CHAPTER
#
#
L
Electric
:
and Fields
Charges
Coulomb 's law of Electrostatics :( In vector form)
distance o !
consider two
+
q and q separated by
charges
where ,
F)2 force exerted on q by 92
on
F.
qz by 9s ma
⊕
Fa force exerted
'
-
=
⑨-Ñ
,
←
=
,
FI
i.
.
1<-181
kq.iq#r?-rI )
=
⇐ ñ=¥ )
¥: ::::
,
=¥÷ :÷÷
>
,k¥÷ 'ñ-ñ Heme PED
i :*
brackets
similarly
,
Force
=
applied by FI
FI
,
=kq;,9÷(É%
I
=%÷
=
FI
÷
Here
,
we
can
clearly
=
:÷÷,
,!%;→cñ-ñ
observe
that
FI
→
# ET
Ñ2Ñt
.
,
E.
I
¥ñ
=%÷f
:
-
>
Hence
=
-
FI
Peᵈ
,
i. e-
3rd law of newton is valid in
electrostatics also
.
Electric Field due
#
to
dipole
on
axis :
→
opposite charges separated by
distance of 21 ; which makes it a dipole
is a random point on axis at a distance
we have two
'
a
.
A
'
r
'
from
centre
dipole
of
Now Field due to ' -9 '
,
similarly
,
Field due to
'
at
+9
'
A
at A
⑦
-
P
l→
g- ate
teacher
.
→
-
E5-
1- +
g.
=
-1T¥,
,k÷e,
,
,
④
-
I
←
-
-
-
→
+721¥
impress
'
'
ayaan sit
IT
1
-
-
-
-
;
ÑN→
f- f-qtftq
So , Net Field
=÷÷ii÷
.
-Ñq( 8+172
=
( (rtl) (8-1)/2
21%21%-2
=
Now
,
if
8 >>
l
,
we
can
2kPr
WI
ignore 12
in the
denominator ,
E-
=2¥
E# Electric
21¥
µenie%¥
field due to dipole
on
"
A
"
equitorial line :
1-+9=4%2
q=¥→
J+E+iE
'
,
"
E-
=
.
-q
=
=
=
=
Enet
T.fr > > d
12
can
be
from ② ,
neglected
in the
denominator
e-
e→
d
Hypo TH (
-1%4%312
=
,
here
%
__
cos ⊖
lose
,
" +nisa ,
÷÷+÷÷:i
JY¥ f+•
11,2%+97,12121050=(71+10520--2105-0)
(2%+1-4152)
putting
,
it
e-e-
=¥÷i¥÷;iÑ
,
i
.
,
C-net
,
'
.
so
,
cos ②
=
¥+2
,
°
pytha)
:
②
-
Hence,
f-
denotes the direction which
direction of dipole / veto +ve)
- here
-kg¥
=
PIM
to
-
resign
is
anti-parallel
-
Hence
Torque
#
Figure
at
a
shows
Dipole
F)
=
qf
dipole
placed
angle
an
force
→
qf
=
2L
=
-
Fi
,
I
=
=
I
q& q
-
uniform electric
=
.
q
q
is equal
acting on dipole
it
will behave
Therefore
it produces torque
dipole
acting
=
-
charge
on
which means the force
direction at the two ends
As couple Ps
on
know
charges
+
with electric field
⊖
charge
on
force
→
FT
we
a
_*÷
.
with
in
_
.
makes
-
field :
in external
electric
an
separation of
field CET
FT
dipole
on
magnitude
like
,
of
(
( magnitude
either force )
✗
to
distance
opposite in
.
.
from
F)
of
action
of
sin 0=0 ; which
means
K=
line
couple
a
.
so
and
in
F ✗ ( BC)
qE
✗
(21 sin ⊖ )
( % P=qke ))
PE Sino
Pᵈ
E=FxÉ
Hence
Caste
:
when
this
②
=
0°
condition
displaced from
fassett :
%
is called stable
this orientation
when ⊖ -480°
this
,Ptam
.
:
condition is
dipole never comes
parallel to the field
.
