Download A uniform magnetic field is applied between two parallel metal

1
Solutions
(a) Electric field strength =
170
V
=
= 21 250 N C–1
d 0.008
Marks
1M
Electric force acting on the electron
= qE
1M
= (1.6  10–19)(21 250)
= 3.4  10–15 N
1A
(b) Since the magnetic force equals the electric force, we have
BQv = FE
B=
1M
FE
3.4  10 15
=
= 2.91  10–4 T
Qv (1.6  10 19 )(7.3  107 )
1A
The magnitude of the magnetic field is 2.91  10–4 T.
2
Solutions
Marks
(a)
(Electron is deflected upwards.)
1A
(Electron moves in circular orbit.)
1A
1
(b) Since KE = mv 2 ,
2
v=
1M
2  3.2  10 16
2KE
=
= 2.65  107 m s–1
31
m
9.1 10
1A
The speed of electron is 2.65  107 m s1.
(c)
Since the magnetic force acting on the electron provides the
centripetal force, we have
B=
mv 2
= qvB.
r
mv (9.1 10 31 )(2.65  10 7 )
=
= 6.03  10–4 T
qr
(1.6  10 19 )(0.25)
The magnitude of the magnetic field is 6.03  10–4 T.
1M
1A
3.
Solutions
(a) The magnetic forces acting on the wires are attractive forces.
Marks
1A
Since the magnetic forces on both wires are an action-and-reaction
pair, they have the same magnitude.
Magnetic force per unit length =
=
μ 0 I1I 2 l
2πr
1A
1M
( 4 π  10 7 )(5)(0.3)(1)
(2π)(8  10 3 )
= 3.75  105 N
(b) The magnetic forces acting on the wires become repulsive forces.
The magnitude of the forces does not change.
1A
1A
1A
4.
Solutions
(a) By F =
IX =
=
μ 0 I1I 2 l
,
2πr
=
1M
2πrF
μ 0 IY l
(2π )(0.3)(4.2  10 4 )
( 4 π  10 7 )(12)(1)
= 52.5 A
IZ =
Marks
1M
1A
2πrF
μ0I X l
(2π )(0.6)(4.2  10 4 )
( 4 π  10 7 )(52.5)(1)
= 24 A
1A
The current in X is 52.5 A and the current in Z is 24 A.
(b) The magnetic force acting on Y due to X is attractive and that due to
Z is repulsive, therefore the resultant force acting on Y is towards X.
1A
Resultant magnetic force per unit length acting on Y
=
( 4 π  10 7 )(52.5)(12)(1) ( 4 π  10 7 )(24 )(12)(1)
+
(2π)(0.3)
(2π)(0.3)
= 6.12  104 N
1A