Chapter 23 - Magnetics (Part - 1)- Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

DC Pandey (Questions & Solutions) of Physics: NEET

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NEET : Chapter 23 - Magnetics (Part - 1)- Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

 Page 1


Introductory Exercise 23.1
1. [ ] [ ] F F
e m
=
[ ] [ ] qE qvB =            
Þ 
E
B
v
é
ë
ê
ù
û
ú
= =
-
[ ] [L T ]
1
2. F v B
®
=
®
´
®
q( ) 
\ F v
®
^
®
 and F B
®
^
®
Because cross product of any two vectors is
always perpendicular to both the vectors.
3. No. As F
m
q
®
=
®
´
®
( ) v B
Þ | | sin F
m
qvB
®
= q
If F
m
= 0, either B = 0 or sin q = 0,
i.e., q = 0
4. F v B
®
=
®
´
®
q( )
= - ´ ´ ´
- - -
4 10 10 10
6 6 2
 [( )
^ ^ ^
2 3 i j k - + 
´ + - ( )]
^ ^ ^
2 5 3 i j k
= - ´ + +
-
4 10 4 8 16
2
( )
^ ^ ^
i j k
= - + + ´
-
16 2 4 10
2
( )
^ ^ ^
i j k N
Introductory Exercise 23.2
1. As mag netic field can ex ert force on charged
par ti cle, it can be ac cel er ated in mag netic
field but its speed can not in creases as
mag ne tic forc e is al way s per p en dic u lar to
the di rec tion of mo tion of charged particle.
2. F v B
m
e
®
= -
®
´
®
( )
By Fleming’s left hand rule, B
®
 must be along 
positive z-axis.
3. As mag netic force pro vides nec es sary
cen trip e tal force to the  particle to de scribe a
circle.
              qvB
mv
r
=
2
Þ               r
mv
qB
=
(a)       r
mv
qB
=
 Þ       r m µ
Hence, electron will describe smaller circle.
(b)     T
r
v
m
qB
= =
2 2 p p
 
       f
T
qB
m
= =
1
2p
Þ      f
m
µ
1
\ electron have greater frequency.
4. Elec trons are re fo cused on x-axis at a
dis tance equal to pitch, i.e.,
n p v T = =
| |
     =
2p q mv
eB
cos
Magnetics
23
Page 2


Introductory Exercise 23.1
1. [ ] [ ] F F
e m
=
[ ] [ ] qE qvB =            
Þ 
E
B
v
é
ë
ê
ù
û
ú
= =
-
[ ] [L T ]
1
2. F v B
®
=
®
´
®
q( ) 
\ F v
®
^
®
 and F B
®
^
®
Because cross product of any two vectors is
always perpendicular to both the vectors.
3. No. As F
m
q
®
=
®
´
®
( ) v B
Þ | | sin F
m
qvB
®
= q
If F
m
= 0, either B = 0 or sin q = 0,
i.e., q = 0
4. F v B
®
=
®
´
®
q( )
= - ´ ´ ´
- - -
4 10 10 10
6 6 2
 [( )
^ ^ ^
2 3 i j k - + 
´ + - ( )]
^ ^ ^
2 5 3 i j k
= - ´ + +
-
4 10 4 8 16
2
( )
^ ^ ^
i j k
= - + + ´
-
16 2 4 10
2
( )
^ ^ ^
i j k N
Introductory Exercise 23.2
1. As mag netic field can ex ert force on charged
par ti cle, it can be ac cel er ated in mag netic
field but its speed can not in creases as
mag ne tic forc e is al way s per p en dic u lar to
the di rec tion of mo tion of charged particle.
2. F v B
m
e
®
= -
®
´
®
( )
By Fleming’s left hand rule, B
®
 must be along 
positive z-axis.
3. As mag netic force pro vides nec es sary
cen trip e tal force to the  particle to de scribe a
circle.
              qvB
mv
r
=
2
Þ               r
mv
qB
=
(a)       r
mv
qB
=
 Þ       r m µ
Hence, electron will describe smaller circle.
(b)     T
r
v
m
qB
= =
2 2 p p
 
       f
T
qB
m
= =
1
2p
Þ      f
m
µ
1
\ electron have greater frequency.
4. Elec trons are re fo cused on x-axis at a
dis tance equal to pitch, i.e.,
n p v T = =
| |
     =
2p q mv
eB
cos
Magnetics
23
5. (a) If L r
mv
qB
³ = ,
(b) The particle will describe a semi-circle.
Hence, q p =
(c)          
L
l
=cos
q
2
Þ      
L
R 2
2
2
sin
cos
q
q
=
            