CEIL
when -0=9-8
i.
dipole
is
configuration
.
means
aligns
itself
_É¥teaur
e✗Pᵗᵈ
af
sin 90=1 ; which means T.is maximum
PEsÉᵈ
II=Pcm→
F-
because when the
back to same
12=07
cause once displaced
equilibrium
backlothientaninsted it
sin 1800--0 ; which
called unstable
the
:
equilibrium
.
.
unstable
1$47
☒
I2&¥
paper
"
G-
Ñ
Miguel
ot ¥%
*
teacher
d- An FEI 34TH /
#
Gauss
law
verification
Coulomb's law :
using
late know, the net electric field
enclosed by the
the net
charge
through
surface
Hosed
Verification :
charge
also
,
will remain
-
( As § ds
constant
according
to
∅e
=
∅ c-
=
means
To
Now ,
times
F. dA→
§
%÷=
=
,
for spherical surface
To electric flux :
¥
=
=
intensity of electric field IÉI
q
.
According to electric flux ,
☒ § Ed? § Edsco so
c.
we know
,
closed surface B. D) Ps
a
∅e
same
distance
;gᵗ
from
,
0=00
C-
Ads
c-
fgds
area
=
at
cos 0°
=
4*82)
f 4-11-82
Coulomb's law 9
C-
-
②
=
{◦
( putting
%
in
②,
we
∅e=¥¥×
Aog
◦
Ole
=
=
get :-
*
c- "
¥9
i.
✗
( enclosed charge)
Pwᵈ
Have
field
# Electric
due to
straight long charged
a
conductor
08
Electric field due to
Consider
conductor
So
,
this
'
length
part of
a
line of density'd!
straight uniformly charged infinite*
a
l
case
will be
be the small areas
conductor is positively
of E- field will
direction
outwards
Now
/ c- DA
02 =/ EDA
∅
,
,
cylindrical
this
on
Hence Netflix
One +
÷
,
"
I
#
|
,
A
1
d
't""
1-
c-
-
!
;
,
G- 90T
=
[⊖
cos ⊖
=
90° )
[
( 0=0]
cos ⊖
ffda
cos 90
0
+
ffdA
cos
.
Law
C-
+
(211-8/1)
curved
=
1¥
[:
2¥74
P¥ʳᵈ
=
Hence
see
,
C- ✗
gin Al]
__
±
^
So
,
1
-
o
Here
A- linear charge den
.
,
12=9/1
SEDA
:
E-
clearly
da
/EDA cost
C- A
=
∅=9÷
from⑦ ,
90° -1
0
+
[ Total
Ace to Gauss
i
×
,
i.
we can
,
1
∅ -102+0}
=
=
∅
I
1
C- DA
=
Here ,
in
the
charged
be
radially
cos ⊖
=
}
,
↑
.
∅ =/
Now
uniform
.
da
As
this
on
.
Gaussian surface
Let
'
Graphically
÷
c-
Eats
>
area
]
of surface =2ñrl
# Electric Field due to
infinite
i
d←§
-
-
-
-
!et
phage
¥
+
-
±
-
-
+
'
+1
+
I
Draw
a
surfaces 'd A
Total
flux
① ②
at
,
,
∅
net
=
=
=
=
& ③
area
gauss
from
law,
② d④
Ida
¥
-
'
o
:
-
T→da
infinitely long charged
with uniform
density
→
( o)
surface
plane sheet
charge
.
of radius
'
8
'
.
Take 3
sample
.
∅ -10/2+03
,
ffd A cos 0°
/ C- DA
C- A
+
∅
+
+
/ C- DA cos 0° / C- DA
0
/ C- DA
+
cos
90°
+
C- A
l∅A→
Acc to
-
'
_
Gaussian cylinder of
'
-
density
1
•
I
of
she
②
=9g÷
0¥
oq.t /C---oy-- Pw¥ᵈ
ZEA
=
_
↳•
independent of ,
small
capacitance
CHAPTER # 2 : Electric Potential and
# Potential
at
point due
a
+a
to
point charge :
ᵗ→p*
--
there be
Let
Electric
to the
point
a
potential
point
%
P
at
P
a
Wip
→
a)
/
=
-
§
=
Wtp
→
✓
ooo
-
a
+
recharge from infinite
=
§¥
f-±
KQ
KQ
[ I 0=180
-
'
cos 180=-1
]
t 's ) ]
1- ± I]
+
=
,µueP¥ᵈ
dipole :
on
cos ⊖
dr
1¥15
¥
4¥ ¥
a)
dr
D
*•
.
