A Level Physics 20.2 Force on a Current-Carrying Conductor MCQs – 30 Practice Questions with Answers (Cambridge 9702)
This page covers Section 20.2 of the Cambridge International AS & A Level Physics (9702) syllabus, an A Level (A2) topic on the motor effect, Fleming’s left-hand rule and the simple d.c. motor. Try all 30 original practice questions, then open each answer to see the full working.
Formula recap
- Force on a conductor: F = B I L sinθ, where θ is the angle between the current direction and the magnetic field
- Conductor at right angles to the field: F = B I L
- B = F / (I L) for a perpendicular conductor: 1 tesla = 1 N A−1 m−1
- Fleming’s left-hand rule: First finger = Field, seCond finger = Current, thuMb = Motion (direction of the force)
Full formula sheet: A Level Physics Formula Sheet
Force on a Conductor
Q1. The motor effect is the observation that
- (A) a compass needle is deflected when it is placed near a current-carrying wire
- (B) a current-carrying conductor placed in a magnetic field experiences a force
- (C) a changing magnetic field induces an e.m.f. in a nearby conductor
- (D) a current is induced in a coil when the coil is rotated in a magnetic field
Show Answer
Answer: B – A conductor carrying a current in a magnetic field experiences a force at right angles to both the current and the field. This is the motor effect, the principle behind electric motors. The other options describe different electromagnetic phenomena.
Q2. In the equation F = B I L sinθ, the angle θ is the angle between
- (A) the conductor and the direction of the force
- (B) the magnetic field and the direction of the force
- (C) the conductor and a line perpendicular to the field lines
- (D) the direction of the current and the direction of the magnetic field
Show Answer
Answer: D – θ is measured between the current (along the conductor) and the field lines. When the conductor is at right angles to the field, θ = 90° and sinθ = 1, giving the maximum force F = B I L.
Q3. A straight wire carrying a fixed current is placed in a uniform magnetic field. The force on the wire is greatest when the wire is
- (A) at right angles to the field
- (B) parallel to the field
- (C) at 30° to the field
- (D) carrying the current in the reverse direction
Show Answer
Answer: A – F = B I L sinθ is maximum when sinθ = 1, i.e. when θ = 90° and the conductor is perpendicular to the field. Reversing the current reverses the force but does not change its size.
Q4. A wire carries a current parallel to the direction of a uniform magnetic field. The force on the wire is
- (A) the maximum possible value B I L
- (B) equal to B I L
- (C) zero
- (D) half of B I L
Show Answer
Answer: C – With the current parallel to the field, θ = 0° and sinθ = 0, so F = B I L sinθ = 0. A current parallel (or anti-parallel) to the field experiences no force.
Q5. A wire of length 0.20 m carries a current of 5.0 A at right angles to a uniform magnetic field of flux density 0.40 T. The force on the wire is
- (A) 0.040 N
- (B) 0.40 N
- (C) 4.0 N
- (D) 10 N
Show Answer
Answer: B – F = B I L = 0.40 × 5.0 × 0.20 = 0.40 N.
Q6. A wire of length 0.50 m carries a current of 4.0 A in a uniform magnetic field of flux density 0.60 T. The wire makes an angle of 30° with the field. The force on the wire is
- (A) 1.2 N
- (B) 0.30 N
- (C) 2.4 N
- (D) 0.60 N
Show Answer
Answer: D – F = B I L sinθ = 0.60 × 4.0 × 0.50 × sin30° = 1.2 × 0.50 = 0.60 N. Forgetting the sinθ factor gives the trap answer 1.2 N.
Q7. The tesla is defined by the force on a current-carrying conductor. One tesla is the flux density of a field that exerts
- (A) a force of 1 N per metre of length on a conductor carrying a current of 1 A placed at right angles to the field
- (B) a force of 1 N per metre of length on a conductor carrying a current of 1 A placed parallel to the field
- (C) a force of 1 N on a conductor of length 1 m carrying a current of 2 A at right angles to the field
- (D) a force of 1 N per metre of length on a conductor carrying a current of 1 A at 30° to the field
Show Answer
Answer: A – From B = F / (I L), B = 1 T when F = 1 N for each metre of a conductor carrying 1 A at right angles to the field, i.e. 1 T = 1 N A−1 m−1.
Q8. A wire of length 0.30 m carrying a current of 10 A is placed at right angles to a uniform magnetic field and experiences a force of 0.60 N. The flux density of the field is
- (A) 0.020 T
- (B) 2.0 T
- (C) 0.20 T
- (D) 18 T
Show Answer
Answer: C – B = F / (I L) = 0.60 / (10 × 0.30) = 0.60 / 3.0 = 0.20 T.
