AB and BC, joined at B.
carrying a current of 2.0 A and subjected to a magnetic field of flux density
001 T direction in the plane ABC at to AB. Both AB and BC are 5_Ocm long. The
angle ABC is 60'. Calculate the forces AB and BC What mmerncnt do tlx two forces
together try to prcxlucc?
The force on AB is
F o.(ns7 x
2.0 x S x 10-2
The force on BC is but. whic the force AB is upwards out Of
the the left-hand rule gives thc force on BC to These forces therefore a couple
the line BD in the diagram. and ngatim this c-sFcted.
x 10-4 N, Rotaú'n aNNt BD. Force on a charged particle moving
through a magnetic field
Here q is the charge carried by the particle, v is the
particle's vekxity and B is the magnetic flux density to v.
Wc arc B to IXrrx•ndÉular
to v. Otherwise B must be replaced in the formula by its perpendicular to v,
since only this is effective. The direction Of F is Of given the left-hand rule
this rule considers current direction, which for the movement
Of particles such as electrons will be opposite to that of the particle
An electron is moving with a speed
of 10 7 ms" I perpendicular to a magnetic field
having a uniform flux density of oun2T.
(a)Calculate the force on the electron.
(b)Calculate the radius of the circular path followed by the (
Electron Ch e = I , 6 I C.
Note: A •uniform' field is a
constant field. i.e., it has the same B value for all pans of the electron•s
path. Such fields can be obtained with magnets, coils or solenoids (see Chapter
(a) F Bqv or Be.' when e is used to dcnotc the inwards
= 7.0 x 10-2 m or 7.0cm
(a) N (b)
Couple on a coil
Couple on a coil
In Fig. 23.5 a force BIL acts on each vertical wire. If there
are n turns of wire on the coil. the total force is BILn on each side Of the
FOR A-LEVfL PHYSICS
due to this pair Of forces (a'uple) is C = 2BlLnR. However. the "'il area
A equals 2RL, so that
the magnetic field is radial (see Fig. 23.6a). then B is always parallel to the
plane Of the even when the coil is allowed to rotate. and C — HAIn still. If
instead the field is uniform (sce Fig. 23.6b and c). then the component of B
which is effective is B cos o (sce diagram). and the torque
an ordinary mosing-coil meter the current to be measured flows through the
coil, the field is radial, and the torque BAIn turns the coil. The turning
tightens a spring which therefore produces an torque of k newlon metre per unit
angle Of rotation.
coil comes to restWhen BAIn = kO, where the angle of
rotation of the coil O is in degrees or radians.
torque BAIn is also used to prtxluce rotation in simple electric motors.
coil mctcr has a coil with 40 turns, cach with an area Of 20cm:. It is
suspended in a vertical plane and its sides are perpendicular to a radial
torque on the coil When a current Of flows through it.
(b)If the has a
resistance of and a sensitivity Of fun What series resistance is necdcd to
convert the rnetcr to a 10mVFSD meter?
The Earth's magnetism
the United Kingdom the direction Of the magnetic flux density due to the
Earth's magnetism makes an angle O (called the •angle Of dip') Of about 70' to
the horizontal. The horizontal coml¾'nent of this flux density B is B cos o, and the vertical component
Bo sin O
= BsinO) = cos = BC, tan O, where O is the angle of
dip as shown in Fig. 23.7.
An electron moving at a steady
speed of 106 ms -I passes txtween two flat, parallcl metal plates
2.0can apart with a PD of
necessary take to tx4z x H m-
long parallel wires m apart in air each carry a alrrcnt Of 50 A. What the
force. in newtons, each metre length of wire?
wire Of length 40cm is moving vertically dm•nwards, with a current of 1.0 A
through it. If the plane in Which the Wire mmes is to a flux of O. I T,
cakulate on the wire duc to the
3A moving coil
meter a 50-turn measuring I .0cm 20cm. It is in a radial magnetK• of flux
density O. 15 T arui its a torsional anstant of 3.0 x 10 N m rad What current
is required to gne a deflection Of
4In Fig. 23.8 a
flat, rectangular coil is fitted symmetrically on an axle and lies in a
horizontal plane. Of 10 turns Of insulated Wire and its dimensions are as sh(Nn
current of ZOA round the coil
(a)What size is the vertical force on side BC caused by
interaction txtween the current and the Earth's magnetic field? (Take the
horizontal of this field to
(b)Cakulate total the due to this to the simuar forœ
on skk DA.
(c)Cakulatc t•al that wtulld expcricnccd by con if its
planc were at an ange of 20' to horizmtaL
tiEm. The electron is kept traveling in a straight line perpendicular to the
electric field between the plates by applying a magnetic field to the
electron's path and to the electric late:
the intensity Of the electric field (b) the magnetic flux density needed.
the electron charge is not needed. It carxcls.)
long, straight, parallel wires in a vacuum are 025 m apart.
(i)The Wires each carry a current Of
240 A in the Same direction. Calculate the force between the wires per metre of
their length. Draw a sketch showing clearly the direction Of the force Ott each
(ii)The current in one of the wires is reduced to 064 A.
Calculate the current needed in the second wire to maintain the same force
per metre Of their length
(Take = 41 x 10-'Hm -l )
23.9, PORS is a rectangular coil consisting of N turns of wire and carrying
current l. The plane of PORS is parallel to a uniform magnetic field of flux
density B. The length of PO is L and the length Of
MAGNETIC FORCES Example 3
Couple on a coil FO. 23.5 Couple on a coil
Calculate the torque on the coil
MAGNETIC FORCES An electron moving at a steady speed of 10 6 ms -I passes txtween two flat, parallcl metal plates 2