Potential due to
kQ¥
KQ
-
=
point
bring
.
Fext dr
-
=
a
charge
unit
from +9
N
=
At
'
.
=
(a)
r
work done to
means
"
#
'
distance
line
axial
:
-9
+
q
④-10
a
-
-
-
-
-
-
-
-
-
-
•p
→
Consider
point
A
So
,
a
P
dipole
dies
with
on
charges
the axial
to
+1 , ✓+
due to q ,
-
potential
-
line at
Potential at P due
% Net
+94
at
Pg
V
a
separated by
distance
'
r
¥ᵈa
=
=
-
Vaxial
q
=
-_a
µ) + ( V )
.
=4¥aHH¥a )
distance of '2A !
from centre of dipole
a
'
.
k9G+aj¥aY*
=
12,91%
¥a2
=
Hence
dipole ( r
short
for
>>
Vaxial
,
a)
=
2s Vania,
proved
KI
=
82
#
p
Her
(b) At
a
point
Equatorial
on
Let there be
'
point
a
P' at
ar y
line :
in
as
Potential
diagram
at
a
distance 's
equatorial
on
:
4-
,
i.
line
'
'
.
'
"
"
'
'
'
q
'
+
↑
a-a
P due to +9
Arta
'
'
'
-
so ,
iii.
,
q
1- a- a-
!÷g
=
,
due to -9
So
Net
,
potential
V.
,
=
,ka¥ᵈ→
P,
at
Veg
(4) ( K)
+
=
=÷→+t÷
Hence , electric
due to
arbitrary point
(c) At any
let
potential
lVeg=07_µenoePnᵈ
at
Observe the
figure carefully
into two
:
If
distance 's
a
⊖ with
,
we
resolve
rectangular components
and
,
.
axis
o
'
.
on
potential
point A
lies
on
but as discussed
will be zero
.
a
due to this
equatorial
line
above as A is
,
.
^
q
Pcos ⊖
Éga
dipole moment pcos -0
component
k(pcg
"
☒
+
q
→
'
=
of dipole with
on
!
P
axial line of dipole with
at
0
,
④
-
e- a →
point A lies
so
eq dine will be
from
'
dipole
.
Then
on
"
an
dipole
point
any
"
arbitrary point
of dipole making
angle
moment(F)
as
shown
at
:
A be any
centre
dipole
eq line
.
i.
dipole
moment
potential
'
p
due to this
sin-0
'
component
Vnet
Hence ,
=kpc
kp
V
Consider two
,
Va
=
know
positive charge from
unit
dw
dw
=
-
=
dw
,
q( Va
-
Bto A :
cos 180°
Fdn
=Ee→j )
C-
i.
dW=
Fdr
=
c-
As
so
we
separated by
Vtdv
displace
to
_y
of surface
'
'
(
Also ,
and Potential :
equipotential surface
da
let the potential
done
Now , Work
field
A and B
distance of
B be VB=V
and of A be
a
O
µeneePIeᵈ
=
between Electric
# Relation
+
F
=
Édse
-
-
②
)
VB
dw-tt.tl/Vtdv-HdW--dv
④
-
from
② &
-
Édx
-
#
Potential
Initially
E-
=
=
dv
date
c-
Energy of system
there
were
charge
no
we'll
9 from
Firstly
bring
done
So
to
work
,
,
=
charge ( in
of two point
.
we'll
,
So ,
%
bring 92 from
potential
at B
Work done to
•
92
9,
•
to
to B
due to % at A
place
A-
.
