DC Pandey Solutions: Magnetics- 2

# DC Pandey Solutions: Magnetics- 2 | DC Pandey Solutions for NEET Physics PDF Download

``` Page 1

AIEEE Corner
Subjective Questions (Level-1)
1. Pos i tive. By Flem ings left hand rule.
2. F evB
m
= sin q
Þ    v
F
eB
e
=
sin q
=
´
´ ´ ´ ´ °
-
- -
4.6
1.6 3.5
10
10 10 60
15
19 3
sin
= ´ 9.46 10
6
m s /
3. F qvB
m
= sin q
= ´ ´ ´ ´ ´
-
( ) 2 10 10 1
19 5
1.6 0.8
= ´
-
2.56 10
14
N
4. (a)  F v B
m
e
®
=
®
´
®
( )
= - ´ ´ + ´
-
16 10 20 10 30 10
19 6 6
. [( . ) ( . ) ]
^ ^
i j
´ + ( . . )
^ ^
003 015 i j
= - ´
-
( )
^
6.24 N 10
4
k
(b) =
® ®
´
®
= - ´
-
F v B k
m
e ( ) ( . )
^
624 10
4
N
5. F v B
m
e
®
=
®
´
®
( )
( ) [( )
^ ^ ^
6.4 1.6 ´ = - ´ +
- -
10 10 2 4
19 19
k i j
´ + ( )]
^ ^
B B
x x
i j 3
6.4 1.6 ´ = - ´
- -
10 10 2
19 19
k k
^ ^
[ ] B
x
B
x
=
´
- ´
= -
-
-
6.4
3.2
2.0 T
10
10
19
19
6.(a) As mag netic force al ways acts
per pen dic u lar to mag netic field, mag netic
field must be along x-axis.
F qvB
1 1 1
= sinq
Þ B
F
qvB
= =
´
´ ´ ´
-
-
1
1 1
3
6 6
5 2 10
1 10 10
1
2
sinq
Þ B=
-
10
3
T
or    B i
®
=
-
( )
^
10
3
T
(b) F qv B
2 2 2
= sin q
= ´ ´ ´ ´ °
- -
1 10 10 10 90
6 6 3
sin
=
-
10
3
N
F
2
1 = mN
7. Let B i j k
®
= + + B B B
x y z
^ ^ ^
(a) F v B
®
=
®
´
®
q( )
7.6 5.2 ´ - ´
- -
10 10
3 3
i k
^ ^
= - ´ ´ ´ -
-
7.8 3.8 10 10
6 3
( )
^ ^
B B
z x
i k
Þ      B
x
= - 0.175 T, B
z
= - 0.256 T
(b) Cannot be determined by this information.
(c) As F v B
®
=
®
´
®
q( )
F B
®
^
®
Hence, B F
®
×
®
= 0
8. B i
®
= B
^
(a) v j
®
= v
^
F v B k
®
=
®
´
®
= - q qvB ( )
^
(b) v j
®
= v
^
F v B j
®
=
®
´
®
= q qvB ( )
^
(c) v i
®
= - v
^

F v B
®
=
®
´
®
= q( ) 0
(d) v i k
®
= ° - ° v v cos cos
^ ^
45 45
F v B j
®
=
®
´
®
= - q
qvB
( )
^
2
(e) v j k
®
= ° - ° v v cos cos
^ ^
45 45
F v B j k
®
=
®
´
®
= - - q
qvB
( ) ( )
^ ^
2
= - +
qvB
2
( )
^ ^
j k
9. r
mv
qB
mk
eB
meV
eB
= = =
2 2
Þ          B
mV
e
r =
2

=
´ ´ ´ ´
´
´
-
-
2 10 2 10
10
31 3
19
9.1
1.6
0.180
88
Page 2

AIEEE Corner
Subjective Questions (Level-1)
1. Pos i tive. By Flem ings left hand rule.
2. F evB
m
= sin q
Þ    v
F
eB
e
=
sin q
=
´
´ ´ ´ ´ °
-
- -
4.6
1.6 3.5
10
10 10 60
15
19 3
sin
= ´ 9.46 10
6
m s /
3. F qvB
m
= sin q
= ´ ´ ´ ´ ´
-
( ) 2 10 10 1
19 5
1.6 0.8
= ´
-
2.56 10
14
N
4. (a)  F v B
m
e
®
=
®
´
®
( )
= - ´ ´ + ´
-
16 10 20 10 30 10
19 6 6
. [( . ) ( . ) ]
^ ^
i j
´ + ( . . )
^ ^
003 015 i j
= - ´
-
( )
^
6.24 N 10
4
k
(b) =
® ®
´
®
= - ´
-
F v B k
m
e ( ) ( . )
^
624 10
4
N
5. F v B
m
e
®
=
®
´
®
( )
( ) [( )
^ ^ ^
6.4 1.6 ´ = - ´ +
- -
10 10 2 4
19 19
k i j
´ + ( )]
^ ^
B B
x x
i j 3
6.4 1.6 ´ = - ´
- -
10 10 2
19 19
k k
^ ^
[ ] B
x
B
x
=
´
- ´
= -
-
-
6.4
3.2
2.0 T
10
10
19
19
6.(a) As mag netic force al ways acts
per pen dic u lar to mag netic field, mag netic
field must be along x-axis.
F qvB
1 1 1
= sinq
Þ B
F
qvB
= =
´
´ ´ ´
-
-
1
1 1
3
6 6
5 2 10
1 10 10
1
2
sinq
Þ B=
-
10
3
T
or    B i
®
=
-
( )
^
10
3
T
(b) F qv B
2 2 2
= sin q
= ´ ´ ´ ´ °
- -
1 10 10 10 90
6 6 3
sin
=
-
10
3
N
F
2
1 = mN
7. Let B i j k
®
= + + B B B
x y z
^ ^ ^
(a) F v B
®
=
®
´
®
q( )
7.6 5.2 ´ - ´
- -
10 10
3 3
i k
^ ^
= - ´ ´ ´ -
-
7.8 3.8 10 10
6 3
( )
^ ^
B B
z x
i k
Þ      B
x
= - 0.175 T, B
z
= - 0.256 T
(b) Cannot be determined by this information.
