A Level Physics 20.4 Magnetic Fields Due to Currents MCQs – 30 Practice Questions with Answers (Cambridge 9702)
These 30 original practice questions cover Cambridge AS & A Level Physics (9702) Section 20.4, an A Level (A2) topic on the magnetic fields produced by electric currents. They cover the field around a straight wire, the fields at the centre of a flat coil and inside a solenoid, and mixed calculations including points where the fields of two wires combine.
Formula recap (full sheet: A Level Physics Formula Sheet)
- Straight wire: B = μ0I / (2πr)
- Flat circular coil, field at the centre: B = μ0NI / (2r)
- Solenoid, field inside: B = μ0nI, where n = number of turns per unit length
- μ0 = 4π × 10−7 H/m, so for a straight wire B = (2 × 10−7 I) / r
- Right-hand grip rule: thumb along the conventional current, curled fingers give the direction of the magnetic field
Field Around a Straight Wire
Q1. A long straight wire carries a steady current. What is the pattern of the magnetic field around the wire?
- (A) Straight lines running parallel to the wire
- (B) Concentric circles centred on the wire
- (C) Straight lines radiating outwards from the wire in all directions
- (D) Lines running along the wire from one end to the other
Show Answer
Answer: B – The field lines form concentric circles around the wire, in planes perpendicular to it. The circles are closer together near the wire, where the field is strongest.
Q2. In the right-hand grip rule for a straight wire, what do the thumb and the curled fingers represent?
- (A) Thumb: direction of the conventional current; curled fingers: direction of the magnetic field
- (B) Thumb: direction of the magnetic field; curled fingers: direction of the conventional current
- (C) Thumb: direction of electron flow; curled fingers: direction of the magnetic field
- (D) Thumb: direction towards the north pole; curled fingers: direction of the conventional current
Show Answer
Answer: A – Grip the wire with the right hand, thumb pointing along the conventional current; the fingers curl in the direction of the field lines around the wire.
Q3. A corkscrew is turned so that it advances in the direction of the conventional current in a wire. According to the corkscrew rule, the direction of the magnetic field is given by
- (A) the direction in which the corkscrew advances
- (B) the direction opposite to the rotation of the corkscrew
- (C) the direction of rotation of the corkscrew
- (D) any direction along the length of the wire
Show Answer
Answer: C – The screw advances along the current direction, and the sense of its rotation gives the direction of the circular field lines around the wire.
Q4. A long straight wire carries a current of 10 A. What is the magnetic flux density at a distance of 0.050 m from the wire?
- (A) 2.0 × 10−6 T
- (B) 4.0 × 10−6 T
- (C) 2.0 × 10−5 T
- (D) 4.0 × 10−5 T
Show Answer
Answer: D – B = μ0I / (2πr) = (2 × 10−7 × 10) / 0.050 = (2.0 × 10−6) / 0.050 = 4.0 × 10−5 T.
Q5. A current of 5.0 A flows in a long straight wire. The flux density at a point 0.10 m from the wire is
- (A) 1.0 × 10−5 T
- (B) 2.0 × 10−5 T
- (C) 1.0 × 10−4 T
- (D) 5.0 × 10−6 T
Show Answer
Answer: A – B = (2 × 10−7 × 5.0) / 0.10 = (1.0 × 10−6) / 0.10 = 1.0 × 10−5 T.
Q6. A wire carries a current coming out of the plane of the page towards the observer, represented by a dot. At a point immediately to the right of the wire, the magnetic field points
- (A) to the right
- (B) to the left
- (C) towards the top of the page
- (D) towards the bottom of the page
Show Answer
Answer: C – With the current coming out of the page, the grip rule gives field lines circling anticlockwise as seen by the observer. At the rightmost point of a circle, the anticlockwise tangent points towards the top of the page.
Q7. A wire carries a current going into the plane of the page, represented by a cross. At a point immediately above the wire, the magnetic field points
- (A) towards the top of the page
- (B) to the right
- (C) to the left
- (D) towards the bottom of the page
Show Answer
Answer: B – With the current going into the page, the field lines circle clockwise as seen by the observer. At the topmost point of a clockwise circle, the tangent points to the right.