L
R
=sinq Þ  sin q =
1
2
Þ           q
p
=
6
6. r
mv
eB
mk
eB
= =
2
      = =
2
1 2
meV
eB B
mV
e
For electron, 
r =
´ ´ ´
´
-
-
1 2 10 100
10
31
19
0.2
9.1
1.6
   
= ´
-
1.67 10
4
 m = 0.0167 cm
For proton
r =
´ ´ ´
´
-
-
1 2 10 100
10
27
19
0.2
1.67
1.6
= ´
-
7 10
3
 m = 0.7  cm           
7. r
mv
qB
m k
qB
= =
2
\ r
m
q
µ    
\ r r r
p d
: : : :
a
=
1
1
2
1
4
2
         
= 1 2 1 : :
Introductory Exercise 23.3
1. Let at any in stant
V i j k
®
= + + V V V
x y z
^ ^ ^
Now, V V V
x y
2 2
0
2
+ = = constant
and V V
qE
m
f
2 0
= -
V
®
 is minimum when V
2
0 =
at            f
mv
qE
=
0
and     V V
min
=
0
2. Af ter one rev o lu tion, y = 0 , 
x p = = pitch of heating
=
2p q mv
qB
sin
Hence, coordination of the particle,
= ( , ) x b =
æ
è
ç
ç
ö
ø
÷
÷
0
2
,
sin p q mv
qB
3. F l B i j k
®
=
®
´
®
= ´ + i ilB ( ) [ ( )]
^ ^ ^
( ) F
®
= 2 ilB
4. No. as i i j k i j i j k
^ ^ ^ ^ ^ ^ ^ ^ ^
( ) ( ) ´ + + = ´ + ´ ´
But i j
^ ^
´ = 0
\  i i j k i j k
^ ^ ^ ^ ^ ^ ^
( ) ( ) ´ + + = ´ ´
 84
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
O
q
q
× × ×
q/2
p
2
–
q
2
L
Page 3


Introductory Exercise 23.1
1. [ ] [ ] F F
e m
=
[ ] [ ] qE qvB =            
Þ 
E
B
v
é
ë
ê
ù
û
ú
= =
-
[ ] [L T ]
1
2. F v B
®
=
®
´
®
q( ) 
\ F v
®
^
®
 and F B
®
^
®
Because cross product of any two vectors is
always perpendicular to both the vectors.
3. No. As F
m
q
®
=
®
´
®
( ) v B
Þ | | sin F
m
qvB
®
= q
If F
m
= 0, either B = 0 or sin q = 0,
i.e., q = 0
4. F v B
®
=
®
´
®
q( )
= - ´ ´ ´
- - -
4 10 10 10
6 6 2
 [( )
^ ^ ^
2 3 i j k - + 
´ + - ( )]
^ ^ ^
2 5 3 i j k
= - ´ + +
-
4 10 4 8 16
2
( )
^ ^ ^
i j k
= - + + ´
-
16 2 4 10
2
( )
^ ^ ^
i j k N
Introductory Exercise 23.2
1. As mag netic field can ex ert force on charged
par ti cle, it can be ac cel er ated in mag netic
field but its speed can not in creases as
mag ne tic forc e is al way s per p en dic u lar to
the di rec tion of mo tion of charged particle.
2. F v B
m
e
®
= -
®
´
®
( )
By Fleming’s left hand rule, B
®
 must be along 
positive z-axis.
3. As mag netic force pro vides nec es sary
cen trip e tal force to the  particle to de scribe a
circle.
              qvB
mv
r
=
2
Þ               r
mv
qB
=
(a)       r
mv
qB
=
 Þ       r m µ
Hence, electron will describe smaller circle.
(b)     T
r
v
m
qB
= =
2 2 p p
 
       f
T
qB
m
= =
1
2p
Þ      f
m
µ
1
\ electron have greater frequency.
4. Elec trons are re fo cused on x-axis at a
dis tance equal to pitch, i.e.,
n p v T = =
| |
     =
2p q mv
eB
cos
Magnetics
23
5. (a) If L r
mv
qB
³ = ,
(b) The particle will describe a semi-circle.
Hence, q p =
(c)          
L
l
=cos
q
2
Þ      
L
R 2
2
2
sin
cos
q
q
=
            