(c) As F v B
®
=
®
´
®
q( )
F B
®
^
®
Hence, B F
®
×
®
= 0
8. B i
®
= B
^
(a) v j
®
= v
^
F v B k
®
=
®
´
®
= - q qvB ( )
^
(b) v j
®
= v
^
F v B j
®
=
®
´
®
= q qvB ( )
^
(c) v i
®
= - v
^

F v B
®
=
®
´
®
= q( ) 0
(d) v i k
®
= ° - ° v v cos cos
^ ^
45 45
F v B j
®
=
®
´
®
= - q
qvB
( )
^
2
(e) v j k
®
= ° - ° v v cos cos
^ ^
45 45
F v B j k
®
=
®
´
®
= - - q
qvB
( ) ( )
^ ^
2
= - +
qvB
2
( )
^ ^
j k
9. r
mv
qB
mk
eB
meV
eB
= = =
2 2
Þ          B
mV
e
r =
2

=
´ ´ ´ ´
´
´
-
-
2 10 2 10
10
31 3
19
9.1
1.6
0.180
88
= ´
-
0.36 10
4
T
B = ´
-
3.6 10
4
T
10. (a) r
mv
qB
= Þ v
qBr
m
=
=
´ ´ ´ ´
´
- -
-
1.6 2.5 6.96
3.34
10 10
10
19 3
27
= ´ 8.33 10
5
ms
-1
(b) t
T m
qB
= =
2
p
=
´ ´
´ ´
-
-
3.14 3.34
1.6 2.5
10
10
27
19
= ´
-
2.62 10
8
s
(c) k eV mv = =
1
2
2
Þ V
mv
e
=
2
2
=
´ ´ ´
´ ´
-
-
3.34 8.33
1.6
10 10
2 10
27 5 2
19
( )
= ´ 7.26 10
3
V
= 7.26 kV
11. (a) - q. As ini tially par ti cle is neu tral, charge
on two par ti cles must be equal and op po site.
(b) The will collide after completing half
rotation, i.e.,
t
T m
qB
= =
2
p
12. Here, r = =
10.0
5.0
2
cm,
(a) r
mv
qB
= Þ B
mv
qr
=
=
´ ´ ´
´ ´ ´
-
- -
9.1 1.41
1.6
10 10
10 5 10
31 6
19 2
= ´
-
1.6 T 10
4
By Fleming’s left hand rule, direction of
magnetic field must be inward.
(b) t
T m
qB
= =
2
p
=
´ ´
´ ´ ´
-
- -
3.14 9.1
1.6 1.6
10
10 10
31
19 4
= ´
-
1.1 10
7
s
13. The com po nent of ve loc ity along the
mag netic field (i.e., v
x
) will re main
un changed and the pro ton will move in a
he li cal path.
At any instant,
Components of velocity of particle along
Y-axis and Z-axis
¢ = = v v v t
y y y
cos cos q w
and   ¢ = - = v v v t
z z z
sin sin q w
where,      w=
qB
m
\  v i j k
®
= + - v v t v t
x y z
^ ^ ^
cos sin w w
14. For the elec tron to hit the tar get, dis tance
G S must be mul ti ple of pitch, i.e.,
GS np =
For minimum distance, n = 1
Þ     GS p
mv
qB
= =
2p q cos
Þ       p
mk
qB
=
° 2 2 60 p cos
(mv mk = 2 )
Þ         B
mk
qp
=
° 2 2 60 p cos
=
´ ´ ´ ´ ´ ´ ´ ´
´ ´
- -
-
2 2 10 2 10
1
2
10
31 16
19
3.14 9.1 1.6
1.6 0.1
Þ B = ´
-
4.73 10
4
T
15. (a) From Ques tion 5 (c)
Introductory Exercise 23.2
L
R
= sin q Þ L R = sin q
R
R
sin 60
2
° =
Þ L
mv
qB
mv
qB
= =
2 2
0
0
(b) Now, L L R ¢ = = 2.1 1.05
As L R ¢ > ,
89
v
y v sinwt
y
v cos wt
y
z
®
B
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
B
–q
q
+q
Page 3

AIEEE Corner
Subjective Questions (Level-1)
1. Pos i tive. By Flem ings left hand rule.
2. F evB
m
= sin q
Þ    v
F
eB
e
=
sin q
=
´
´ ´ ´ ´ °
-
- -
4.6
1.6 3.5
10
10 10 60
15
19 3
sin
= ´ 9.46 10
6
m s /
3. F qvB
m
= sin q
= ´ ´ ´ ´ ´
-
( ) 2 10 10 1
19 5
1.6 0.8
= ´
-
2.56 10
14
N
4. (a)  F v B
m
e
®
=
®
´
®
( )
= - ´ ´ + ´
-
16 10 20 10 30 10
19 6 6
. [( . ) ( . ) ]
^ ^
i j
´ + ( . . )
^ ^
003 015 i j
= - ´
-
( )
^
6.24 N 10
4
k
(b) =
® ®
´
®
= - ´
-
F v B k
m
e ( ) ( . )
^
624 10
4
N
5. F v B
m
e
®
=
®
´
®
( )
( ) [( )
^ ^ ^
6.4 1.6 ´ = - ´ +
- -
10 10 2 4
19 19
k i j
´ + ( )]
^ ^
B B
x x
i j 3
6.4 1.6 ´ = - ´
- -
10 10 2
19 19
k k
^ ^
[ ] B
x
B
x
=
´
- ´
= -
-
-
6.4
3.2
2.0 T
10
10
19
19
6.(a) As mag netic force al ways acts
per pen dic u lar to mag netic field, mag netic
field must be along x-axis.