Q8. The flux density at a distance r from a long straight wire is X. What is the flux density at a distance of 3r from the same wire?
- (A) 3X
- (B) 9X
- (C) X / 9
- (D) X / 3
Show Answer
Answer: D – For a straight wire B is proportional to 1/r (not 1/r2). Tripling the distance reduces B to X / 3.
Q9. The magnetic field around a current-carrying straight wire is strongest at points where
- (A) the field lines are closest together, nearest to the wire
- (B) the field lines are furthest apart, far from the wire
- (C) the point is closest to one of the ends of the wire
- (D) the circles of field are largest
Show Answer
Answer: A – The spacing of field lines shows the strength of the field. Near the wire the circular lines are closely spaced and B is largest; B falls as 1/r with distance.
Q10. A long straight wire carries a current of 20 A. At what distance from the wire is the flux density 8.0 × 10−5 T?
- (A) 0.010 m
- (B) 0.025 m
- (C) 0.050 m
- (D) 0.10 m
Show Answer
Answer: C – r = μ0I / (2πB) = (2 × 10−7 × 20) / (8.0 × 10−5) = (4.0 × 10−6) / (8.0 × 10−5) = 0.050 m.
Fields of Coils & Solenoids
Q11. A flat circular coil has N turns, radius r, and carries a current I. The magnetic flux density at the centre of the coil is
- (A) μ0NI / (2πr)
- (B) μ0NI / (2r)
- (C) μ0nI, where n is the number of turns per unit length
- (D) μ0I / (2πr)
Show Answer
Answer: B – At the centre of a flat coil B = μ0NI / (2r). Option (D) is the field of a straight wire, and option (C) is the field inside a solenoid.
Q12. A flat circular coil of 50 turns and radius 0.050 m carries a current of 2.0 A. The flux density at the centre of the coil is
- (A) 1.3 × 10−4 T
- (B) 6.3 × 10−4 T
- (C) 2.5 × 10−3 T
- (D) 1.3 × 10−3 T
Show Answer
Answer: D – B = μ0NI / (2r) = (4π × 10−7 × 50 × 2.0) / (2 × 0.050) = (4π × 10−5) / 0.10 = 1.26 × 10−3 T, about 1.3 × 10−3 T.
Q13. Inside a long solenoid carrying a steady current, at points away from the ends, the magnetic field is
- (A) strong, approximately uniform, and parallel to the axis of the solenoid
- (B) zero, because the fields of the opposite sides of the coils cancel
- (C) circular, centred on the axis of the solenoid
- (D) strongest at the two ends and zero at the centre
Show Answer
Answer: A – The fields of the many turns add together inside the solenoid to give a strong, nearly uniform field directed along its axis. Outside, the field is much weaker.
Q14. A solenoid has 1000 turns per metre and carries a current of 2.0 A. The flux density inside the solenoid is
- (A) 4.0 × 10−4 T
- (B) 1.3 × 10−3 T
- (C) 2.5 × 10−3 T
- (D) 5.0 × 10−3 T
Show Answer
Answer: C – B = μ0nI = 4π × 10−7 × 1000 × 2.0 = 8π × 10−4 = 2.5 × 10−3 T.
Q15. One end of a solenoid is viewed from outside that end. The current in the windings at this end flows anticlockwise as seen by the observer. This end of the solenoid behaves as
- (A) a south pole
- (B) a north pole
- (C) neither pole, because the field inside a solenoid is uniform
- (D) a pole only when an iron core is inserted
Show Answer
Answer: B – By the grip rule, the end where the current appears anticlockwise is the end the field lines leave, which is the north pole of the solenoid.
Q16. The magnetic field pattern outside a current-carrying solenoid is most similar to the field pattern of
- (A) a bar magnet
- (B) a single long straight wire
- (C) a positive point charge
- (D) two oppositely charged parallel plates
Show Answer
Answer: A – Outside the solenoid the field lines spread out from the north end and return to the south end, just as for a bar magnet, although the external field is much weaker than the field inside.
Q17. Which change would NOT increase the magnetic flux density inside a solenoid?