L
R
=sinq Þ  sin q =
1
2
Þ           q
p
=
6
6. r
mv
eB
mk
eB
= =
2
      = =
2
1 2
meV
eB B
mV
e
For electron, 
r =
´ ´ ´
´
-
-
1 2 10 100
10
31
19
0.2
9.1
1.6
   
= ´
-
1.67 10
4
 m = 0.0167 cm
For proton
r =
´ ´ ´
´
-
-
1 2 10 100
10
27
19
0.2
1.67
1.6
= ´
-
7 10
3
 m = 0.7  cm           
7. r
mv
qB
m k
qB
= =
2
\ r
m
q
µ    
\ r r r
p d
: : : :
a
=
1
1
2
1
4
2
         
= 1 2 1 : :
Introductory Exercise 23.3
1. Let at any in stant
V i j k
®
= + + V V V
x y z
^ ^ ^
Now, V V V
x y
2 2
0
2
+ = = constant
and V V
qE
m
f
2 0
= -
V
®
 is minimum when V
2
0 =
at            f
mv
qE
=
0
and     V V
min
=
0
2. Af ter one rev o lu tion, y = 0 , 
x p = = pitch of heating
=
2p q mv
qB
sin
Hence, coordination of the particle,
= ( , ) x b =
æ
è
ç
ç
ö
ø
÷
÷
0
2
,
sin p q mv
qB
3. F l B i j k
®
=
®
´
®
= ´ + i ilB ( ) [ ( )]
^ ^ ^
( ) F
®
= 2 ilB
4. No. as i i j k i j i j k
^ ^ ^ ^ ^ ^ ^ ^ ^
( ) ( ) ´ + + = ´ + ´ ´
But i j
^ ^
´ = 0
\  i i j k i j k
^ ^ ^ ^ ^ ^ ^
( ) ( ) ´ + + = ´ ´
 84
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
O
q
q
× × ×
q/2
p
2
–
q
2
L
Introductory Exercise 23.4
1. Con sider the disc to be made up of large
num ber of elementary concentric rings.
Con sider one such ring of ra dius x and
thick ne ss d x .
Charge on this ring
dq
q
R
dA
q
R
x dx = × + ´
p p
p
2 2
2
       dq
qxdx
R
=
2
2
Current in this ring,
di
dq
T
dq qx dx
R
= + =
w
p
w
p 2
2
\  Magnetic moment of this ring,
dM di A
qx dx
R
x = ´ = ´
w
p
p
2
2
=
wq
R
x dx
2
3
\ Magnetic moment of entire disc,
M dM
q
R
x dx
R
= =
ò ò
w
2
3
0
=
é
ë
ê
ê
ù
û
ú
ú
=
w
w
q
R
R
qR
2
4
2
4
1
4
2. M OA AB
®
= ´
®
´
®
i [( )]
OA j k
®
= + OA OA cos sin
^ ^
q q
AB i
®
= AB
^
\ M j k i
®
= ´ × + ´ i OA AB [(cos sin ) ]
^ ^ ^
q q
= ´ ´ +
æ
è
ç
ç
ö
ø
÷
÷
´
é
ë
ê
ê
ù
û
ú
ú
4
3
2
1
2
0.2 0.1 j k i
^ ^ ^
= - ( )
^ ^
0.04 0.07 j k A-m
2
Introductory Exercise 23.5
1. (a) B B B B
1 2 3 4
= = =
= × ° + °
m
p
0
4 2
45 45
i
l /
[sin sin ]
= ×
m
p
0
4
2 2i
l
Net magnetic field at the centre of the
square,
B B B B B
i
l
= + + + = ×
1 2 3 4
0
4
8 2 m
p
= =
2 2
0
m
p
m
i
l
28.3 T (inward)
(b)If the conductor is converted into a
circular loop, then
2 4 pr l = Þ r
l
=
2
p
B
i
r
i
l
= = =
m p m
m
0 0
2 4
24.7 T (inward)
2. B
i
x
= ×
m
p
0
4
(As P is lying near one end of conductor 1)
B
2
0 = (Magnetic field on the axis of a current 
carrying conductor is zero)
       B B =
1
               = ×
m
p
0
4
i
x
By right hand thumb rule, direction of
magnetic field at P is inward.
85 
1
2
45°
i
3
l
l
2
4
2 i
x
P
i
1
x
Page 4