F qvB
1 1 1
= sinq
Þ B
F
qvB
= =
´
´ ´ ´
-
-
1
1 1
3
6 6
5 2 10
1 10 10
1
2
sinq
Þ B=
-
10
3
T
or    B i
®
=
-
( )
^
10
3
T
(b) F qv B
2 2 2
= sin q
= ´ ´ ´ ´ °
- -
1 10 10 10 90
6 6 3
sin
=
-
10
3
N
F
2
1 = mN
7. Let B i j k
®
= + + B B B
x y z
^ ^ ^
(a) F v B
®
=
®
´
®
q( )
7.6 5.2 ´ - ´
- -
10 10
3 3
i k
^ ^
= - ´ ´ ´ -
-
7.8 3.8 10 10
6 3
( )
^ ^
B B
z x
i k
Þ      B
x
= - 0.175 T, B
z
= - 0.256 T
(b) Cannot be determined by this information.
(c) As F v B
®
=
®
´
®
q( )
F B
®
^
®
Hence, B F
®
×
®
= 0
8. B i
®
= B
^
(a) v j
®
= v
^
F v B k
®
=
®
´
®
= - q qvB ( )
^
(b) v j
®
= v
^
F v B j
®
=
®
´
®
= q qvB ( )
^
(c) v i
®
= - v
^

F v B
®
=
®
´
®
= q( ) 0
(d) v i k
®
= ° - ° v v cos cos
^ ^
45 45
F v B j
®
=
®
´
®
= - q
qvB
( )
^
2
(e) v j k
®
= ° - ° v v cos cos
^ ^
45 45
F v B j k
®
=
®
´
®
= - - q
qvB
( ) ( )
^ ^
2
= - +
qvB
2
( )
^ ^
j k
9. r
mv
qB
mk
eB
meV
eB
= = =
2 2
Þ          B
mV
e
r =
2

=
´ ´ ´ ´
´
´
-
-
2 10 2 10
10
31 3
19
9.1
1.6
0.180
88
= ´
-
0.36 10
4
T
B = ´
-
3.6 10
4
T
10. (a) r
mv
qB
= Þ v
qBr
m
=
=
´ ´ ´ ´
´
- -
-
1.6 2.5 6.96
3.34
10 10
10
19 3
27
= ´ 8.33 10
5
ms
-1
(b) t
T m
qB
= =
2
p
=
´ ´
´ ´
-
-
3.14 3.34
1.6 2.5
10
10
27
19
= ´
-
2.62 10
8
s
(c) k eV mv = =
1
2
2
Þ V
mv
e
=
2
2
=
´ ´ ´
´ ´
-
-
3.34 8.33
1.6
10 10
2 10
27 5 2
19
( )
= ´ 7.26 10
3
V
= 7.26 kV
11. (a) - q. As ini tially par ti cle is neu tral, charge
on two par ti cles must be equal and op po site.
(b) The will collide after completing half
rotation, i.e.,
t
T m
qB
= =
2
p
12. Here, r = =
10.0
5.0
2
cm,
(a) r
mv
qB
= Þ B
mv
qr
=
=
´ ´ ´
´ ´ ´
-
- -
9.1 1.41
1.6
10 10
10 5 10
31 6
19 2
= ´
-
1.6 T 10
4
By Fleming’s left hand rule, direction of
magnetic field must be inward.
(b) t
T m
qB
= =
2
p
=
´ ´
´ ´ ´
-
- -
3.14 9.1
1.6 1.6
10
10 10
31
19 4
= ´
-
1.1 10
7
s
13. The com po nent of ve loc ity along the
mag netic field (i.e., v
x
) will re main
un changed and the pro ton will move in a
he li cal path.
At any instant,
Components of velocity of particle along
Y-axis and Z-axis
¢ = = v v v t
y y y
cos cos q w
and   ¢ = - = v v v t
z z z
sin sin q w
where,      w=
qB
m
\  v i j k
®
= + - v v t v t
x y z
^ ^ ^
cos sin w w
14. For the elec tron to hit the tar get, dis tance
G S must be mul ti ple of pitch, i.e.,
GS np =
For minimum distance, n = 1
Þ     GS p
mv
qB
= =
2p q cos
Þ       p
mk
qB
=
° 2 2 60 p cos
(mv mk = 2 )
Þ         B
mk
qp
=
° 2 2 60 p cos
=
´ ´ ´ ´ ´ ´ ´ ´
´ ´
- -
-
2 2 10 2 10
1
2
10
31 16
19
3.14 9.1 1.6
1.6 0.1
Þ B = ´
-
4.73 10
4
T
15. (a) From Ques tion 5 (c)
Introductory Exercise 23.2
L
R
= sin q Þ L R = sin q
R
R
sin 60
2
° =
Þ L
mv
qB
mv
qB
= =
2 2
0
0
(b) Now, L L R ¢ = = 2.1 1.05
As L R ¢ > ,
89
v
y v sinwt
y
v cos wt
y
z
®
B
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
B
–q
q
+q
Particle will describe a semicircle and move
out of the magnetic field moving in opposite
direction, i.e.,
v v v ¢ = - = -
0
i
^
and t
T m
qB
= =
2
0
p
16. v i
®
=
-
( )
^
50
1
ms , B j
®
= ( )
^
2.0 mT
As particle move with uniform velocity,
F E v B
®
=
®
+
®
´
®
= q( ) 0
Þ E B v
®
=
®
´
®
= -( 0.1 N/C)k
^
17. If v be the speed of par ti cle at point ( , , ) 0 y z
then by work-en ergy the o rem,
W K mv = = D
1
2
2
But work done by magnetic force is zero,
hence, network done = work done by electric
force
= qEZ
\ qE Z mv
0
2
1
2
=
Þ          v
qE Z
m
=
2
0
As the magnetic field is along Y-axis,
particle will move in XZ-plane.
The path of particle will be a cycloid. In this
case, instantaneous centre of curvature of
the particle will move along X-axis.
