1. show an appreciation that a force might act on a cutent'carrying conductor placed in amagnetic field.recall and solve problems using the equation F=B sing, with directions as interyreted byFleming's left.hand rule.

A straight, horizontal, current-carrying wire AB lies at right angles to a horizontal magnetic field.The field exerts a vertical force of 8.0 mN on the wire.The wire is rotated, in its horizontal plane, through 30" as shown in Figure 1. The flux density ofthe magnetic field is halved.

(\v)

2.

rf

bWhat is the

Overtical force on the wire? [3.5 mN]

Let original field-be B.F=B//=B ). 10- N

New force on wire F' = B'l/

=!Bcos3ooll

=Ly!3s11= 3.s i ro* r.r:35mN

A piece of wire carrying a current of 6 A is placed in a uniform magnetic field as shown below. lfthe magnitude of the magnetic field is 0.4 T, what is the force acting on each of the segments WX,XY and Yz?

Force acting on WX = BIL (sin 0")=BlLx0=0N

--- B- 12 cm

= BIL (sin 90").= (0 a x 6 x0.11)x (1)

= 0.2EN

= BIL (sin 60')= (0 4 x 6 x0.14) x (sin 60")= 0.29 N

Force acting on XY

Force acting on Yz

A wire 60 cm long and of massdensity '1.2 T. Determine thetension in the supporting leads.

10 g is suspended by amaqnitude and direction

leads in a magnetic field of fluxcurrent required to remove the

pair ofof the

Direction of current: to the right-

F= BIL sin emg = BIL sin 6

(]0*q, o.et; = 1t.Z " I x 0.60) x (sin 90")

The diagram below show a cylindrical aluminium bar AB of length 5.0 cm resting on tvvo horizontalaluminium rails. The rails can be connected to a battery to drive a cunent through A.

(a) A magnetic field, of flux density 0.10 T, acts perpendicularly to thedirection will AB move if the current flows?

Ljsing Fleming's Left Hand Rule, AB will move to ihe right

paper and into it. ln which

(b) Calculate the angle to the horizontal to which the rails must be tilted to keep AB stationary ifits mass is 5.0 g, the current in it is 4.0 A and the direction of the field remains unchanged.

122'l

To ensure that AB remains stationary,mg sing = FE cosqmg sinq = Bll cos9si0_Btlcosg trq

- o o^4.u.0.15lanH = 0005x931B=22.2'=22'

3.4.

define magnetic flux density and the tesla.show an undershnding of how the force on a current.carrying conductor can be used tomeasure the flux densily of a magnetic field using a current balance.

A wire frame, balanced on a pivot, is placed inside a solenoid as shown. The solenoid has 600turns m_r and carries a current of 4.0 A.

aa@D

ooooooooooooooo

(a) Given that the magnetic flux density B on the axis of a long solenoid is B=(4n x 107)nl,where r, is the number of turns of the solenoid and / is the current passing through thesolenoid, calculate the magnetic flux density in the region of side CD of the frame. I3.0 x1Or Tl

BcD = 4n x 1Ot nl= 4n x 10/)< 600 x 4= 3.02 x 1O'3

=3.0x1037

(b) Side CD has length 3 cm. Given that a current of 5.0 A now flows in the wire frame,calculate the force acting on CD due to the magnetic field in the solenoid. t4.5 x 104 Nl

Fco = BII= 3.02 ). 10-3x 5.0x 0 03=4.53x104=4.5x104N

(c) A cotton gauze of mass 0.05 g is placed on the side BA and positioned so as to keep theframe horizontal. Given that XD is of length 10 cm, how far from the pivot must the gauzebe positioned? [9.2 cm]

Let d be the distance between the cotton gauze and the pivot.Taking moments about the pivot, when the frame is horizontal,

mgd = FXXD

mg 0.00005x9.41

=00917=92cm

5.6.

6

predicl the direction of the force on a charge moving in a magnetic field.recall and solve problems using F=Bqv sine.

q-particles, each of speed v and mass m, are travelling in a narrow beam in a vacuum, as shownin the figure below.