- (A) Increasing the current in the coils
- (B) Winding more turns into the same length of solenoid
- (C) Inserting a soft iron core
- (D) Making the solenoid longer while keeping the same total number of turns and the same current
Show Answer
Answer: D – B = μ0nI depends on the turns per unit length, n. Stretching the same number of turns over a longer solenoid reduces n, so B decreases rather than increases.
Q18. A solenoid of length 0.25 m is wound with 500 turns and carries a current of 1.5 A. The flux density inside it is
- (A) 9.4 × 10−4 T
- (B) 1.9 × 10−3 T
- (C) 3.8 × 10−3 T
- (D) 7.5 × 10−3 T
Show Answer
Answer: C – n = 500 / 0.25 = 2000 turns per metre. B = μ0nI = 4π × 10−7 × 2000 × 1.5 = 1.2π × 10−3 = 3.8 × 10−3 T.
Q19. Inserting a soft iron core into a current-carrying solenoid greatly increases the flux density inside the solenoid because
- (A) most of the current then flows through the iron instead of the coils
- (B) the iron becomes magnetised and its field adds to the field of the coils, the iron having a high permeability
- (C) the iron core reduces the electrical resistance of the coils
- (D) the iron core increases the number of turns per unit length
Show Answer
Answer: B – The field of the coils magnetises the soft iron, and the field of the magnetised iron adds to the coil field. A material of high permeability can increase the total flux density by a large factor.
Q20. A flat circular coil lies in the plane of the page and its current flows clockwise as seen by the observer. The magnetic field at the centre of the coil points
- (A) into the plane of the page
- (B) out of the plane of the page
- (C) towards the top of the page, in the plane of the page
- (D) towards the right, in the plane of the page
Show Answer
Answer: A – Curl the fingers of the right hand in the direction of the current (clockwise as seen by the observer); the thumb then points away from the observer, so the field at the centre points into the page.
Magnetic Field Calculations & Comparisons
Q21. The flux density at a distance of 0.040 m from a long straight wire is 6.0 × 10−5 T. At what distance from the wire is the flux density 3.0 × 10−5 T?
- (A) 0.010 m
- (B) 0.020 m
- (C) 0.080 m
- (D) 0.16 m
Show Answer
Answer: C – B is inversely proportional to distance. To halve B, the distance must double: 2 × 0.040 = 0.080 m.
Q22. Two long parallel wires are 0.20 m apart. One carries a current of 6.0 A vertically upwards and the other carries 2.0 A vertically upwards. What is the resultant flux density at the point midway between the wires?
- (A) 1.6 × 10−5 T, into the page
- (B) 1.6 × 10−5 T, out of the page
- (C) 8.0 × 10−6 T, out of the page
- (D) 8.0 × 10−6 T, into the page
Show Answer
Answer: D – The point is 0.10 m from each wire. Field of the 6.0 A wire: (2 × 10−7 × 6.0) / 0.10 = 1.2 × 10−5 T, directed into the page (the point is to the right of this wire if it is the left-hand wire). Field of the 2.0 A wire: (2 × 10−7 × 2.0) / 0.10 = 4.0 × 10−6 T, directed out of the page. The fields oppose, so the resultant is 1.2 × 10−5 − 4.0 × 10−6 = 8.0 × 10−6 T, into the page.
Q23. Two long parallel wires, 0.20 m apart, carry currents of 3.0 A and 6.0 A, both vertically upwards. Point P lies on the line joining the wires, on the far side of the 6.0 A wire from the 3.0 A wire, at a distance of 0.10 m from the 6.0 A wire and 0.30 m from the 3.0 A wire. The resultant flux density at P is
- (A) 1.0 × 10−5 T, into the page
- (B) 1.4 × 10−5 T, into the page
- (C) 1.4 × 10−5 T, out of the page
- (D) 1.0 × 10−5 T, out of the page
Show Answer
Answer: B – P is on the same side of both wires, so both fields point into the page and add. From the 3.0 A wire: (2 × 10−7 × 3.0) / 0.30 = 2.0 × 10−6 T. From the 6.0 A wire: (2 × 10−7 × 6.0) / 0.10 = 1.2 × 10−5 T. Total = 2.0 × 10−6 + 1.2 × 10−5 = 1.4 × 10−5 T, into the page.