As magnetic force provides centripetal force
to the particle,
qvB
mv
R
0
2
=
v
qB R
m
=
0
v v
qB R
m
x
= = cos
cos
q
q
0
=
qB Z
m
0
( Q R Z cos q = )
Now, v v v
qE Z
m
q B Z
m
z x
= - = -
2 2 0
2
0
2 2
2
2
18. Given, E j
®
= E
^
, B k
®
= B
^
,
v j k
®
= + v v cos sin
^ ^
q q
As protons are moving undeflected,
F
®
= 0 Þ e ( ) E v B
®
+
®
´
®
= 0
Þ       e E vB ( cos )
^ ^
j j - = q 0
or        v
E
B
=
cosq
Now, if electric field is switched off
p
mv
qB
mE
qB
= =
2 2
2
p q p q sin tan
(Component of velocity along magnetic field
= = v v
z
sin q)
19. F I lB = sin q
I
F
lB
= =
´ ´ ° sin sin q
0.13
0.2 0.067 90
= 9.7 A
20. For no ten sion in springs
F mg
m
=
Þ   I lB mg =
I
mg
lB
= =
´ ´
´ ´
-
-
13.0
62.0 0.440
10 10
10
3
2
=0.48 A
By Fleming left hand rule, for magnetic force
to act in upward direction, current in the
wire must be towards right.
21. (a) FBD of metal bar is shown in fig ure, for
metal to be in equi lib rium,
F N mg
m
+ =
Þ         F mg N
m
= -
Þ        IlB m N = -
Þ
V
R
lB mg N = -
Þ          V
R
lB
mg N = - ( )
90
q
q
z
v
z
v
v
x
R
X
v
y
z
q
O
Z
E = EK.
v
v
x
x
B = –B
®
j
^
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
mg
F
m
I I
F
m
N
mg
Page 4

AIEEE Corner
Subjective Questions (Level-1)
1. Pos i tive. By Flem ings left hand rule.
2. F evB
m
= sin q
Þ    v
F
eB
e
=
sin q
=
´
´ ´ ´ ´ °
-
- -
4.6
1.6 3.5
10
10 10 60
15
19 3
sin
= ´ 9.46 10
6
m s /
3. F qvB
m
= sin q
= ´ ´ ´ ´ ´
-
( ) 2 10 10 1
19 5
1.6 0.8
= ´
-
2.56 10
14
N
4. (a)  F v B
m
e
®
=
®
´
®
( )
= - ´ ´ + ´
-
16 10 20 10 30 10
19 6 6
. [( . ) ( . ) ]
^ ^
i j
´ + ( . . )
^ ^
003 015 i j
= - ´
-
( )
^
6.24 N 10
4
k
(b) =
® ®
´
®
= - ´
-
F v B k
m
e ( ) ( . )
^
624 10
4
N
5. F v B
m
e
®
=
®
´
®
( )
( ) [( )
^ ^ ^
6.4 1.6 ´ = - ´ +
- -
10 10 2 4
19 19
k i j
´ + ( )]
^ ^
B B
x x
i j 3
6.4 1.6 ´ = - ´
- -
10 10 2
19 19
k k
^ ^
[ ] B
x
B
x
=
´
- ´
= -
-
-
6.4
3.2
2.0 T
10
10
19
19
6.(a) As mag netic force al ways acts
per pen dic u lar to mag netic field, mag netic
field must be along x-axis.
F qvB
1 1 1
= sinq
Þ B
F
qvB
= =
´
´ ´ ´
-
-
1
1 1
3
6 6
5 2 10
1 10 10
1
2
sinq
Þ B=
-
10
3
T
or    B i
®
=
-
( )
^
10
3
T
(b) F qv B
2 2 2
= sin q
= ´ ´ ´ ´ °
- -
1 10 10 10 90
6 6 3
sin
=
-
10
3
N
F
2
1 = mN
7. Let B i j k
®
= + + B B B
x y z
^ ^ ^
(a) F v B
®
=
®
´
®
q( )
7.6 5.2 ´ - ´
- -
10 10
3 3
i k
^ ^
= - ´ ´ ´ -
-
7.8 3.8 10 10
6 3
( )
^ ^
B B
z x
i k
Þ      B
x
= - 0.175 T, B
z
= - 0.256 T
(b) Cannot be determined by this information.
(c) As F v B
®
=
®
´
®
q( )
F B
®
^
®
Hence, B F
®
×
®
= 0
8. B i
®
= B
^
(a) v j
®
= v
^
F v B k
®
=
®
´
®
= - q qvB ( )
^
(b) v j
®
= v
^
F v B j
®
=
®
´
®
= q qvB ( )
^
(c) v i
®
= - v
^

F v B
®
=
®
´
®
= q( ) 0
(d) v i k
®
= ° - ° v v cos cos
^ ^
45 45
F v B j
®
=
®
´
®
= - q
qvB
( )
^
2
(e) v j k
®
= ° - ° v v cos cos
^ ^
45 45
F v B j k
®
=
®
´
®
= - - q
qvB
( ) ( )
^ ^
2
= - +
qvB
2
( )
^ ^
j k
9. r
mv
qB
mk
eB
meV
eB
= = =
2 2
Þ          B
mV
e
r =
2

=
´ ´ ´ ´
´
´
-
-
2 10 2 10
10
31 3
19
9.1
1.6
0.180
88
= ´
-
0.36 10
4
T
B = ´
-
3.6 10
4
T
10. (a) r
mv
qB
= Þ v
qBr
m
=
=
´ ´ ´ ´
´
- -
-
1.6 2.5 6.96
3.34
10 10
10
19 3
27
= ´ 8.33 10
5
ms
-1
(b) t
T m
qB
= =
2
p
=
´ ´
´ ´
-
-
3.14 3.34
1.6 2.5
10
10
27
19
= ´
-
2.62 10
8
s
(c) k eV mv = =
1
2
2
Þ V
mv
e
=
2
2
=
´ ´ ´
´ ´
-
-
3.34 8.33
1.6
10 10
2 10
27 5 2
19
( )
= ´ 7.26 10
3
V
= 7.26 kV
11. (a) - q. As ini tially par ti cle is neu tral, charge
on two par ti cles must be equal and op po site.
(b) The will collide after completing half
rotation, i.e.,
t
T m
qB
= =
2
p
12. Here, r = =
10.0
5.0
2
cm,
(a) r
mv
qB
= Þ B
mv
qr
=
=
´ ´ ´
´ ´ ´
-
- -
9.1 1.41
1.6
10 10
10 5 10
31 6
19 2
= ´
-
1.6 T 10
4
By Fleming’s left hand rule, direction of
magnetic field must be inward.