Uniform magneticfield, flux density B

The q-particles enter a region where there is a uniform magnetic fleld of flux densityparticles in the magnetic field move along the arc of a circle of radius r.

(a) State the direction ofthe magneticfield.

out of the page

Show that, for an q-particle in this beam,

(i) the magnitude p of its momentum is given by the expression

p = 2Berwhere e is the elementary charge.

F= Bgv sin e

F=Br.2er. ysin90' =a4zEief=-=mvP=2Bet

(ii) the kinetic energy EK is given by the expression

8. The q-

(b)

2(B er)z

=P'

',,-;_ 2(Ber)z

_ (Ber)z

(c) On the figure above, sketch the path of an o-particle that has a speed greater ihan v.

(the initial path is in dotted lines)Explanationr The particle with a larger speed would be deflected less, or move along an arc with a

larqer radius r.

Uniform magneticfield, flux density B

describe and analyze deflections of beams or charged pafticles by uniform electric anduniform magnetic fields,

A collision between elementary particles in a magnetic field of strength '1 mT produces othercharged particles which continue to move as shown in Figure 5. lt is known that particles A and Bare either an electron or proton. (l\,4ass of electron=9.11x1031)(a) lf both particles A and B are moving with the same speed, determine which is the proton and

which is the electron.

Distinguish by direction of magnetic force: Using Fleming's Lefi Hand Rule, A is the proton and Bis the electron (net force acts towards the centre of the circular path).Distinguish by radius of path: The particle with larger momentum (proton, as it has a larger massand same speed) would trace out an arc with larger radius.

(b) Given that the radius of the circular path of the electron is 8 pm, what is its speed?11.4 \ 103 msrl

For circular motion,

= t.i"' to' rs'

o-particles

I A proton and q-particle are injected with the same velocity at right angles into a uniform magneticfield. What is the ratio of the radius of the circular path of the proton to that of the q-particle? [0.5](Assume that the mass of the o-particle is approximately 4 times that of the proton.)

q-particle = 1fleTherefore q, = 2

lL =!t!2 +!e!.s

:-)\-Rq.p 4mr

= !s!4qp

t24l

exptain how electric and magnetic fields can be used in velocity selection for chargedparticles.

The figure below shows a region PQRS of a uniform magnetic field directed downwards into theplane of the paper. Electrons, all having the same speed, enter the region of the magnetic field inthe direction shown.

(a) On the figure below, show the path of the electrons as they pass through the magneiicfield, emerging from side RS.

(b) A uniform electric field is also applied in the region PQRS such that the electrons now passundeflected through this region. On Figure B, mark with an arrow labeled E, the direction of theelectric field.

(a)and (b)PS

path of electrons

T'Qeld ob*].

Bqu = Eq

EB

(i) lf speed = 2v, FE =Bqvwill be doubled;FE = Eq will remain unchanged.Therefore Fs > FEi.e. downward force > upward forceThe elecirons are deflected downwards

r ii) From\-f . charoeof *e or -e does not afleol drrectron dellecLron.'' B

A$ul,o^ fu *"

(c) The undeflected electrons in (b)each have charge -e, mass m and speed v. State and explainthe effect, if any, on particles entering the region PQRS of the same magnetic and electricfields as in (b) if the parlicles each have

(i) charge -e, mass m and speed 2v, and

(ii) charge +e, mass m and speed v-

For zero deflection,

9.

10.

11.

10

sketch flux palterns due to a long straightwire, a flat circular coil and a long solenoid.

show an understanding that the field due to a solenoid may be influenced by the presenceof a ferrous core.explain the forces between cufient-carrying conductors and predict the direction of theforces.

Four parallel conductors, carrying equal currents, pass vertically through the 4 corners of a squareWXYZ. ln 2 conductors, the current is directed into the page and, in the other 2, jt is directed out ofthe page. lf the resultant magnetic field at P is in the direction shown, determine the direction ofcuffent in each of the wires.

resultant B

a

c

11 Four wires in a horizontal plane each carry a current / in the directions shown below. They crosseach other without touching to form a square. Point P is at the centre of the square.

