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Date May 2018 Marks available 3 Reference code 18M.2.HL.TZ1.4
Level Higher level Paper Paper 2 Time zone Time zone 1
Command term Show that Question number 4 Adapted from N/A

Question

An ohmic conductor is connected to an ideal ammeter and to a power supply of output voltage V.

M18/4/PHYSI/SP2/ENG/TZ1/04

The following data are available for the conductor:

                    density of free electrons     = 8.5 × 1022 cm−3

                    resistivity                          ρ = 1.7 × 10−8 Ωm

                    dimensions           w × h × l = 0.020 cm × 0.020 cm × 10 cm.

 

The ammeter reading is 2.0 A.

The electric field E inside the sample can be approximated as the uniform electric field between two parallel plates.

Calculate the drift speed v of the electrons in the conductor in cm s–1.

[2]
b.

Determine the electric field strength E.

[2]
c.i.

Show that \(\frac{v}{E} = \frac{1}{{ne\rho }}\).

[3]
c.ii.

Markscheme

v «= \(\frac{I}{{neA}}\)» = \(\frac{2}{{8.5 \times {{10}^{22}} \times 1.60 \times {{10}^{ - 19}} \times {{0.02}^2}}}\)

0.37 «cms–1»

 

[2 marks]

b.

VRI = 0.086 «V»

«\(\frac{V}{d} = \frac{{0.086}}{{0.10}} = \)» 0.86 «V m–1»

 

Allow ECF from 4(a).

Allow ECF from MP1.

[2 marks]

c.i.

ALTERNATIVE 1

clear use of Ohm’s Law (IR)

clear use of R = \(\frac{{\rho L}}{A}\)

combining with I = nAve and V = EL to reach result.

 

ALTERNATIVE 2

attempts to substitute values into equation.

correctly calculates LHS as 4.3 × 109.

correctly calculates RHS as 4.3 × 109.

 

For ALTERNATIVE 1 look for:

VIR

R = \(\frac{{\rho L}}{A}\)

VEL

InAve

VI\(\frac{{\rho L}}{A}\)

EL = I\(\frac{{\rho L}}{A}\)

EI\(\frac{\rho }{A}\)

EnAve\(\frac{\rho }{A}\) = nveρ

\(\frac{v}{E} = \frac{1}{{ne\rho }}\)

[3 marks]

c.ii.

Examiners report

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

Syllabus sections

Core » Topic 5: Electricity and magnetism » 5.1 – Electric fields
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