(b) t
T m
qB
= =
2
p
=
´ ´
´ ´ ´
-
- -
3.14 9.1
1.6 1.6
10
10 10
31
19 4
= ´
-
1.1 10
7
s
13. The com po nent of ve loc ity along the
mag netic field (i.e., v
x
) will re main
un changed and the pro ton will move in a
he li cal path.
At any instant,
Components of velocity of particle along
Y-axis and Z-axis
¢ = = v v v t
y y y
cos cos q w
and   ¢ = - = v v v t
z z z
sin sin q w
where,      w=
qB
m
\  v i j k
®
= + - v v t v t
x y z
^ ^ ^
cos sin w w
14. For the elec tron to hit the tar get, dis tance
G S must be mul ti ple of pitch, i.e.,
GS np =
For minimum distance, n = 1
Þ     GS p
mv
qB
= =
2p q cos
Þ       p
mk
qB
=
° 2 2 60 p cos
(mv mk = 2 )
Þ         B
mk
qp
=
° 2 2 60 p cos
=
´ ´ ´ ´ ´ ´ ´ ´
´ ´
- -
-
2 2 10 2 10
1
2
10
31 16
19
3.14 9.1 1.6
1.6 0.1
Þ B = ´
-
4.73 10
4
T
15. (a) From Ques tion 5 (c)
Introductory Exercise 23.2
L
R
= sin q Þ L R = sin q
R
R
sin 60
2
° =
Þ L
mv
qB
mv
qB
= =
2 2
0
0
(b) Now, L L R ¢ = = 2.1 1.05
As L R ¢ > ,
89
v
y v sinwt
y
v cos wt
y
z
®
B
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
B
–q
q
+q
Particle will describe a semicircle and move
out of the magnetic field moving in opposite
direction, i.e.,
v v v ¢ = - = -
0
i
^
and t
T m
qB
= =
2
0
p
16. v i
®
=
-
( )
^
50
1
ms , B j
®
= ( )
^
2.0 mT
As particle move with uniform velocity,
F E v B
®
=
®
+
®
´
®
= q( ) 0
Þ E B v
®
=
®
´
®
= -( 0.1 N/C)k
^
17. If v be the speed of par ti cle at point ( , , ) 0 y z
then by work-en ergy the o rem,
W K mv = = D
1
2
2
But work done by magnetic force is zero,
hence, network done = work done by electric
force
= qEZ
\ qE Z mv
0
2
1
2
=
Þ          v
qE Z
m
=
2
0
As the magnetic field is along Y-axis,
particle will move in XZ-plane.
The path of particle will be a cycloid. In this
case, instantaneous centre of curvature of
the particle will move along X-axis.
As magnetic force provides centripetal force
to the particle,
qvB
mv
R
0
2
=
v
qB R
m
=
0
v v
qB R
m
x
= = cos
cos
q
q
0
=
qB Z
m
0
( Q R Z cos q = )
Now, v v v
qE Z
m
q B Z
m
z x
= - = -
2 2 0
2
0
2 2
2
2
18. Given, E j
®
= E
^
, B k
®
= B
^
,
v j k
®
= + v v cos sin
^ ^
q q
As protons are moving undeflected,
F
®
= 0 Þ e ( ) E v B
®
+
®
´
®
= 0
Þ       e E vB ( cos )
^ ^
j j - = q 0
or        v
E
B
=
cosq
Now, if electric field is switched off
p
mv
qB
mE
qB
= =
2 2
2
p q p q sin tan
(Component of velocity along magnetic field
= = v v
z
sin q)
19. F I lB = sin q
I
F
lB
= =
´ ´ ° sin sin q
0.13
0.2 0.067 90
= 9.7 A
20. For no ten sion in springs
F mg
m
=
Þ   I lB mg =
I
mg
lB
= =
´ ´
´ ´
-
-
13.0
62.0 0.440
10 10
10
3
2
=0.48 A
By Fleming left hand rule, for magnetic force
to act in upward direction, current in the
wire must be towards right.
21. (a) FBD of metal bar is shown in fig ure, for
metal to be in equi lib rium,
F N mg
m
+ =
Þ         F mg N
m
= -
Þ        IlB m N = -
Þ
V
R
lB mg N = -
Þ          V
R
lB
mg N = - ( )
90
q
q
z
v
z
v
v
x
R
X
v
y
z
q
O
Z
E = EK.
v
v
x
x
B = –B
®
j
^
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
mg
F
m
I I
F
m
N
mg
For largest voltage,
N =0
V
Rmg
lB
= =
´ ´ ´
´ ´
-
-
25 750 10
10
3
2
9.8
50.0 0.450
= 817.5 V
(b) If I lB mg >
I lB mg ma - =
a
I lB mg
m
V lB
Rm
g =
-
= -
=
´ ´ ´
´ ´
-
-
-
817.5 0.45
9.8
50 10
2 750 10
2
3
=112.8 m/s
2
22. I = 3.50 A, l = - ( )
^
1.00 cm i
Þ l = - ´
-
( )
^
1.00 m 10
2
i
(a) B j
®
= - (
^
0.65 T)
F l B k
m
I
®
=
®
´
®
= - ( ) (
^
0.023 N)
(b) B k
®
= + (
^
0.56 T)
F l B j
m
I
®
=
®
´
®
= ( ) (
^
0.0196 N)
(c) B i
®
= - (
^
0.33 T)
F l B
m
I
®
=
®
´
®
= ( ) 0
(d) B i k
®
= -
®
( ( )
^
0.33 T) 0.28T
F l B j
m
I
®
=
®
´
®
= - ( ) (
^
0.0098 N)
(e) B j k
®
= + - ( ( )
^ ^
0.74 T) 0.36 T
F l B k j
m
I
®
=
®
´
®
= - + ( ) ( ( )
^ ^
0.0259 N) 0.0126 N
= - ( ) ( )
^ ^
0.0126 N 0.0259 N j K
23. B j
®
=(
^
0.020T)
l ab j
1
®
=
®
=-(
^
40.0cm)
=- ´
-
( )
^
40.0 m 10
2
j
F l B
1 1
0
®
=
®
´
®
= I( )
l bc k
2
®
=
®
=(
^
40.0cm)
=- ´
-
( )
^
400 m 10
2
k
F l B i
2 2
®
=
®
´
®
= I( ) ( )
^
0.04N
l cd i j
3
2 2
40 10 40 10
®
=
®
=- ´ + ´
- -
( ) ( )
^ ^
m
F l B k
3 3
®
=
®
´
®
=- I( ) (
^
0.04N)
l da i k
4
2 2
40 10 40 10
®
=
®
= ´ - ´
- -
( ) ( )
^ ^
m m
F l B i k
4 4
®
=
®
´
®
= + I( ) ( (
^ ^
0.04N) 0.04N)
24. M M
®
=IA
^
= ´ ´ -
-
0.20 8.0 0.60 0.80 p( ) ( )
^ ^
10
2 2
i j
= ´ -
-
( . )( . . )
^ ^
40 2 10 060 080
4
i j A-m
2
B i k
®
= + ( (
^ ^
0.25T) 0.30T)
(a) t
®
=
®
´
®
M B
= ´ - - +
-
( )( )
^ ^ ^
40.2 0.24 0.18 0.2 10
4
i j k
= - - + ´
-
( )
^ ^ ^
9.6 7.2 8.0 i j k 10
4
N-m.
