Download SUGGESTED SOLUTIONS FOR TUTORIAL 6

SUGGESTED SOLUTIONS FOR TUTORIAL 6
(MAGNETIC FIELD)
1. An electron passes through a region where an electric field, E is perpendicular to a
magnetic field, B with uniform velocity straight line. If E= 2x105 Vm-1 and B=0.02T.
a) Sketch the motion of this electron in the fields,
b)Calculate the magnitude of the velocity of the electron.
Ans: (1.0 x 10 7 ms -1. )
Solution:
a)
+++++++++++++
x
x
x

I
x
Fex
x
x
x
E
v
x
xB
Fm
----------------
The electrostatic force on the electron is acting downward due to the electric field,
E. When the electron is moving to the right, the current is directed to the left. By
Fleming Left hand’s rule, a magnetic force, Fm is exerted downward on the
electron.
b) At equilibrium, Fm = Fe
Bev = eE
v = E/B=2x10 5/0.02 =1.0 x 10 7 ms -1.
2. An electron with energy 600 eV is moving in a direction perpendicular to a magnetic
field 2.0x10-3 T. If the specific charge of electron is 1.60x10-19 C and the mass of
electron is 9.11x10-31 kg,
a) Explain why the electron is moving in a circular path,
b) Calculate the radius of the circular path
Ans: (4.18 cm.)
Solution:
(a) As the electron is moving, a current, I is produced. By Fleming left hand’s
rule, a magnetic force, Fm will exert on the electron. However, B, I and F
must be perpendicular to each others. The electron must move in circular path
so that it moves in a plane perpendicular to B and the force is tangential to the
direction of its velocity which is the opposite direction of I.
(b) Kinetic energy of electron, ½ mv2 = eV
Velocity, v 
2eV
2 x1.6 x1019 x600

m
9.1x1031
= 1.45x10 7 ms -1.
Centripetal force = magnetic force
mv 2
mv
 Bqv and r 
r
Bq
9.1x1031 x1.45 x107
Radius, r=
2 x103 x1.6 x1019
= 0.0418 m
= 4.18 cm.
3. (a) A proton is moving with velocity 3x105 ms-1 vertically across a magnetic field
0.02T. Mass of proton= 1.67x10 -27 kg, mass of electron=9.1x10 -31 kg.
Ans: ( 7.53x10 -17 J)
Calculate (i) kinetic energy of the proton
(ii) the magnetic force exerted on the proton
(iii) the radius of the circular path of the proton
Ans: (9.6x10 -16 N )
Ans: (15.66 cm)
(b) Calculate the force and acceleration experienced by an electron which is moving
with a velocity 3x107 ms-1 across a magnetic field 0.005 T
Ans: (2.64 x 10 16 ms -1)
Solution:
(a) (i) K.E. = ½ mv2
= ½ x1.67 x 10 -27 x( 3x10 5) 2.
= 7.53x10 -17 J
(ii) Magnetic force, Fm= Bqv
= 0.02x1.6x10 -19 x 3x10 -5
= 9.6x10 -16 N
(iii) Centripetal force = magnetic force
mv 2
mv
 Bqv and r 
r
Bq
1.67 x1027 x3x105
Radius, r=
0.02 x1.6 x1019
= 0.1566 m
= 15.66 cm.
(b) Magnetic force, Fm= Bqv
= 0.005x1.6x10 -19 x 3x10 7
= 2.4x10 -14 N
F=ma
F 2.4 x1014
The acceleration, a= 
m 9.1x1031
= 2.64 x 10 16 ms -1.
4. Figure below shows a proton with velocity, v is entering perpendicularly into a
uniform magnetic field, B
x
x
x
x
x
x
x
x
x
xB
v
x
x
x
x
x
x
x
x
x
x
(a) An electric field, E is added to the magnetic field so that the proton can
move in straight line without any deflection. Show the direction of E.
(b) Derive an equation for the electric field, E.
(c) If an electron with the same velocity, v enters the same magnetic field and
electric field, what will happen to the electron?
Solution:
(a) For proton the direction of the current, I is the same as that of its velocity. Using
Fleming left hand rule, a magnetic force, Fm is exerted upward on the proton. To
balance, an electrostatic force, Fe must act downward as shown.
x
x
x
x
Fm
x
x
x
B
x
x
x
x
x
x
x
x
x
v
x
x
Fe
x
x
(b) Electrostatic force =magnetic force
Fe=Fm
qE=Bqv
The electric field, E= Bv
(c)From (b) above, the velocity of a charged particle, v = E/B is independent of the
mass and charge of the particle. Thus the electron will also move in straight line.
5. A moving-coil galvanometer has a square coil with 800 turns with cross sectional
area 1.2x10-3 m2. The coil is placed in a radial magnetic field, B= 0.15T.
(a) What is the torque exerted on the coil if a current 100 mA is flowing through the
coil? Ans: (1.44 x10 -2 Nm)
(b) If the magnet has lost part of its magnetic field, what will happen to the reading
recorded by the galvanometer?
Solution:
(a) The torque,  = BIAN
= 0.15 x (100x10 -3 ) x (1.2x10-3) x 800
= 1.44 x10 -2 Nm
(b) If B decreases, the torque  also decreases. The deflection of the pointer will
be less. The reading of the meter is less than its actual value of the current, I.
6. A cupper strip of thickness 1.20 m is placed in a magnetic field B= 0.80 T. If the
current flowing through this strip, I is 20A and the density of electron in cupper,
n=8.50x1028 m-3
(a) Sketch a diagram to show the Hall voltage,
(b) Calculate the value of this Hall voltage.
Ans:(98 mV )
Solution:
(a) The diagram below shows the Hall voltage.
+++++++++
Fe