Introductory Exercise 23.1
1. [ ] [ ] F F
e m
=
[ ] [ ] qE qvB =            
Þ 
E
B
v
é
ë
ê
ù
û
ú
= =
-
[ ] [L T ]
1
2. F v B
®
=
®
´
®
q( ) 
\ F v
®
^
®
 and F B
®
^
®
Because cross product of any two vectors is
always perpendicular to both the vectors.
3. No. As F
m
q
®
=
®
´
®
( ) v B
Þ | | sin F
m
qvB
®
= q
If F
m
= 0, either B = 0 or sin q = 0,
i.e., q = 0
4. F v B
®
=
®
´
®
q( )
= - ´ ´ ´
- - -
4 10 10 10
6 6 2
 [( )
^ ^ ^
2 3 i j k - + 
´ + - ( )]
^ ^ ^
2 5 3 i j k
= - ´ + +
-
4 10 4 8 16
2
( )
^ ^ ^
i j k
= - + + ´
-
16 2 4 10
2
( )
^ ^ ^
i j k N
Introductory Exercise 23.2
1. As mag netic field can ex ert force on charged
par ti cle, it can be ac cel er ated in mag netic
field but its speed can not in creases as
mag ne tic forc e is al way s per p en dic u lar to
the di rec tion of mo tion of charged particle.
2. F v B
m
e
®
= -
®
´
®
( )
By Fleming’s left hand rule, B
®
 must be along 
positive z-axis.
3. As mag netic force pro vides nec es sary
cen trip e tal force to the  particle to de scribe a
circle.
              qvB
mv
r
=
2
Þ               r
mv
qB
=
(a)       r
mv
qB
=
 Þ       r m µ
Hence, electron will describe smaller circle.
(b)     T
r
v
m
qB
= =
2 2 p p
 
       f
T
qB
m
= =
1
2p
Þ      f
m
µ
1
\ electron have greater frequency.
4. Elec trons are re fo cused on x-axis at a
dis tance equal to pitch, i.e.,
n p v T = =
| |
     =
2p q mv
eB
cos
Magnetics
23
5. (a) If L r
mv
qB
³ = ,
(b) The particle will describe a semi-circle.
Hence, q p =
(c)          
L
l
=cos
q
2
Þ      
L
R 2
2
2
sin
cos
q
q
=
            
L
R
=sinq Þ  sin q =
1
2
Þ           q
p
=
6
6. r
mv
eB
mk
eB
= =
2
      = =
2
1 2
meV
eB B
mV
e
For electron, 
r =
´ ´ ´
´
-
-
1 2 10 100
10
31
19
0.2
9.1
1.6
   
= ´
-
1.67 10
4
 m = 0.0167 cm
For proton
r =
´ ´ ´
´
-
-
1 2 10 100
10
27
19
0.2
1.67
1.6
= ´
-
7 10
3
 m = 0.7  cm           
7. r
mv
qB
m k
qB
= =
2
\ r
m
q
µ    
\ r r r
p d
: : : :
a
=
1
1
2
1
4
2
         