A
.
to A
g
WA
=
92 at
(and
,
B,
VB
in this
=
WB
k¥
=
=
WB
=
%
VA
0
[%
case
q, is
=
Now
absence of C- f.)
at A and B
place charge 9,
,
¥×µuP¥ᵈ
-
-
Va
=
0
94k¥ )
k9jI
ie
already
②
Ez VB
,
( from②)
-
]
potential energy
charge
of static
at A)
And
as
we
know
,
sum
of
done is equal
work
Potential Energy A)
%
U
Potential Energy of
#
Let
potential
Now ,
Wa
9,
=
Work done to
WB
=
k9i9÷
respectively
an
electric field
%
.
,
,
92
was
)
not there
92 at B
place
external
,
( % inHally
②
-
in
at A
,
energy of system
k9÷µwePYᵈ
=
be Va and Vis
place q
Va
1-
system of two charges
a
A and B
at
done to
work
0
=
potential
to the
%
.
.
◦
92 VB t 1<9191
④
-
r
%
Net work done
W
,
=
=
And
as
we
Capacitance of
Consider
a
separation
+
WB
9
Va
+
,
9243+1<9%1
know , this work done is
U
#
WA
a
=
q
,
VA
equal
potential energy of system
to
proved
-192%+1<9%1
Her#
parallel plate capacitor [ without
parallel plate capacitor
of
d.
plate
area
dielectric] :
A , and
-
1¥:|
-
→
-
+
charge densityis
Let
±o be the surface
Electric field outside capacitor
Now
,
the electric
field inside
EPotential
difference
=
.
plates
the
Ig
between the
zero
plates
1-
f-
Ed
✓
Ed
=
V=
-
-
-
d
→
,
application of gauss
,
→
-
.
capacitor plate
( by
.
d- d-②
A Eo
law
of charge plate)
:
-
we know that
G-
,
c=¥n↳d
µµP¥ᵈ
Aeg
c=
#
(from② )
Capacitance of parallel plate capacitor [ with dielectric]
consider a parallel plate capacitor 4- Plate area
'
A
'
and
separation
d.
Let to be the surface charge density The gap
between the plate is filled with dielectric
dielectric constant K
substance
.
having
The electric
field
.
between
Cc-
plates
will be
ETK
¥,
=
f- Ed
"
-
( :o=-)
②
°
d
-
the
plate
/ using -0)
④
<
Now
,
F-d -1-0
:
0
=
Potential difference between
:O
;-g
+
Capacitance C'
=
,
C'
c
C
org
where
,
C
'
=
'
'
C
K
¥
=
=
=
=
=
AE
C K
,
or
C
capacitance
capacitance
'
=
KC
with dielectric
without dielectric
dielectric constant of the medium
Conclusion : After inserting
it's
capacitor
(torn
.
dielectric medium in between the plates of
capacitance
increases
by
'
K' times
initial
a
capacitance
.
Capacitance
#
Consider
Parallel :
Pn
figure
combination
parallel
in parallel
combination
-
+
-
+
-
1-
_
+
.
+
potential difference
c,
all
across
of charge
=
-9
+02
+
:
+
.
+
◦
0=9
:
CV
=
Qz
(
µ + Czv
c✗=X
,
+
(%
(4+5)
↳
The
in
Capacitors
a
algebric
shown in
,
capacitor is different but
remains same
each
charge
figure
✓
V1
=
+
Which
means
in series is :
,
effective
the
d- ±
=
+
,
Energy stored
dq
plate
%
=
+
-02
Az
+
-
+
-
-
+
+
-
-
+
-
+
-
C,
across
+
Cz
-
+ v. →
←v. →
of
-
+
(E) 0--1%+1=10
÷
±
'
¥
-
+
distribution
V2
+
let
-0 ,
.
%
#
som
'
.
potential difference
combination the
In series
→
of n' capacitors
of capacitance of
Q,
as
-
.
in series :
respectively
✓
combination
Consider two capacitors are connected ion series
combination in a circuit with capacitance
C , and Cz
Cz
←
combination Ps
capacitors
each
#
)
effective capacitance of
parallel
:
,
+
µene¥ᵈ
c=c , + Cz
-
±
a- CV
Qi
-
+
capacitors remains same but distribution
across each capacitor will be different
the
-
+
capacitors connected
two
shown in the
as
In
+Qi
'
in
capacitor
capacitance of a
E. E.
+
AND
be the small amount
to
-
ve
plate
±
=
+
-
-
-
+
+
÷
combination of n'
'
In
Expression for Energy density :
of charge transferred by the
source
.
