Complete module assessment with 15 multiple choice questions and 5 written-response questions grounded in this module's lesson content.
Q1. Coulomb's Law shows that electric force is proportional to:
Q2. A positive charge released between parallel plates accelerates:
Q3. The work done moving charge q through potential difference V is:
Q4. The magnetic force on a moving charge is maximum when velocity is:
Q5. A charged particle moving perpendicular to a uniform magnetic field follows a circular path because the magnetic force:
Q6. The motor effect is the force on:
Q7. Two parallel wires carrying currents in the same direction:
Q8. Torque on a current-carrying coil is greatest when:
Q9. A split-ring commutator in a DC motor:
Q10. Faraday's Law states that induced emf depends on:
Q11. Lenz's Law gives the direction of induced current so that it:
Q12. An ideal step-up transformer has:
Q13. In a velocity selector, crossed electric and magnetic fields select particles where:
Q14. Eddy currents are induced loops of current in conductors that often:
Q15. High-voltage transmission reduces energy loss because for the same power it:
Q16. Compare the motion of charged particles in uniform electric and magnetic fields.
Q17. Explain how a DC motor produces continuous rotation.
Q18. Use Faraday's and Lenz's Laws to explain transformer operation.
Q19. Explain how electric and magnetic fields are used in a mass spectrometer to separate ions by mass-to-charge ratio.
Q20. Explain why eddy currents can be useful in braking but wasteful in power transmission, and why high voltage helps transmission.
Q1: B
Q2: C
Q3: D
Q4: A
Q5: B
Q6: D
Q7: C
Q8: A
Q9: B
Q10: D
Q11: C
Q12: A
Q13: D
Q14: B
Q15: C
An electric field exerts force F = qE in the field direction for positive charges and opposite for negative charges, so it can change a particle's speed and kinetic energy. A magnetic field exerts force F = qvB sin theta perpendicular to velocity and field. When velocity is perpendicular to the magnetic field, this force acts as centripetal force and changes direction but not speed, because it does no work.
Marks: 1, electric force | 1, electric field changes energy | 1, magnetic force formula/condition | 1, circular motion | 1, magnetic force no work
A current-carrying coil in a magnetic field experiences forces on opposite sides due to the motor effect, F = BIL sin theta. These forces form a torque that rotates the coil. After each half-turn the split-ring commutator reverses current in the coil, so the forces reverse relative to the coil and the torque remains in the same rotational direction. Brushes maintain electrical contact with the rotating commutator.
Marks: 1, motor effect force | 1, opposite forces/torque | 1, commutator reversal | 1, continuous same-direction torque | 1, brushes/contact
An AC current in the primary coil creates a changing magnetic flux in the iron core. This changing flux links the secondary coil and induces an emf according to Faraday's Law, with magnitude depending on the rate of change of flux linkage. Lenz's Law determines the direction of the induced emf so it opposes the flux change that produced it. The voltage ratio follows Vs/Vp = Ns/Np for an ideal transformer.
Marks: 1, AC changing flux | 1, flux linkage secondary | 1, Faraday magnitude | 1, Lenz direction | 1, turns ratio
Ions are accelerated by an electric potential difference so they gain kinetic energy. A velocity selector can use crossed electric and magnetic fields so only particles with qE = qvB pass undeflected. In the analyser, a magnetic field provides centripetal force, qvB = mv squared/r, so r = mv/qB. For the same speed and charge, heavier ions have larger radii, allowing separation by mass-to-charge ratio.
Marks: 1, acceleration | 1, velocity selector | 1, magnetic centripetal force | 1, radius relationship | 1, mass-to-charge separation
Eddy currents are induced in conductors when magnetic flux changes. By Lenz's Law they oppose the change, so in electromagnetic braking they oppose motion and provide non-contact braking while converting kinetic energy to heat. In transformer cores or transmission hardware, unwanted eddy currents waste energy as heat. High-voltage transmission is used because P = VI, so for the same power a higher voltage means lower current. Since line loss is I squared R, reducing current greatly reduces heating losses.
Marks: 1, eddy current induction | 1, braking use | 1, heating loss | 1, high voltage lowers current | 1, I squared R losses
I have completed this module assessment and reviewed the answers.