= 1 2 1 : :
Introductory Exercise 23.3
1. Let at any in stant
V i j k
®
= + + V V V
x y z
^ ^ ^
Now, V V V
x y
2 2
0
2
+ = = constant
and V V
qE
m
f
2 0
= -
V
®
 is minimum when V
2
0 =
at            f
mv
qE
=
0
and     V V
min
=
0
2. Af ter one rev o lu tion, y = 0 , 
x p = = pitch of heating
=
2p q mv
qB
sin
Hence, coordination of the particle,
= ( , ) x b =
æ
è
ç
ç
ö
ø
÷
÷
0
2
,
sin p q mv
qB
3. F l B i j k
®
=
®
´
®
= ´ + i ilB ( ) [ ( )]
^ ^ ^
( ) F
®
= 2 ilB
4. No. as i i j k i j i j k
^ ^ ^ ^ ^ ^ ^ ^ ^
( ) ( ) ´ + + = ´ + ´ ´
But i j
^ ^
´ = 0
\  i i j k i j k
^ ^ ^ ^ ^ ^ ^
( ) ( ) ´ + + = ´ ´
 84
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
O
q
q
× × ×
q/2
p
2
–
q
2
L
Introductory Exercise 23.4
1. Con sider the disc to be made up of large
num ber of elementary concentric rings.
Con sider one such ring of ra dius x and
thick ne ss d x .
Charge on this ring
dq
q
R
dA
q
R
x dx = × + ´
p p
p
2 2
2
       dq
qxdx
R
=
2
2
Current in this ring,
di
dq
T
dq qx dx
R
= + =
w
p
w
p 2
2
\  Magnetic moment of this ring,
dM di A
qx dx
R
x = ´ = ´
w
p
p
2
2
=
wq
R
x dx
2
3
\ Magnetic moment of entire disc,
M dM
q
R
x dx
R
= =
ò ò
w
2
3
0
=
é
ë
ê
ê
ù
û
ú
ú
=
w
w
q
R
R
qR
2
4
2
4
1
4
2. M OA AB
®
= ´
®
´
®
i [( )]
OA j k
®
= + OA OA cos sin
^ ^
q q
AB i
®
= AB
^
\ M j k i
®
= ´ × + ´ i OA AB [(cos sin ) ]
^ ^ ^
q q
= ´ ´ +
æ
è
ç
ç
ö
ø
÷
÷
´
é
ë
ê
ê
ù
û
ú
ú
4
3
2
1
2
0.2 0.1 j k i
^ ^ ^
= - ( )
^ ^
0.04 0.07 j k A-m
2
Introductory Exercise 23.5
1. (a) B B B B
1 2 3 4
= = =
= × ° + °
m
p
0
4 2
45 45
i
l /
[sin sin ]
= ×
m
p
0
4
2 2i
l
Net magnetic field at the centre of the
square,
B B B B B
i
l
= + + + = ×
1 2 3 4
0
4
8 2 m
p
= =
2 2
0
m
p
m
i
l
28.3 T (inward)
(b)If the conductor is converted into a
circular loop, then
2 4 pr l = Þ r
l
=
2
p
B
i
r
i
l
= = =
m p m
m
0 0
2 4
24.7 T (inward)
2. B
i
x
= ×
m
p
0
4
(As P is lying near one end of conductor 1)
B
2
0 = (Magnetic field on the axis of a current 
carrying conductor is zero)
       B B =
1
               = ×
m
p
0
4
i
x
By right hand thumb rule, direction of
magnetic field at P is inward.
85 
1
2
45°
i
3
l
l
2
4
2 i
x
P
i
1
x
3. Mag netic field due to straight con duc tor at O
B
i
R
1
0
4
2
= ×
m
p
Magnetic field at O due to circular loop
B
i
R
2
0
2
=
m
By right hand thumb rule, both the filds are
acting inward.
Hence,
B B B
i
R
= + = +
1 2
0
2
1
m
p
p ( )
       =
´ ´
´ ´
+
æ
è
ç
ö
ø
÷
-
-
4 10 7
2 10 10
1
22
7
7
2
p
p
               = ´ =
-
58 10 58
6
T T m (inward).
4. B B
1 2
0 = = (Mag netic field on the axis of
cur rent car ry ing con duc t or is zero)
B
i
R
i
R
3
0 0
1
4 2 8
= × =
m m
    =
´ ´
´ ´
-
-
4 10 5
8 3 10
7
2
p
               = ´
-
2.62 10
5
 T
       =26.2 T m (inward).
5. B B
1 2
0 = = (Mag netic field on the axis of
straight con duc tor is zero)
B
i
b
i
b
2
0 0
60
360 2 12
=
°
°
× =
m m
 (inward)
  B
i
a a
3
0 0
60
360 2 12
=
°
°
= =
m m
 (outward)
As B B
3 2
> ,
Net magnetic field at P,
        B B B = -
3 2
          = -
é
ë
ê
ù
û
ú
m
0
12
1 1 i
a b
6. AB AP , and BP  from Pythagorus trip let, 
hence Ð = ° APB 90 
\  B PB
1
0 1
1
4
2 ®
= ×
m
p
i
r
^
B AP
2
0 2
2
4
2 ®
= ×
m
p
i
r
^
  B B B = +
1
2
2
2
     =
æ
è
ç
ç
ö
ø
÷
÷
+
æ
è
ç
ç
ö
ø
÷
÷
m
p
0 1
1
2
2
2
2
2
i
r
i
r
=
´
´
æ
è
ç
ö
ø
÷
+
æ
è
ç
ö
ø
÷
-
4 10
2
3 3
7 2 2
p
p 0.05 0.12
     = ´
-
1.3 10
5
 T
     =13 mT
7. t q = NIABcos
= ´ ´ ´ ´ ´ ° 100 3 30 1.2 0.4 0. 0.8 cos
        = 9.98 N-m
Rotation will be clockwise as seen from
above.
 86
60°
2 3
1
4
X
X
B
A
i
1
i
2
P ®
B
2 ®
B
1
5.0cm
13.0cm
12.0 cm
i
O
R
3
O
i
1
2
Page 5


