DP Physics (first assessment 2025)

Question EXE.2.SL.TZ0.14

Date	Example questions Example questions	Marks available	[Maximum mark: 7]	Reference code	EXE.2.SL.TZ0.14
Level	SL	Paper	2	Time zone	TZ0
Command term	Calculate, Deduce, Explain, State	Question number	14	Adapted from	N/A

14.

[Maximum mark: 7]

EXE.2.SL.TZ0.14

Two long parallel current-carrying wires P and Q are separated by 0.10 m. The current in wire P is 5.0 A.

The magnetic force on a length of 0.50 m of wire P due to the current in wire Q is 2.0 × 10⁻⁵ N.

(a)

State and explain the magnitude of the force on a length of 0.50 m of wire Q due to the current in P.

[2]

Markscheme

From Newton’s third law, the force on a length of Q is equal but opposite to the force on the same length of P ✓

$2.0 \times 10^{- 5} N$ ✓

(b)

Calculate the current in wire Q.

[2]

Markscheme

$\frac{2.0 \times 10^{- 5}}{0.50} = \frac{4 π \times 10^{- 7} \times 5.0 \times I_{Q}}{2 π \times 0.10}$ ✓

$I_{Q} = 4.0$ «A» ✓

(c)

Another current-carrying wire R is placed parallel to wires P and Q and halfway between them as shown.

The net magnetic force on wire Q is now zero.

(c.i)

State the direction of the current in R, relative to the current in P.

[1]

Markscheme

Opposite ✓

(c.ii)

Deduce the current in R.

[2]

Markscheme

The force on Q due to R must have the same magnitude «but opposite direction» as the force on Q due to P ✓

The distance is halved therefore one half of the current is needed to produce the same force, so 2.5 A ✓

Syllabus sections