Wb-turn.
Marks: 1 product, 1 unit/explanation.

Q16 [2 marks]
$P = I^2 r = (2.0)^2(0.5) = 2.0\ \text{W}$.
Marks: 1 formula, 1 answer.

Q17 [3 marks]
Total $R = 300+600 = 900\ \Omega$.
Current $I = 9/900 = 0.01\ \text{A}$.
$V_{\text{out}} = IR_2 = 0.01\times600 = 6.0\ \text{V}$.
Marks: 1 total, 1 current, 1 Vout.

Q18 [3 marks]
$\Delta U = q\Delta V = (3.0\times10^{-6})(20 - 50) = -9.0\times10^{-5}\ \text{J}$.
Answer: decrease of $90\ \mu\text{J}$.
Marks: 1 formula, 1 sub, 1 sign/answer.

Q19 [2 marks]
Example: Connect LED with correct polarity (anode to positive) to avoid reverse breakdown; or use current-limiting resistor to prevent excess current.
Marks: 1 precaution, 1 reason.

Q20 [3 marks]
$\varepsilon = BLv = (0.25)(0.40)(3.0) = 0.30\ \text{V}$.
Marks: 1 formula, 1 sub, 1 answer.