Introductory Exercise 23.1
1. [ ] [ ] F F
e m
=
[ ] [ ] qE qvB =            
Þ 
E
B
v
é
ë
ê
ù
û
ú
= =
-
[ ] [L T ]
1
2. F v B
®
=
®
´
®
q( ) 
\ F v
®
^
®
 and F B
®
^
®
Because cross product of any two vectors is
always perpendicular to both the vectors.
3. No. As F
m
q
®
=
®
´
®
( ) v B
Þ | | sin F
m
qvB
®
= q
If F
m
= 0, either B = 0 or sin q = 0,
i.e., q = 0
4. F v B
®
=
®
´
®
q( )
= - ´ ´ ´
- - -
4 10 10 10
6 6 2
 [( )
^ ^ ^
2 3 i j k - + 
´ + - ( )]
^ ^ ^
2 5 3 i j k
= - ´ + +
-
4 10 4 8 16
2
( )
^ ^ ^
i j k
= - + + ´
-
16 2 4 10
2
( )
^ ^ ^
i j k N
Introductory Exercise 23.2
1. As mag netic field can ex ert force on charged
par ti cle, it can be ac cel er ated in mag netic
field but its speed can not in creases as
mag ne tic forc e is al way s per p en dic u lar to
the di rec tion of mo tion of charged particle.
2. F v B
m
e
®
= -
®
´
®
( )
By Fleming’s left hand rule, B
®
 must be along 
positive z-axis.
3. As mag netic force pro vides nec es sary
cen trip e tal force to the  particle to de scribe a
circle.
              qvB
mv
r
=
2
Þ               r
mv
qB
=
(a)       r
mv
qB
=
 Þ       r m µ
Hence, electron will describe smaller circle.
(b)     T
r
v
m
qB
= =
2 2 p p
 
       f
T
qB
m
= =
1
2p
Þ      f
m
µ
1
\ electron have greater frequency.
4. Elec trons are re fo cused on x-axis at a
dis tance equal to pitch, i.e.,
n p v T = =
| |
     =
2p q mv
eB
cos
Magnetics
23
5. (a) If L r
mv
qB
³ = ,
(b) The particle will describe a semi-circle.
Hence, q p =
(c)          
L
l
=cos
q
2
Þ      
L
R 2
2
2
sin
cos
q
q
=
            
L
R
=sinq Þ  sin q =
1
2
Þ           q
p
=
6
6. r
mv
eB
mk
eB
= =
2
      = =
2
1 2
meV
eB B
mV
e
For electron, 
r =
´ ´ ´
´
-
-
1 2 10 100
10
31
19
0.2
9.1
1.6
   
= ´
-
1.67 10
4
 m = 0.0167 cm
For proton
r =
´ ´ ´
´
-
-
1 2 10 100
10
27
19
0.2
1.67
1.6
= ´
-
7 10
3
 m = 0.7  cm           
7. r
mv
qB
m k
qB
= =
2
\ r
m
q
µ    
\ r r r
p d
: : : :
a
=
1
1
2
1
4
2
         