Q9. A conductor at right angles to a magnetic field experiences a force F. Which change, on its own, makes the force 4F?
- (A) Doubling B and halving I
- (B) Doubling B and doubling I
- (C) Halving B and doubling L
- (D) Doubling L and halving I
Show Answer
Answer: B – F = B I L is proportional to each of B, I and L. Doubling B and doubling I multiplies F by 2 × 2 = 4. In each other option the two changes cancel, leaving F unchanged.
Q10. A wire of length 0.40 m is placed at right angles to a magnetic field of flux density 0.30 T and experiences a force of 0.36 N. The current in the wire is
- (A) 0.30 A
- (B) 0.043 A
- (C) 30 A
- (D) 3.0 A
Show Answer
Answer: D – I = F / (B L) = 0.36 / (0.30 × 0.40) = 0.36 / 0.12 = 3.0 A.
Direction: Fleming’s Left-Hand Rule
Q11. In Fleming’s left-hand rule, the first finger represents
- (A) the magnetic field
- (B) the current
- (C) the force on the conductor
- (D) the induced e.m.f.
Show Answer
Answer: A – First finger = Field, seCond finger = Current, thuMb = Motion, i.e. the direction of the force (thrust) on the conductor.
Q12. A horizontal wire runs across the page and carries a current from left to right. A uniform magnetic field points vertically downwards, from the top of the page towards the bottom. The force on the wire acts
- (A) upwards, towards the top of the page
- (B) to the left
- (C) away from the observer, into the page
- (D) towards the observer, out of the page
Show Answer
Answer: C – Point the first finger downwards (field) and the second finger to the right (current); the thumb then points into the page, away from the observer.
Q13. A vertical wire carries a current upwards, towards the top of the page. A uniform magnetic field is directed into the page. The force on the wire acts
- (A) to the right
- (B) out of the page, towards the observer
- (C) downwards, towards the bottom of the page
- (D) to the left
Show Answer
Answer: D – Point the first finger into the page (field) and the second finger upwards (current); the thumb points to the left.
Q14. A wire carries a current into the page. It is placed in a uniform magnetic field directed to the right. The force on the wire acts
- (A) upwards
- (B) downwards
- (C) to the left
- (D) out of the page
Show Answer
Answer: B – Point the first finger to the right (field) and the second finger into the page (current); the thumb points downwards.
Q15. The current in a wire in a magnetic field is reversed, while the field is left unchanged. The force on the wire
- (A) acts in the opposite direction to before
- (B) doubles in magnitude
- (C) acts in the same direction as before
- (D) becomes zero
Show Answer
Answer: A – Reversing the current reverses the second finger in Fleming’s left-hand rule, so the thumb (force) points the opposite way. The magnitude B I L sinθ is unchanged.
Q16. The current in a wire is reversed and, at the same time, the magnetic field is also reversed. The force on the wire
- (A) acts in the opposite direction to before
- (B) becomes zero
- (C) acts in the same direction as before
- (D) is halved in magnitude
Show Answer
Answer: C – Reversing the field flips the force once and reversing the current flips it again. The two reversals cancel, so the force keeps its original direction and magnitude.
Q17. A wire carries a current out of the page, towards the observer. A uniform magnetic field points vertically downwards. The force on the wire acts
- (A) to the left
- (B) upwards
- (C) into the page
- (D) to the right
Show Answer
Answer: D – Point the first finger downwards (field) and the second finger out of the page (current); the thumb points to the right.
Q18. A horizontal wire carries a current from right to left. A uniform magnetic field is directed out of the page, towards the observer. The force on the wire acts
- (A) downwards
- (B) upwards
- (C) to the right
- (D) into the page
Show Answer
Answer: B – Point the first finger out of the page (field) and the second finger to the left (current); the thumb points upwards.
Q19. A wire is placed between a north pole on its left and a south pole on its right, so the field runs from left to right. The current in the wire also flows from left to right. The force on the wire is
- (A) upwards
- (B) downwards
- (C) zero, because the current is parallel to the field
- (D) into the page
Show Answer
Answer: C – The current and the field are parallel, so θ = 0° and F = B I L sinθ = 0. Fleming’s left-hand rule cannot be applied because the first and second fingers point the same way.
Q20. A vertical wire carries a current upwards. A uniform magnetic field is directed to the right. The force on the wire acts
- (A) into the page, away from the observer
- (B) out of the page, towards the observer
- (C) to the left
- (D) downwards
Show Answer
Answer: A – Point the first finger to the right (field) and the second finger upwards (current); the thumb points into the page.