Then work done by
capacitors
the source is
dw=vdq
[%
dw=dq
%
r=d¥ ]
g- cv ]
from
+ve
%
Total work done
by
the
is
source
W
transferring
of charge
.
=/ dw
If dq
w=
{ fqdq
± (E)
w=±¥
w
amount
=
w=
w=
w=
W
Now , the work done is in the
U
or
{(
=
1-2 C. v2
,
i. e.
:
-
lzqv
v21
µewePIʳᵈ
:
-
potential energy
density
%
Energy density
The
0¥
v2
±
U=
ENERGY DENSITY
±
form of potential energy
U=
,
=
( : q=cv]
Iz I¥
per unit volume of
a
capacitor
.
µ=¥me
,
µ
ᵗfaI
=
=¥:÷
tEA÷¥
A
±
=
=
±
E÷¥dA
Fd
is known
as
?
→
Energy
µ
If
any
=
£
Eo C-
medium is there between
14=21
£ Er
2
plates of
2
C-
a
capacitor
,
CHAPTER ☒ 3
# Obtain
current
:
Electricity
expression for Drift velocity of
an
Electrons
5-
velocity
Drift
is the velocity with which electrons in a conductor
towards the positive terminals of the potential source
drifted
We know that Pn
Initially
'
,
there
with
field the electrons
velocity ( ri )
some
ie
,
±
.
N number
are
without any electric
randomly
move
conductor
a
0
=
-
of electrons
are
.
.
in the conductor
②
,
Now, when
applied
an
on
electric field is applied across the conductor ;
a electron by the electric field is :
ma
a
where
a=
,
m
take
we
'
E'
to
-
0
=
Vd
=
motion
a
A
'
and
a
conductor of
n' be the
no
.
E
-
%
F- QE]
:
f : F- ma)
¥
-
+ve
terminal
-
length
femi ) (2)
eEIm
-
where Vd
drift velocity
l
=
and
=
drift
velocity
area
-
:
-
of cross-section
unit volume
Current in the conductor ,
E-
nA¥
Hug
Iced
of e⊖ present per
=
time ( the time interval
,
N=nAl
Total
charge , Q=n Ale
%
°
+
between current and
'
e
-
[
(4) avg
Hence
'
=
E
collision)
successive
Vd
consider
=
e
average relaxation
be the
by first equation of
Relation
-
acceleration of e⊖ towards
mass of the electron
H¥gCig
#
=
.
between any two
then
force
-
F
It
The
IAk ⇐ he ¥ )
.
E
combination of Resistance :
# Series
resistors of resistance Ri
Two
Pn
As
series
and Ra
are
connected
voltage
%
V= V,
R= R ,
#
'
'
n
no
.
Parallel
of
,
orients
v
I Rz
+
✗ ( Rit
=
,
law ,
IR ,
=
☒
comp
V2-1
1-
V2
+
using IR0hm's
for
↳ -1 _
1- V, -1
•
.
we know, in series combination , current Ps
is different across the
but
same
To
→Ñm-nnÉ
+
R2)
Rz
resistance
Peᵈ
Hence
in
R
series,
=
R , + Rat R}
t
-
-
-
combination of Resistance :
Two
resistors R and Rz
with a battery of
are
,
voltage
'
ⁿmʰ→¥
connected in parallel
V'
.
-mÉm_¥f
1-11-1
1- ✓
2
^
I
0hm 's
¥
law
n'
=
Consider
+
¥ ¥
E.)
+
,
resistors in parallel ,
= Relation
✓
Hence
¥ ¥
=
P¥ᵈ
¥
+
,
+
+
between Internal resistance , terminal
-
-
cell
of emf
'
E
'
.