= 1 2 1 : :
Introductory Exercise 23.3
1. Let at any in stant
V i j k
®
= + + V V V
x y z
^ ^ ^
Now, V V V
x y
2 2
0
2
+ = = constant
and V V
qE
m
f
2 0
= -
V
®
 is minimum when V
2
0 =
at            f
mv
qE
=
0
and     V V
min
=
0
2. Af ter one rev o lu tion, y = 0 , 
x p = = pitch of heating
=
2p q mv
qB
sin
Hence, coordination of the particle,
= ( , ) x b =
æ
è
ç
ç
ö
ø
÷
÷
0
2
,
sin p q mv
qB
3. F l B i j k
®
=
®
´
®
= ´ + i ilB ( ) [ ( )]
^ ^ ^
( ) F
®
= 2 ilB
4. No. as i i j k i j i j k
^ ^ ^ ^ ^ ^ ^ ^ ^
( ) ( ) ´ + + = ´ + ´ ´
But i j
^ ^
´ = 0
\  i i j k i j k
^ ^ ^ ^ ^ ^ ^
( ) ( ) ´ + + = ´ ´
 84
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
×
O
q
q
× × ×
q/2
p
2
–
q
2
L
Introductory Exercise 23.4
1. Con sider the disc to be made up of large
num ber of elementary concentric rings.
Con sider one such ring of ra dius x and
thick ne ss d x .
Charge on this ring
dq
q
R
dA
q
R
x dx = × + ´
p p
p
2 2
2
       dq
qxdx
R
=
2
2
Current in this ring,
di
dq
T
dq qx dx
R
= + =
w
p
w
p 2
2
\  Magnetic moment of this ring,
dM di A
qx dx
R
x = ´ = ´
w
p
p
2
2
=
wq
R
x dx
2
3
\ Magnetic moment of entire disc,
M dM
q
R
x dx
R
= =
ò ò
w
2
3
0
=
é
ë
ê
ê
ù
û
ú
ú
=
w
w
q
R
R
qR
2
4
2
4
1
4
2. M OA AB
®
= ´
®
´
®
i [( )]
OA j k
®
= + OA OA cos sin
^ ^
q q
AB i
®
= AB
^
\ M j k i
®
= ´ × + ´ i OA AB [(cos sin ) ]
^ ^ ^
q q
= ´ ´ +
æ
è
ç
ç
ö
ø
÷
÷
´
é
ë
ê
ê
ù
û
ú
ú
4
3
2
1
2
0.2 0.1 j k i
^ ^ ^
= - ( )
^ ^
0.04 0.07 j k A-m
2
Introductory Exercise 23.5
1. (a) B B B B
1 2 3 4
= = =
= × ° + °
m
p
0
4 2
45 45
i
l /
[sin sin ]
= ×
m
p
0
4
2 2i
l
Net magnetic field at the centre of the
square,
B B B B B
i
l
= + + + = ×
1 2 3 4
0
4
8 2 m
p
= =
2 2
0
m
p
m
i
l
28.3 T (inward)
(b)If the conductor is converted into a
circular loop, then
2 4 pr l = Þ r
l
=
2
p
B
i
r
i
l
= = =
m p m
m
0 0
2 4
24.7 T (inward)
2. B
i
x
= ×
m
p
0
4
(As P is lying near one end of conductor 1)
B
2
0 = (Magnetic field on the axis of a current 
carrying conductor is zero)
       B B =
1
               = ×
m
p
0
4
i
x
By right hand thumb rule, direction of
magnetic field at P is inward.
85 
1
2
45°
i
3
l
l
2
4
2 i
x
P
i
1
x
3. Mag netic field due to straight con duc tor at O
B
i
R
1
0
4
2
= ×
m
p
Magnetic field at O due to circular loop
B
i
R
2
0
2
=
m
By right hand thumb rule, both the filds are
acting inward.
Hence,
B B B
i
R
= + = +
1 2
0
2
1
m
p
p ( )
       =
´ ´
´ ´
+
æ
è
ç
ö
ø
÷
-
-
4 10 7
2 10 10
1
22
7
7
2
p
p
               = ´ =
-
58 10 58
6
T T m (inward).
4. B B
1 2
0 = = (Mag netic field on the axis of
cur rent car ry ing con duc t or is zero)
B
i
R
i
R
3
0 0
1
4 2 8
= × =
m m
    =
