practice test - magnetic fields motion in magnetic fields with solutions

5
Le Fevre High School SACE Stage 2 Physics Practice Test - Magnetic Fields & Motion in Magnetic Fields 1. A small current carrying wire of length 2 cm is  placed in a uniform magnetic field as shown. The magnetic field is perpendicular and directed into the plane of the page. (a) Determine the direction of the force on the current carrying wire if the current flows from west to east. (1 mark) (b) The magnitude of the force on the wire is measured as 2 x 10 !  "# when the current flowing through the wire is 1.$ A. Determine the magnitude of the magnetic field. (2 marks) 2. %&etch the magnetic field in each of the following situations. (a) %t ra ight cond uc ti ng wi re. (b) 'r oss se ct io n of a sole no id (3 marks) . A particle of mass m# charge * is initially stationary in an electric field of strength +. + is in the plane of the  page as shown. (a) +xplain the resulting motion of the  positi,e charge. (2 marks) (b) The electric field is now turned off and a magnetic field of strength - directed at right angles to the plane of the page is now switched on. +xplain the resulting motion of the mo,ing charged particle. (2 marks) (c) %how that the period of re,olution of the resultant motion is gi,en by E 'ross sectional cut through centre of the solenoid.

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Page 1: Practice Test - Magnetic Fields Motion in Magnetic Fields With Solutions

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Le Fevre High School

SACE Stage 2 PhysicsPractice Test - Magnetic Fields & Motion in Magnetic Fields

1. A small current carrying wire of length 2 cm is placed in a uniform magnetic field as shown.The magnetic field is perpendicular anddirected into the plane of the page.

(a) Determine the direction of theforce on the current carrying wire ifthe current flows from west to east.

(1 mark)(b) The magnitude of the force on the

wire is measured as 2 x 10 ! "#when the current flowing through

the wire is 1.$ A. Determine themagnitude of the magnetic field.

(2 marks)

2. %&etch the magnetic field in each of thefollowing situations.(a) %traight conducting wire. (b) 'ross section of a solenoid

(3 marks). A particle of mass m# charge * is

initially stationary in an electric fieldof strength +. + is in the plane of the

page as shown.

(a) +xplain the resulting motion of the positi,e charge.(2 marks)

(b) The electric field is now turned offand a magnetic field of strength -directed at right angles to the planeof the page is now switched on.+xplain the resulting motion of themo,ing charged particle.

(2 marks)(c) %how that the period of re,olution

of the resultant motion is gi,en by

E

'ross sectionalcut throughcentre of thesolenoid .

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Le Fevre High School

-*m2

T π

=

(3 marks)/. (a) Draw a schematic diagram of a mo,ing coil loudspea&er. (4 marks)

(b) -riefly explain the action of the mo,ing coil loudspea&er. (3 marks)

$. (a) hat *uantities determine the time an ion ta&es to complete a semi circle in thedee of a cyclotron.

(2 marks)(b) n an experiment a magnetic field of strength 2.$ T was used# and protons were

extracted at a radius of $ cm. hat was the period of the protons in thecyclotron

(2 marks). ons tra,elling with the same ,elocity enter a uniform magnetic field initially at

right angles to the field.

(a) f two of the ions ha,e e*ual masses but different charges# explain which ionwill mo,e in the path with the larger radius

(2 marks)(b) f an ion is tra,elling with a ,elocity of $.0 x 10 ! m s 1 and the magnetic field

is of intensity.$ x 10 T# determine the magnitude of the force per unit charge acting on

the ion.(2 marks)

TOTAL MARKS: 28

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Le Fevre High School

SACE Stage 2 PhysicsPractice Test - Magnetic Fields & Motion in Magnetic Fields

Solution

1. (a) 3sing the right hand rule

4orce will be in the plane of the page and directed up the page.

(b) l 5 2 cm 5 0.02 m#

4 - 5 2 x 10!

"# 5 1.$ A

4 5 - l sinθ at θ 5 60 o 2 x 10 ! 5 -(1.$)(0.02)- 5 .! x 10 T

2.

.(a) %ince in an electric field and charge experiences a force# then this force on a

positi,e charge is directed the same as that of the field lines. 7ence the chargemo,es in a straight line at a constant acceleration in the direction of the electricfield lines. (to the left)

(b) 3sing the right hand rule# with ,elocity (thumb) to the left# the - field into the page (extended fingers)# then the resulting force is in the plane of the page and,ertically down.

This then changes the direction of the mo,ing particle. 7owe,er since the force is perpendicular to the ,elocity# it does not alter its speed. 7owe,er as the charge

changes direction# then the force also changes to remain mutually perpendicular.This results in the particle again changing direction but not speed. This process

I

FB

B

4ield lines circular# nonuniformly spaced (closertogether near the wire)

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Le Fevre High School

continues resulting in the particle mo,ing in a circular path (or arc of a circle) asthe force is always directed perpendicular to the ,elocity and hence is ancentripetal force.

(c) %ince the centripetal force is supplied by the magnetic force

4 - 5 -*,4 c 5

mv R

2

Asmv

R

2

5 m Rω 2 5 m

T R

2 2

π

-*, 5 BqT

R2 π

mT

R2

5 BqT

R2 π

8e arranging T 5

2 π m

Bq

/.(a)

/ (b) two ends of the ,oice coil are connected to the output terminals of an amplifier.

Across these terminals is a changing potential difference that oscillates in proportion to the sound wa,eform to be reproduced. The changing potentialdifference produces an oscillating current in the ,oice coil.

$.(a) the time is affected by the charge 9 mass of the ion# and the applied magnetic

field strength.

(b) T 52 π m Bq (from : )

5)$.2)(10x.1(

)10x!.1(216

2!

−π

5 2. 2 x 10 ; s

<uter suspension

inner suspension

Air gap

=oice coil

'entre cap 9dust dome

8ing >agnet

%oft iron pole pieces

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Le Fevre High School

. hen they enter the uniform magnetic field then magnetic force supplies thecentripetal force.

4 c 5 4 - mv

R

2

5 -* v

hence 8 5 mv Bq

since 8 ∝ 1?* and they ha,e the same m# v the particle with the smaller charge will tra,el in the larger radius path.

(b)4 5 -* v

4?* 5 - v 5 ( .$ x 10 )($.0 x 10 ! )

5 1.!$ x 10 $ " ' 1