(b) U = -
®
×
®
= - ´
-
M B ( )( 40.2 0.15) J 10
4
» - ´
-
6.0 J 10
4
25. Con sider the wire is bent in the form of a
loop of N turns,
L
N
=
2p
Magnetic dipole moment associated with the
loop
M NiA Ni r
i L
N
= = ´ = p
p
2
2
2
4
t
p
= ° = MB
iL B
N
sin90
4
2
Clearly t is maximum, when N = 1
and the maximum torque is given by
t
p
m
i L B
=
2
4
26. Con sider the disc to be made up of large
num ber of el e men tary rings. Con sider on
such ring of ra dius x and thick ness dx.
Charge on this ring,
dq
q
R
x dx
q
R
x dx = ´ =
p
p
2 2
2
2
91
x
dx
Page 5

AIEEE Corner
Subjective Questions (Level-1)
1. Pos i tive. By Flem ings left hand rule.
2. F evB
m
= sin q
Þ    v
F
eB
e
=
sin q
=
´
´ ´ ´ ´ °
-
- -
4.6
1.6 3.5
10
10 10 60
15
19 3
sin
= ´ 9.46 10
6
m s /
3. F qvB
m
= sin q
= ´ ´ ´ ´ ´
-
( ) 2 10 10 1
19 5
1.6 0.8
= ´
-
2.56 10
14
N
4. (a)  F v B
m
e
®
=
®
´
®
( )
= - ´ ´ + ´
-
16 10 20 10 30 10
19 6 6
. [( . ) ( . ) ]
^ ^
i j
´ + ( . . )
^ ^
003 015 i j
= - ´
-
( )
^
6.24 N 10
4
k
(b) =
® ®
´
®
= - ´
-
F v B k
m
e ( ) ( . )
^
624 10
4
N
5. F v B
m
e
®
=
®
´
®
( )
( ) [( )
^ ^ ^
6.4 1.6 ´ = - ´ +
- -
10 10 2 4
19 19
k i j
´ + ( )]
^ ^
B B
x x
i j 3
6.4 1.6 ´ = - ´
- -
10 10 2
19 19
k k
^ ^
[ ] B
x
B
x
=
´
- ´
= -
-
-
6.4
3.2
2.0 T
10
10
19
19
6.(a) As mag netic force al ways acts
per pen dic u lar to mag netic field, mag netic
field must be along x-axis.
F qvB
1 1 1
= sinq
Þ B
F
qvB
= =
´
´ ´ ´
-
-
1
1 1
3
6 6
5 2 10
1 10 10
1
2
sinq
Þ B=
-
10
3
T
or    B i
®
=
-
( )
^
10
3
T
(b) F qv B
2 2 2
= sin q
= ´ ´ ´ ´ °
- -
1 10 10 10 90
6 6 3
sin
=
-
10
3
N
F
2
1 = mN
7. Let B i j k
®
= + + B B B
x y z
^ ^ ^
(a) F v B
®
=
®
´
®
q( )
7.6 5.2 ´ - ´
- -
10 10
3 3
i k
^ ^
= - ´ ´ ´ -
-
7.8 3.8 10 10
6 3
( )
^ ^
B B
z x
i k
Þ      B
x
= - 0.175 T, B
z
= - 0.256 T
(b) Cannot be determined by this information.
(c) As F v B
®
=
®
´
®
q( )
F B
®
^
®
Hence, B F
®
×
®
= 0
8. B i
®
= B
^
(a) v j
®
= v
^
F v B k
®
=
®
´
®
= - q qvB ( )
^
(b) v j
®
= v
^
F v B j
®
=
®
´
®
= q qvB ( )
^
(c) v i
®
= - v
^

F v B
®
=
®
´
®
= q( ) 0
(d) v i k
®
= ° - ° v v cos cos
^ ^
45 45
F v B j
®
=
®
´
®
= - q
qvB
( )
^
2
(e) v j k
®
= ° - ° v v cos cos
^ ^
45 45
F v B j k
®
=
®
´
®
= - - q
qvB
( ) ( )
^ ^
2
= - +
qvB
2
( )
^ ^
j k
9. r
mv
qB
mk
eB
meV
eB
= = =
2 2
Þ          B
mV
e
r =
2

=
´ ´ ´ ´
´
´
-
-
2 10 2 10
10
31 3
19
9.1
1.6
0.180
88
= ´
-
0.36 10
4
T
B = ´
-
3.6 10
4
T
10. (a) r
mv
qB
= Þ v
qBr
m
=
=
´ ´ ´ ´
´
- -
-
1.6 2.5 6.96
3.34
10 10
10
19 3
27
= ´ 8.33 10
5
ms
-1
(b) t
T m
qB
= =
2
p
=
´ ´
´ ´
-
-
3.14 3.34
1.6 2.5
10
10
27
19
= ´
-
2.62 10
8
s
(c) k eV mv = =
1
2
2
Þ V
mv
e
=
2
2
=
´ ´ ´
´ ´
-
-
3.34 8.33
1.6
10 10
2 10
27 5 2
19
( )
= ´ 7.26 10
3
V
= 7.26 kV
11. (a) - q. As ini tially par ti cle is neu tral, charge
on two par ti cles must be equal and op po site.
