DP Physics (first assessment 2025)

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Question 21N.2.HL.TZ0.5

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Date November 2021 Marks available [Maximum mark: 11] Reference code 21N.2.HL.TZ0.5
Level HL Paper 2 Time zone TZ0
Command term Calculate, Estimate, Show that, Sketch Question number 5 Adapted from N/A
5.
[Maximum mark: 11]
21N.2.HL.TZ0.5

A square loop of side 5.0 cm enters a region of uniform magnetic field at t = 0. The loop exits the region of magnetic field at t = 3.5 s. The magnetic field strength is 0.94 T and is directed into the plane of the paper. The magnetic field extends over a length 65 cm. The speed of the loop is constant.

(a)

Show that the speed of the loop is 20 cm s−1.

[1]

Markscheme

703.5 ✓

(b.i)

Sketch, on the axes, a graph to show the variation with time of the magnetic flux linkage Φ in the loop.

[1]

Markscheme

shape as above ✓

(b.ii)

Sketch, on the axes, a graph to show the variation with time of the magnitude of the emf induced in the loop.

[1]

Markscheme

shape as above ✓

 

Vertical lines not necessary to score.

Allow ECF from (b)(i).

(c.i)

There are 85 turns of wire in the loop. Calculate the maximum induced emf in the loop.

[2]

Markscheme

ALTERNATIVE 1

maximum flux at «5.0×5.0×10-4×85×0.94» =0.19975≈0.20 «Wb» ✓

emf = «0.200.25=» 0.80 «V» ✓


ALTERNATIVE 2

emf induced in one turn = BvL = 0.94×0.20×0.05=0.0094 «V» ✓

emf =85×0.0094=0.80 «V» ✓

 

Award [2] marks for a bald correct answer.

Allow ECF from MP1.

(c.ii)

The resistance of the loop is 2.4 Ω. Calculate the magnitude of the magnetic force on the loop as it enters the region of magnetic field.

[2]

Markscheme

I=«VR=»0.82.4  OR  0.33 «A» ✓

F=«NBIL=85×0.94×0.33×0.05=»=1.3 «N» ✓

 

Allow ECF from (c)(i).

Award [2] marks for a bald correct answer.

(d.i)

Show that the energy dissipated in the loop from t = 0 to t = 3.5 s is 0.13 J.

[2]

Markscheme

Energy is being dissipated for 0.50 s ✓


E=Fvt=1.3×0.20×0.50=«0.13 J»

OR

E=Vlt=0.80×0.33×0.50=«0.13 J» ✓

 

Allow ECF from (b) and (c).

Watch for candidates who do not justify somehow the use of 0.5 s and just divide by 2 their answer.

(d.ii)

The mass of the wire is 18 g. The specific heat capacity of copper is 385 J kg−1 K−1. Estimate the increase in temperature of the wire.

[2]

Markscheme

∆T=0.130.018×385 ✓

∆T=1.9×10-2 «K» ✓

 

Allow [2] marks for a bald correct answer.

Award [1] for a POT error in MP1.

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