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Date November 2021 Marks available 1 Reference code 21N.2.HL.TZ0.5
Level Higher level Paper Paper 2 Time zone 0 - no time zone
Command term Show that Question number 5 Adapted from N/A

Question

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.

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

[1]
a.

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

[1]
b.i.

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

[1]
b.ii.

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

[2]
c.i.

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]
c.ii.

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

[2]
d.i.

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]
d.ii.

Markscheme

703.5

a.

shape as above ✓

b.i.

shape as above ✓

 

Vertical lines not necessary to score.

Allow ECF from (b)(i).

b.ii.

ALTERNATIVE 1

maximum flux at «5.0×5.0×10-4×85×0.94» =0.199750.20«Wb» ✓

emf = «0.200.25=» 0.80«V» ✓


ALTERNATIVE 2

emf induced in one turn = BvL0.94×0.20×0.05=0.0094«V» ✓

emf =85×0.0094=0.80«V» ✓

 

Award [2] marks for a bald correct answer.

Allow ECF from MP1.

c.i.

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.

c.ii.

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.i.

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.

d.ii.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.i.
[N/A]
c.ii.
[N/A]
d.i.
[N/A]
d.ii.

Syllabus sections

Additional higher level (AHL) » Topic 11: Electromagnetic induction » 11.1 – Electromagnetic induction
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Additional higher level (AHL) » Topic 11: Electromagnetic induction
Additional higher level (AHL)

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