(b) The will collide after completing half
rotation, i.e.,
t
T m
qB
= =
2
p
12. Here, r = =
10.0
5.0
2
cm,
(a) r
mv
qB
= Þ B
mv
qr
=
=
´ ´ ´
´ ´ ´
-
- -
9.1 1.41
1.6
10 10
10 5 10
31 6
19 2
= ´
-
1.6 T 10
4
By Fleming’s left hand rule, direction of
magnetic field must be inward.
(b) t
T m
qB
= =
2
p
=
´ ´
´ ´ ´
-
- -
3.14 9.1
1.6 1.6
10
10 10
31
19 4
= ´
-
1.1 10
7
s
13. The com po nent of ve loc ity along the
mag netic field (i.e., v
x
) will re main
un changed and the pro ton will move in a
he li cal path.
At any instant,
Components of velocity of particle along
Y-axis and Z-axis
¢ = = v v v t
y y y
cos cos q w
and   ¢ = - = v v v t
z z z
sin sin q w
where,      w=
qB
m
\  v i j k
®
= + - v v t v t
x y z
^ ^ ^
cos sin w w
14. For the elec tron to hit the tar get, dis tance
G S must be mul ti ple of pitch, i.e.,
GS np =
For minimum distance, n = 1
Þ     GS p
mv
qB
= =
2p q cos
Þ       p
mk
qB
=
° 2 2 60 p cos
(mv mk = 2 )
Þ         B
mk
qp
=
° 2 2 60 p cos
=
´ ´ ´ ´ ´ ´ ´ ´
´ ´
- -
-
2 2 10 2 10
1
2
10
31 16
19
3.14 9.1 1.6
1.6 0.1
Þ B = ´
-
4.73 10
4
T
15. (a) From Ques tion 5 (c)
Introductory Exercise 23.2
L
R
= sin q Þ L R = sin q
R
R
sin 60
2
° =
Þ L
mv
qB
mv
qB
= =
2 2
0
0
(b) Now, L L R ¢ = = 2.1 1.05
As L R ¢ > ,
89
v
y v sinwt
y
v cos wt
y
z
®
B
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
B
–q
q
+q
Particle will describe a semicircle and move
out of the magnetic field moving in opposite
direction, i.e.,
v v v ¢ = - = -
0
i
^
and t
T m
qB
= =
2
0
p
16. v i
®
=
-
( )
^
50
1
ms , B j
®
= ( )
^
2.0 mT
As particle move with uniform velocity,
F E v B
®
=
®
+
®
´
®
= q( ) 0
Þ E B v
®
=
®
´
®
= -( 0.1 N/C)k
^
17. If v be the speed of par ti cle at point ( , , ) 0 y z
then by work-en ergy the o rem,
W K mv = = D
1
2
2
But work done by magnetic force is zero,
hence, network done = work done by electric
force
= qEZ
\ qE Z mv
0
2
1
2
=
Þ          v
qE Z
m
=
2
0
As the magnetic field is along Y-axis,
particle will move in XZ-plane.
The path of particle will be a cycloid. In this
case, instantaneous centre of curvature of
the particle will move along X-axis.
As magnetic force provides centripetal force
to the particle,
qvB
mv
R
0
2
=
v
qB R
m
=
0
v v
qB R
m
x
= = cos
cos
q
q
0
=
qB Z
m
0
( Q R Z cos q = )
Now, v v v
qE Z
m
q B Z
m
z x
= - = -
2 2 0
2
0
2 2
2
2
18. Given, E j
®
= E
^
, B k
®
= B
^
,
v j k
®
= + v v cos sin
^ ^
q q
As protons are moving undeflected,
F
®
= 0 Þ e ( ) E v B
®
+
®
´
®
= 0
Þ       e E vB ( cos )
^ ^
j j - = q 0
or        v
E
B
=
cosq
Now, if electric field is switched off
p
mv
qB
mE
qB
= =
2 2
2
p q p q sin tan
(Component of velocity along magnetic field
= = v v
z
sin q)
19. F I lB = sin q
I
F
lB
= =
´ ´ ° sin sin q
0.13
0.2 0.067 90
= 9.7 A
20. For no ten sion in springs
F mg
m
=
Þ   I lB mg =
I
mg
lB
= =
´ ´
´ ´
-
-
13.0
62.0 0.440
10 10
10
3
2
=0.48 A
By Fleming left hand rule, for magnetic force
to act in upward direction, current in the
wire must be towards right.
21. (a) FBD of metal bar is shown in fig ure, for
metal to be in equi lib rium,
F N mg
m
+ =
Þ         F mg N
m
= -
Þ        IlB m N = -
Þ
V
R
lB mg N = -
Þ          V
R
lB
mg N = - ( )
90
q
q
z
v
z
v
v
x
R
X
v
y
z
q
O
Z
E = EK.