=
¥
1-
-
,
Terminal
potential
difference
V= IR
-
⑤
-
-
-
+
En
potential diff and
with internal resistance 's
connected to the external resistance ( R) The current
in the circuit Ps :
: 5[
②
a
-
+
,
¥
for
)
,
¥ ¥2
=
1--41 E.
%
,
11+12
=
using
→
.
.
I,
.
→
As we know in parallel combination voltage remains
,
and current is different across the
same
components of the circuit
'
Rn
-
-
¥¥Éaⁿd
EMF :
-
.
'
-
⑥
-
jE_ jj
i.
-
-
-
-
-
-
-
if
my→
Now
,
②
be written
can
as
I(R+r )
→
IR
V
✓
too
#
series
Cells in
consider
Is
+
=
Ir
-1
(
E
E
=
=
/ For
µ=E+2
V > C-
emf Gandfz
internal resistance
8, and rz
connected in
series
and
multiply)
v5 E)
which is the relation b/w Errands
,
:
two cells with
-
( from ⊕ )
E
f- 28
=
cross
having
respectively
.
÷ii÷÷÷÷
ii.
mnn→,
:
-
-
[
.
-
-
-
-
-
-
-
-
-
'
n
R
.
V1
also
V2
,
C- ,
=
Ez
=
( for
28 ,
-
-
✓<
e)
I 82
We know , in series current is same but
%
V , + V2
Veg
( E , In) + (Fa Ird
Veg
=
=
Now
Veg
Veq
,
=
=
know
we
,
,
(Gtf) (8,1-82) I
Veg Eeg Ing
-
-
=
②
-
-
⑤
② &④ ,
E,
=
req
Parallel
+
Eat
8, -182-1
=
-
-
-
-
-
-
-
-
proved
- -
flank
÷
:
consider
two cells of emf G and ta with internal
connected in
resistance o, and ra
,
respectively
parallel
-
diff
is same
across
combination
%
current will
but
components
potential
be
-
-
-
,_ʰ?f.
'
-
-
parallel
-
,
-
'
we know, in
diff
-
feg
Cells in
is
Eat ( Ir -1182)
+
Comparing
#
components
across
-
-
(E
potential
_
different
-
-
-
-
-
E-miii-i.fr
-
-
:
-
man
-
.
I
I
=
=
2-
I,
+
Iz
(
4¥ 4¥
G- ¥ -4£ -1¥)
+
%
)=ᵗ%¥
E- Ir
I=E¥
+
v1 "¥
V=
-2
)
.
✓
Comparing
"%¥÷ -47%-1
=
Eeg
and
We
can
also write these
Eeg Ing
-
=
Er¥¥
=
he
,
✓
with
this
simple ways
in
equations
¥g=¥+§÷+
and
Teg
,
¥
=
+
,
+
µenuPÉᵈ
2%-2
=
¥
-
-
-
-
Bridge :
is an
arrangement * to
bridge
used to determine
resistance of
Wheatstone
Wheatstone
resistance
one
For
Now
"
resistors in terms of
a
,
balanced
applying
( as
Kirchoff
rule
-
I, P
,
applying
-
=
Pn
Kirchoff
-
rule
s
Ill -225--0
I ,Q
dividingeg②&④i
=
we
Ias
get
-
:
resistors
.
-2
loop
ADBA .÷
G%
tiny
R
figure)
on
1,10=0
IZR
three
_A€Éᵈʰ%¥
^
,
VB=V☐
Iz R
Now
bridge
other
-
-
-
-
,
-1
I
-11-1 )
-
②
on
loop
BCDB :
-
④
⊕②_ {¥=%÷
↳
G-
=
%
Proud
part
his is the condition
for balanced
wheatstone
bridge
.
#
Finding
Principle of
Pinup
using
⑧
unknown resistance
:
meter
As shown in
figure
¥¥÷÷I÷¥i
bridge
wheat stone
,
R= Unknown
bridge :
unknown resistance
finding
and
bridge
slide wire
resistance
-11-1
I
s
resistance
known
.
=
Move the
reading
As the
Jockey
of
the
bridge
G)
on
wire AC
galvanometer)
is balanced
¥a☐
,
.
length
Let
point
,
☐ be null
wheat sone
¥00 e)
-
=
point
on
point lire
wire AC
bridge principle
.
zero
.