Applications & Interactions
Q21. In a simple d.c. motor, the coil is placed in a magnetic field and carries a current. The forces on the two long sides of the coil
- (A) act in the same direction and push the coil sideways
- (B) are both zero because the sides are parallel to the field
- (C) cancel each other, so the coil does not turn
- (D) act in opposite directions and together produce a turning effect on the coil
Show Answer
Answer: D – The currents in the two sides of the coil flow in opposite directions, so by Fleming’s left-hand rule the forces on them are opposite. The pair of forces forms a couple that rotates the coil.
Q22. The split-ring commutator in a simple d.c. motor is needed because it
- (A) increases the current supplied to the coil
- (B) reverses the current in the coil every half turn, so the turning effect stays in the same direction
- (C) increases the strength of the magnetic field
- (D) prevents the coil from overheating
Show Answer
Answer: B – Each half of the split ring swaps contact with the brushes every half turn, reversing the current in the coil just as the coil passes the vertical position. The forces therefore keep turning the coil the same way and rotation continues.
Q23. A coil is connected directly to a d.c. supply by plain sliding contacts, with no commutator, and is free to turn in a magnetic field. The coil will
- (A) rotate faster and faster in one direction
- (B) rotate continuously in the opposite direction
- (C) swing to and fro with decreasing amplitude and come to rest with its plane perpendicular to the field
- (D) not move at all
Show Answer
Answer: C – With the current fixed in direction, the turning effect reverses after each half turn, pushing the coil back the way it came. The coil oscillates about, and settles in, the position where the forces act in line and produce no turning effect.
Q24. Which change increases the turning effect of a simple d.c. motor?
- (A) Increasing the number of turns on the coil
- (B) Decreasing the area of the coil
- (C) Using a weaker magnet
- (D) Reducing the current in the coil
Show Answer
Answer: A – Each turn of the coil experiences the force F = B I L, so more turns give a larger total force and a larger turning effect. A larger current, a stronger field and a larger coil area also increase it; the other options all decrease it.
Q25. Two long parallel wires carry currents in the same direction. The wires
- (A) repel each other
- (B) attract each other
- (C) exert no force on each other
- (D) rotate about each other
Show Answer
Answer: B – Each wire sits in the magnetic field produced by the other. Applying Fleming’s left-hand rule shows the force on each wire is directed towards the other wire, so parallel currents in the same direction attract.
Q26. Two long parallel wires carry currents in opposite directions. The wires
- (A) attract each other
- (B) exert no force on each other
- (C) experience a force on only one of the wires
- (D) repel each other
Show Answer
Answer: D – Reversing one current reverses the force on that wire compared with the same-direction case. Each wire is pushed away from the other, so anti-parallel currents repel. By Newton’s third law the forces on the two wires are equal and opposite.
Q27. Wire X carries a current vertically upwards. It produces a magnetic field which, at a point to the right of X, is directed into the page. Wire Y is placed to the right of X and also carries a current vertically upwards. The force on wire Y acts
- (A) to the right, away from X
- (B) out of the page
- (C) to the left, towards X
- (D) downwards
Show Answer
Answer: C – Wire Y carries a current upwards in a field (due to X) that points into the page. Fleming’s left-hand rule gives a force to the left, towards X, confirming that the two wires attract.
Q28. In a current balance, a wire of length 0.080 m is held at right angles to a magnetic field and carries a current of 3.0 A. The extra force measured on the balance is 0.24 N. The flux density of the field is
- (A) 1.0 T
- (B) 0.10 T
- (C) 0.58 T
- (D) 10 T
Show Answer
Answer: A – B = F / (I L) = 0.24 / (3.0 × 0.080) = 0.24 / 0.24 = 1.0 T. This is the working principle of the current balance: a measured force gives B directly.
Q29. The coil of a d.c. motor has 200 turns. Each side of the coil is 0.040 m long and lies in a magnetic field of flux density 0.25 T. When the current in the coil is 1.5 A, the force on one side of the coil, in the position where the force is greatest, is
- (A) 0.015 N
- (B) 3.0 N
- (C) 1.5 N
- (D) 30 N
Show Answer
Answer: B – Force on one turn = B I L = 0.25 × 1.5 × 0.040 = 0.015 N. For 200 turns, F = 200 × 0.015 = 3.0 N.
Q30. In a loudspeaker, an alternating current passes through a coil placed in the field of a permanent magnet. The paper cone attached to the coil vibrates because
- (A) the coil is permanently attracted to the magnet
- (B) the current in the coil is constant
- (C) the force on the coil reverses direction repeatedly as the current alternates
- (D) the cone itself produces the magnetic field
Show Answer
Answer: C – The force on the current-carrying coil follows F = B I L. Each time the alternating current reverses, the force reverses, driving the coil and cone backwards and forwards at the frequency of the signal.
Keep practising
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