Q24. Two long parallel wires are 0.40 m apart and carry currents of 2.0 A and 8.0 A in the same direction. The point between the wires where the resultant magnetic field is zero is at a distance of
- (A) 0.080 m from the wire carrying 2.0 A
- (B) 0.20 m from the wire carrying 2.0 A
- (C) 0.32 m from the wire carrying 2.0 A
- (D) there is no such point between the wires
Show Answer
Answer: A – Between the wires the two fields point in opposite directions. At distance x from the 2.0 A wire, the fields are equal when 2.0 / x = 8.0 / (0.40 − x), so 2.0(0.40 − x) = 8.0x, giving 0.80 = 10x and x = 0.080 m. The null point is closer to the wire carrying the smaller current.
Q25. Two long parallel wires carry currents in opposite directions. At which points can the resultant magnetic field be zero?
- (A) Exactly midway between the wires
- (B) At points anywhere between the wires
- (C) Only at points outside the pair of wires, on the side of the wire carrying the smaller current
- (D) Nowhere, because the two fields always add together
Show Answer
Answer: C – Between wires carrying opposite currents, the two fields point in the same direction and reinforce, so there is no null point between them. Outside the pair the fields oppose, and they can cancel on the side of the smaller current, where that wire’s field falls from being dominant close up to being weaker far away.
Q26. The magnetic field of the Earth has a flux density of about 5.0 × 10−5 T. A long straight wire carries a current of 15 A. At what distance from the wire is its magnetic field equal to 5.0 × 10−5 T?
- (A) 0.015 m
- (B) 0.030 m
- (C) 0.12 m
- (D) 0.060 m
Show Answer
Answer: D – r = (2 × 10−7 × 15) / (5.0 × 10−5) = (3.0 × 10−6) / (5.0 × 10−5) = 0.060 m. Within a few centimetres of the wire, its field is stronger than the Earth’s field.
Q27. A solenoid has 500 turns per metre and carries a current of 0.40 A. A second point is 0.020 m from a long straight wire carrying a current of 8.0 A. Which field is the larger?
- (A) The field inside the solenoid, about 2.5 × 10−4 T, is larger than the field near the wire, 8.0 × 10−5 T
- (B) The field near the wire, about 8.0 × 10−4 T, is larger than the field inside the solenoid, 2.5 × 10−5 T
- (C) The two fields are equal
- (D) The field inside the solenoid is zero, so only the wire produces a field
Show Answer
Answer: A – Solenoid: B = μ0nI = 4π × 10−7 × 500 × 0.40 = 2.5 × 10−4 T. Wire: B = (2 × 10−7 × 8.0) / 0.020 = 8.0 × 10−5 T. The solenoid field is about three times larger.
Q28. Two long parallel wires are 0.10 m apart. Each carries a current of 5.0 A, the currents being in opposite directions. The resultant flux density at the point midway between the wires is
- (A) zero
- (B) 4.0 × 10−5 T
- (C) 2.0 × 10−5 T
- (D) 8.0 × 10−5 T
Show Answer
Answer: B – The point is 0.050 m from each wire, so each field is (2 × 10−7 × 5.0) / 0.050 = 2.0 × 10−5 T. For opposite currents the two fields point in the same direction between the wires and add: total = 4.0 × 10−5 T.
Q29. A solenoid has 2000 turns per metre. What current is needed to produce a flux density of 5.0 × 10−3 T inside the solenoid?
- (A) 0.50 A
- (B) 1.0 A
- (C) 2.0 A
- (D) 4.0 A
Show Answer
Answer: C – I = B / (μ0n) = (5.0 × 10−3) / (4π × 10−7 × 2000) = (5.0 × 10−3) / (2.5 × 10−3) = 2.0 A.
Q30. The current in a long straight wire is doubled, and the flux density is measured at half the original distance from the wire. Compared with the original flux density, the new flux density is
- (A) unchanged
- (B) twice as large
- (C) half as large
- (D) four times as large
Show Answer
Answer: D – B is proportional to I / r. Doubling I multiplies B by 2, and halving r multiplies B by 2 again, so the new flux density is 2 × 2 = 4 times the original.
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