:
-
Hms
←
¥13
=
R
I to obtain the null
therefore by
=
Fe
of
¥
proved
¥eµ•=
⑥
}tÉÉ←
RAD
•
RDB
CHAPTER # 4
#
Magnetic
field
consider
current I
We have to
.
Consider
a
small
loop
this
Applying
find magnetic
Field
Magnetic
1dB
-
-
-
B
=
B
=
14¥ ¥
=
MET
of
a
-
✗
(
211-8
"
:
fdl
means
total
circumference
loop :
current
-
!dñ°ˢ° dB→
-
>
I
^
dB→sPn
⊖
t.rs#si-no-----
I
-
_
_
-ÉÉÉ
distort
sdB→
,
circular loop of radius
have to
According
°
µencePᵈ
circular
r
-
:
'
a' the axis
the circular
to the circular
of
calculate the magnetic field due
the distance between the loop and the point
we
→↑d2
is 90?
o
ME ¥ fall
-
Iq
a
-
>
-
I
Consider
and
g
-
2
get
we
,
"
of
-
-
circumference
get
we
-
1M¥ Idgaf
=
the axis
on
↑
-
d
-
-
1%+1%4
=
B
-
between
y
both sides
integrating
-
on
←
dB=M÷( 1¥90 )
dB
#
I"
current element d
savant 's law ,
Biot
loop
.
Clearly angle
.
magnetism
and
carrying current :
carrying loop carrying
field at the
current
of this loop
centre
of
at the centre of a circular
circular
a
Moving charges
:
to
Biot savants law ,
dB=¥¥Iᵈ¥irE
'
P'
.
loop
loop
at which
and ✗ is
So the
,
magnetic field
at P due to current element
dB=%ˢ Idl{In9I
dB
ME ,É+→
[:
°
r
Idf
:
-1dL ]
=
)
Magnetic field
at P due to current element
dB '
dB
we can see
Resolving
→
Here
=
'
=
14¥ Idlgi29I
ME {aᵈ÷×z )
dB =D B
,
'
components we find that cos ⊖ component for
opposite elements cancel each other
dB in two
diametrically
So that
,
therefore
,
total
B-
=
B-
sin ⊖
47%4×5*+7 fall
=
¥Y%¥×→a≠×
%a?pk
=
then
⑤
if
a
=
=
Ps
Mz÷s
's
31k
#a
%¥
Hence
neg legible
Mz¥ñ)
only
to the whole coil
µ%a?→fᵈl
=
⑤
at P will be
11¥ IdlaÉ+n-
=
B-
,
5dB
=
☒
n> >> a
field due
magnetic
B→
it
two
.
magnetic field intensity
,
Idt
.
due to sin 0
component
Ampere 's
#
circuital Law :
It states that the line integral of
is µo times the total current
field
magnetic
the
threading
↳
Proof :
i.
intensity
loop
over
of
straight
conductor
Consider a
as shown in the
carrying
Consider a circular American loop of radius r
around the conductor
-
figure
-
loop
.
§ B- .de?--1UoI
e.
closed
a
!
.
"ÉqÑ↑B
"
'
-
-
-
-
-
-
.
As B- and
DÑ
in
are
same
/ B- diBdl
%
direction
so
angle
fdl
means
between the miso
•
/
=
=
/Bdl
cos 0°
fBdl
Bfdl
=
(
=M¥j#
Not
=
peen
:
circumference
Application of Ampere's circuital
Magnetic field
we've
given
current I
Now
But
,
.
due to
an
=2#r)
Pw¥ᵈ
MF due to •
straight current
conductor
carrying
[ solenoid
.
#
^
.
law
long
→
[
Toroid
infinitely long straight
current
carrying conductor
of a cross-sectional radius a' carrying steady
straight
along
This
current is uniformly distributed across this cross-section
'
wire
.
have to calculate
field at a distance
magnetic
here we 'll have 3 cases :
④ r > a ; i. e. point lies outside wire
E) r a ; ie point lies on the wire
ie
r< a
point lies inside the wire
;
we
o
from
-
=
.