Given that the magnetic flux density at P due to one wire alone is B, calculate the strength and

direction of the magnetic field at P due to all 4 wires.

The strength of the magnetic field at P due to all 4 wires is 28, and the direction of the field is intothe page.

(the 2 parallel wires carrying current in the same direction will produce magnetic fields withopposite direction but the same magnitude that cancel out)(the 2 parallel wires carrying current in the opposite direction will produce magnetic field with same

direction (into the page)

There the magnetic field at P due to all 4 wires is 28, jnto the page.

12 W is a long straight wire that is carrying a current out of the plane of the page. lf it experiences a

uniform external B field as shown in the diagram below, redraw ihe resultant field lines near W.

By examining the resultant field lines,force, and if so, in which direction.

discuss whether the current-carrying wire experiences any

wo)

3 wires, X, Y and Z, are placed normally to the plane of the paper. Each of them carry curent 11, 12

and /3 into the plane of the paper and each wire is separated by a distance r as shown below.Cuffents /r and /2 cause magnetic fields of flux density 87and 82 at Z respectively.

Copy the figure and draw the direction of the magnetic field which /r causes at Z and labelit 8,. AIso draw Br, the direction of the magnetic field which /2 causes at Z.

Draw, on the same diagram, the approximate net force F actjng on the wire Z.

B

13

(a)

(b)

(c) lf wire Z is replaced with a stationary positive charge, explain the efiect on the charge,

any.

There would be no force acting on the charge because there is no current flow since thecharge is stationary, therefore there is no magnetic force.

14 A small coil lies inside a larger coil. The two coils are horizontal, concentric and carry currents in

the same direction as shown in the figure below.

Describe and explain the force/torque, if any, acting on the small coil.

There is no resultant force acting on the smaller coil.

At each point on the circumference of the smaller coil, there is a force that acts radially ouhvards(current in 2 wires in the same direction). However, this force is balanced by another force of equalmagnitude, acting on the coil on a point directly opposite.

The figure below shows the presence of a current / flowing downwards through a long straightwire. A current of / also flows through the coil in a clockwise direction-Describe and explain the motion of the rectangular coil.

current-carryingcoil

long straight wire

Rectangular coil wjll move to the right. The side of the .ectangular coil nearer to the wire will beattracted to the wire while the opposite side of the rectangular coil will be repelled due to thedirection of the current. However, though boih sides have the aame current flowing through them,the side closer to the wire will experience a bigger force and hence, the resultant force on therectangular coil wrll be towards the right.

15

Additional Question

16 A U-shaped magnet is resting on a weighing machine as shown. Between the poles of the magnetis suspended a conducting wire such that 2.5 cm of the wire lies perpendicularly to the field lines ofthe magnetic poles. When the wire is not carrying any current, the weighing machine reads 100 g-

When a current of 5.0 A passes through the wire, the reading on the weighing machine changes.The B-field between the poles of the magnets has a magnitude of 0.50 T.

(a) Calculate the magnetic force acting on the wire as a result of the current flow.

The magnetic force is upwards.

F= BIL sin e

= (o 50 x 5.0 x 1oo?:)

x (sin 90')= 0.0625 N= 0.063 N (to 2 s.f.)

Hence deduce the additional force acting on the magnet. Justify your answer.

The additionalforce acting on the magnet is 0.063 N.This is an action - reaction force. The magnet exerts a force on the current carrying wireupwards, and the wire exerts an equal and opposite force on the magnet downwards-

Draw the free-body diagram of the magnet.

ighing machine on a magnet

Force by wire on maq

(b)

c)

By considering the forces acting on theweighing machine.

The new readino = 100 + l!!!l x 10001

=1069

Force by

Weight of magnet

(d) maqnet, find the new readinq, in grams, on the