´ ´
´ ´
-
-
4 10 5
8 3 10
7
2
p
               = ´
-
2.62 10
5
 T
       =26.2 T m (inward).
5. B B
1 2
0 = = (Mag netic field on the axis of
straight con duc tor is zero)
B
i
b
i
b
2
0 0
60
360 2 12
=
°
°
× =
m m
 (inward)
  B
i
a a
3
0 0
60
360 2 12
=
°
°
= =
m m
 (outward)
As B B
3 2
> ,
Net magnetic field at P,
        B B B = -
3 2
          = -
é
ë
ê
ù
û
ú
m
0
12
1 1 i
a b
6. AB AP , and BP  from Pythagorus trip let, 
hence Ð = ° APB 90 
\  B PB
1
0 1
1
4
2 ®
= ×
m
p
i
r
^
B AP
2
0 2
2
4
2 ®
= ×
m
p
i
r
^
  B B B = +
1
2
2
2
     =
æ
è
ç
ç
ö
ø
÷
÷
+
æ
è
ç
ç
ö
ø
÷
÷
m
p
0 1
1
2
2
2
2
2
i
r
i
r
=
´
´
æ
è
ç
ö
ø
÷
+
æ
è
ç
ö
ø
÷
-
4 10
2
3 3
7 2 2
p
p 0.05 0.12
     = ´
-
1.3 10
5
 T
     =13 mT
7. t q = NIABcos
= ´ ´ ´ ´ ´ ° 100 3 30 1.2 0.4 0. 0.8 cos
        = 9.98 N-m
Rotation will be clockwise as seen from
above.
 86
60°
2 3
1
4
X
X
B
A
i
1
i
2
P ®
B
2 ®
B
1
5.0cm
13.0cm
12.0 cm
i
O
R
3
O
i
1
2
Introductory Exercise 23.6
1. By right hand thumb rule, di rec tion of
magnetic field due to con duc tor A B C , , and 
D are as shown in fig ure.
B B B B
I
r
A B C D
= = = = ×
m
p
0
4
2
Here, I = 5 A
r
a
= = =
2 2
0.2
0.14
\ Net magnetic field at P
B B B B B
A D B C
= + + + ( ) ( )
2 2
                = ×
m
p
0
4
4 2 I
r
       =
´ ´
= ´
-
-
10 4 2 5
20 10
7
6
0.2 / 2
T
                = 20 mT
Clearly resultant magnetic field is
downward.
2. At point A
B
I
r
1
0 1
1
4
= ×
m
p
B
2
0 = (Magnetic field inside a current
carrying hollow cylinder is zero)
\  B B B
I
r
a
= + = ×
1 2
0 1
1
4
m
p
   =
´
´
-
-
-
-
10 1
1 10
10
7
3
4
 T
  =100mT (upward)
At point B
B
I
r
1
0
0
1
2
4
= × ­
m
pe
, B
I
r
2
0 2
2
4
= × ¯
m
p
Net field at B
B B B
r
I I = - = × -
2 1
0
2
2 1
4
1 m
p
( )
                =
´
´ - = ´
-
-
-
10
3 10 3
3 2 10
7
4
( ) 0.67 T
                  = 67 mT
3. Con sider the cyl in der to be made up of large
num ber of el e men tary hol low cyl in ders.
Consider one such cylinder of radius r and
thickness dr.
Current passing through this hollow
cylinder,
di jdA j rdr brdr = = ( ) 2 2
2
p p
(a) Total current inside the portion of radius r
1
,
     I di b r dr
r
1
2
0
2
1
= =
ò ò
p
          =
é
ë
ê
ê
ù
û
ú
ú
2
3
1
3
0
1
pb
r
r
       =
2
3
1
3
pbr
By ampere’s circuital law,
       B dl i
r
× =
ò
2
01
1
p
m
      B r br
1 1 0 1
3
2
2
3
´ =
æ
è
ç
ö
ø
÷
p m p
Þ              B
br
1
0 1
2
3
=
m
(b) Total current inside the cylinder
          i b r dr
R
=
ò
2
2
0
p
           =
2
3
3
pbR
         B
i
r
bR
r
2
0
2
0
3
2
4
2
3
= =
m
p
m
87 
R
r
I
2
× ×
I
1
b
B
1
B
2
q
m
®
B
®
B
A
®
B
D
®
B
B
®
B
C
A X
2m
X
B
D
C
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