VH
v
- - - - Fm- - - - - - - B
(b) VH 
IB
20 x0.80

28
net (8.50 x10 x(1.6 x1019 ) x(1.20 x106 )
= 98 mV.
7. A long straight wire carries a current I=20.0 A flowing vertically downward. If the
horizontal component of the earth magnetic field is 2.0x10-5 T, calculate the resultant
magnetic field at a point
Ans: 6x10 -5 T(To North)
a) 10 cm to the East,
Ans: -2x10 -5 T( To South)
b)10 cm to the West of the wire.
Solution:
The magnetic fields due to the flow of current in the wire at the two given points are
shown below
N
X
WEST
EAST
2x10 -5 T
S
(a) The magnitude of the magnetic field produced,
B=
o I
o x 20
=
= 4x10 -5 T
2 r
2 (10 x102 )
For point at the WEST, resultant B = 4x10 -5 + 2x10 -5
= 6x10 -5 T(To North)
(b)
For point at the EAST, resultant B= -4x10 -5 + 2x10 -5
= -2x10 -5 T( To South)
8. Two circular coils with radii 20.0 cm and 40.0 cm are placed with their centres at a
point P. If the currents flowing in the same clockwise direction through these coils are
0.5 A and 0.8 A respectively, calculate the resultant magnetic field at P.
Ans: 2.83x10 -6 T ( into the paper)
I2=0.8A
P
I1=0.5A
Solution:
Using Maxwell cock-screw rule, I1 and I2 produce magnetic fields with direction into
the plane of the paper.
X
P
For I1, B1==
I2=0.8A
o I
o x0.5
=
2r
2(20 x102 )
= 1.57x10 -6 T
For I2, B2==
o I
o x0.8
=
2r
2(40 x102 )
= 1.257x10 -6 T
Total magnetic field at P= B1 + B2
=1.57x10 -6 +1.257x10 -6
= 2.83x10 -6 T ( into the paper)
9. The electron at X is moving with speed 2.80x10
6
ms
-1
to Y in a uniform magnetic
field. Mass of electron= 9.1x10 -31 kg
v
X
Y
4 cm
Determine,
(a) The magnitude and direction of the magnetic field which causes the electron to
move in the semi-circular path from X to Y.
(b) The time taken for the electron to move from X to Y
Ans: (7.97x10 -4 T)
Ans: (2.24x10 4 s )
Solution:
The current due to the flow of electrons is opposite to the direction of the velocity of
the electrons. Using Fleming left hand rule, a centripetal force is produced by a
magnetic force.
x x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
v
F
x
x
x
I
(a) Centripetal force = magnetic force
mv 2
 Bev
r
B
mv 9.1x1031 x2.8x106
=7.97x10 -4 T

er 1.6 x1019 x2 x102
(b) To move in a circle,
v
2 r
T
T
2 r
v
The time taken to move in the semi-circle=1/2 T=
r
v

 x 2 x102
2.80 x106
= 2.24x10 4 s
10. A long, straight conducting wire X carries a current I1 = 10.0A. Another wire Y
carrying current I2= 20.0A is placed parallel to X and a distance d from X. If the
resultant magnetic field at point O is zero.
Wire X
I2
I1
O
R
d
(a) What is the direction of the current I2?
(b)Calculate the value of d in term of R.
Ans: (3R)
(c)Calculate the force per unit length on these wires in terms of o and R
Ans: (
100  o
Nm-1 . )
3 R
Solution:
(a) Using Maxwell’s cock-screw rule, B1 at O is directed into the paper. The current
I2 must produce a magnetic field B2 at O which is directed out of the paper. Thus I2
must flow upward.
Wire X
I1
I2
R
d
(b) The magnetic field B1 due to I1 at O
B1=
B2=
o I
 x10
= o
2 R
2 r
 x 20
o I
= o
2 r2
2 r
The resultant magnetic field at O=0
B1= B2
o x10 o x 20
=
2 R
2 r2
r2 = 2R
d = R + 2R= 3R.
O
(c) The force per unit length, F=
 o I1 I 2
2 r
=
o x10 x 20
2 (3R)
=
100  o
Nm-1 .
3 R