v
v
x
x
B = –B
®
j
^
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
mg
F
m
I I
F
m
N
mg
For largest voltage,
N =0
V
Rmg
lB
= =
´ ´ ´
´ ´
-
-
25 750 10
10
3
2
9.8
50.0 0.450
= 817.5 V
(b) If I lB mg >
I lB mg ma - =
a
I lB mg
m
V lB
Rm
g =
-
= -
=
´ ´ ´
´ ´
-
-
-
817.5 0.45
9.8
50 10
2 750 10
2
3
=112.8 m/s
2
22. I = 3.50 A, l = - ( )
^
1.00 cm i
Þ l = - ´
-
( )
^
1.00 m 10
2
i
(a) B j
®
= - (
^
0.65 T)
F l B k
m
I
®
=
®
´
®
= - ( ) (
^
0.023 N)
(b) B k
®
= + (
^
0.56 T)
F l B j
m
I
®
=
®
´
®
= ( ) (
^
0.0196 N)
(c) B i
®
= - (
^
0.33 T)
F l B
m
I
®
=
®
´
®
= ( ) 0
(d) B i k
®
= -
®
( ( )
^
0.33 T) 0.28T
F l B j
m
I
®
=
®
´
®
= - ( ) (
^
0.0098 N)
(e) B j k
®
= + - ( ( )
^ ^
0.74 T) 0.36 T
F l B k j
m
I
®
=
®
´
®
= - + ( ) ( ( )
^ ^
0.0259 N) 0.0126 N
= - ( ) ( )
^ ^
0.0126 N 0.0259 N j K
23. B j
®
=(
^
0.020T)
l ab j
1
®
=
®
=-(
^
40.0cm)
=- ´
-
( )
^
40.0 m 10
2
j
F l B
1 1
0
®
=
®
´
®
= I( )
l bc k
2
®
=
®
=(
^
40.0cm)
=- ´
-
( )
^
400 m 10
2
k
F l B i
2 2
®
=
®
´
®
= I( ) ( )
^
0.04N
l cd i j
3
2 2
40 10 40 10
®
=
®
=- ´ + ´
- -
( ) ( )
^ ^
m
F l B k
3 3
®
=
®
´
®
=- I( ) (
^
0.04N)
l da i k
4
2 2
40 10 40 10
®
=
®
= ´ - ´
- -
( ) ( )
^ ^
m m
F l B i k
4 4
®
=
®
´
®
= + I( ) ( (
^ ^
0.04N) 0.04N)
24. M M
®
=IA
^
= ´ ´ -
-
0.20 8.0 0.60 0.80 p( ) ( )
^ ^
10
2 2
i j
= ´ -
-
( . )( . . )
^ ^
40 2 10 060 080
4
i j A-m
2
B i k
®
= + ( (
^ ^
0.25T) 0.30T)
(a) t
®
=
®
´
®
M B
= ´ - - +
-
( )( )
^ ^ ^
40.2 0.24 0.18 0.2 10
4
i j k
= - - + ´
-
( )
^ ^ ^
9.6 7.2 8.0 i j k 10
4
N-m.
(b) U = -
®
×
®
= - ´
-
M B ( )( 40.2 0.15) J 10
4
» - ´
-
6.0 J 10
4
25. Con sider the wire is bent in the form of a
loop of N turns,
L
N
=
2p
Magnetic dipole moment associated with the
loop
M NiA Ni r
i L
N
= = ´ = p
p
2
2
2
4
t
p
= ° = MB
iL B
N
sin90
4
2
Clearly t is maximum, when N = 1
and the maximum torque is given by
t
p
m
i L B
=
2
4
26. Con sider the disc to be made up of large
num ber of el e men tary rings. Con sider on
such ring of ra dius x and thick ness dx.
Charge on this ring,
dq
q
R
x dx
q
R
x dx = ´ =
p
p
2 2
2
2
91
x
dx
Current associated with this ring,
di
dq
T
dq q
R
x dx = = =
w
p
w
p 2
2
Magnetic moment of this ring
dM x di
q
R
x dx = = p
w
2
2
3
Magnetic moment of entire disc,
M dM
q
R
x dx qR
R
= = =
ò ò
w
w
2
3
0
2
1
4
…(i)
Magnetic field at the centre of disc due to the
elementary ring under consideration
dB
di
x
q
R
dx = =
m m w
p
0 0
2
2
2 2
Net magnetic field at the centre of the disc,
B dB
q
R
dx
q
R
R
= = =
ò ò
m w
p
m w
p
0
2
0
0
2 2
\
M
B
R
=
p
m
3
0
2
27. (a) By prin ci ple of con ser va tion of en ergy,
Gain in KE = Loss in PE
KE PE ME = - + cosq
f
K
cosq= - = -
´
´ ´
-
-
1 1
10
52 10
3
3
ME
0.80
0.02
=
10
13
q= = °
-
cos
1
10
13
76.7
(b) q= = °
-
cos
1
10
13
76.7
Entire KE will again get converted into PE
28. DU U U MB MB = - = - - +
2 1
( )
= -2MB
= - ´ ´ = - 2 1.45 0.835 2.42J
29. (a) T
r
v
= =
´ ´ ´
´
-
2 2 10
10
11
6
p 3.14 5.3
2.2
= ´
-
1.5 10
16
s
(b) i
e
T
= =
´
´
= ´
-
-
-
1.6
1.5
1.1 A
10
10
10
19
16
3
= 1.1 mA
(c) M r i = p
2
= ´ ´ ´ ´
- -
3.14 5.3 1.1 ( ) 10 10
11 2 3
= ´
-
9.3 10
24
A-m
2
30. Sup pose equal and op po site cur rents are
flow ing in sides a d and eh, so that three
com plete cur rent car ry ing loops are formed,
M k
®
= -
abcd
i l
2 ^
M k
®
=
efgh
i l
2 ^
M j
®
=
i l
2 ^
\ Total magnetic moment of the closed path,
M M M M j
®
=
®
+
®
+
®
=
i l
2 ^
31. Cir cuit is same as in Q.30
M j j
®
= = i l
2 ^ ^
B j
®
= 2
^
t
®
=
®
´
®
= M B 0
32. B
I
r
1
0
4
= ×
m
p
B
I
r
2
0
4
= ×
m
p
Here, B
1
and B
2
are perpendicular to each
other, hence,
B B B = +
1
2
2
2
= × =
´ ´
´
-
-
m
p
0
7
2
4
2 10 2 5
35 10
I
r
= ´
-
2.0 T 10
6
=2.0 T m
33. Clearly D D BOC AOB ~
\
r
r
BC
2
6
=
Þ           r r
2
2 =
=100 mm
92
y
f
c
b
g
h
e
x
d
a
z
2
l
l
1
```

122 docs

## DC Pandey Solutions for NEET Physics

122 docs

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