.
CA:
-
8 >a
at
point Pi
.
Now , to find the magnetic field at point P,
make a circular loop
outside the wire
'
made of radius o as shown in figure
.
-
'
.
-
-
-
•
Pl
centre
.
:
Using
Ampere's law ,
ftp.di
µ I
=
§ Bdl
B Jodl
.
cos 0°
B
=
(2*8)
=
=
No I
:
Now to find
,
Make
a
%
cased :
-
at
r=a
-
the
To find the
of
Now in this
,
than the
is ,
at
G-
②st
we'll
→
radius
surface of
make
case the
value
.
a
circular
(rea)
8
r
is the distance
of point from]
centre
at point B.
on the
surface
of the wire
.
B- MI
→
-
21T A
at
loop
-
-
-
§
-
enclosed current Ie is not I
Since
Ie
-
but less
the current distribution is uniform the current enclosed
,
=
I✗
2
at
using Ampere's
.
Pz
point
the
wire
( where
circumference -211-8)
a)
get
magnetic field intensity
the
point B inside
cylindrical
made
loop of
like
means
Pz
point
similarly
,
o>
②
intensity
magnetic field
radius
a)
circular
Ra
°
-
( for
Bttg
( : Jodl
MOI
B=Y¥-g
Cased
Not
law
§ B. DI Mo Ie
§ Bdl =M°Ia¥
=
,
§ de
B
B
=
(2*4)
1321T
Mo
=
=
Ia¥
MoIa¥
Nigg
B=
µz¥%
↳
Bar
#
Field due to solenoid :
Magnetic
Mmm
←
@£%①①☆
MM
⊕⊕É⊕⊕⊕⊕ק
"
2
"
g-I →
No
.
Of
turns
=
<
N
<
D
⑥
<
C
③ ③ ① ③ ⑨ ③ ①③ ①
:
A
-13
I
④ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ⊕ ?⃝
let
solenoid consists of
of turns per unit length and carry current
inside
the solenoid is uniform and strong
Magnetic field
M F
outside the solenoid is weak Ialmost zero)
a
'
n'
no
.
.
.
.
Consider
a
loop
close
ABCD
.
§ Bode fB→•de→ JEDI fB?dT
%
+
=
BC
AB
B. all
Here
,
+
[B
0
=
CD
+
fB?dT
DA
outside -0]
dt-afB.de?0fiB.tdeT
B-
Hence
§ # DI
,
=
§ B- di +0+0+0
-
AB
§ B.DI / Edt
=
cos 0°
AB
§ B- di
to
Here ,
Ampere's
§
Jdt
AB
§ Bill
According
B-
=
BTL )
=
law
B. all
-
:
=
No 7-
N number of turns
To
B. dl
MONI
BCL )
MON I
§
②
,
are
present
=
=
B=
1401¥
-
from ②
I.
⑦
where ,n=
no
of
.
=
Mon I
length
turns per unit
B=M
4¥
"
B= Mon
i.e.
NI
n=
,
I
i
i.
>
corner
centre
#
Using
Case)
Ampere 's circuital
law , obtained
magnetic
the
a
-
Inside
=
-
-
_
,
,
??;
,
from Ampere 's
law
:
-
§BdÑ=
here
toroid
Outside / Between) :
inside
field
Iin
,
[at Pi
Motion
)
1
,
,
0
=
'
.
t
•
'
:
'
'
§ B- di
Casey )
0
=
'
13--0
"
,
-4 if
'
Between the turns :
§ Edt
-
§ Bdl
B
=
cos
fall
B (211-8)
fat
No 1in
②
=
=
=
B-
-
Case
_
Mo
B
-
-
-
from Ampere 's law :
*
_
B)
Lin
MONI
MONI
M◦¥÷
or
[ ñn=N_
B- Mon I
-
=
Egg ]
Outside :-( at B)
§ Bdl
=
Nitin
13--0
# Force
acting on
Consider
carrying
a
⊖
angle
conductor
current
a
conductor of
lengthinland
placed
current I
shown
Ps n then
,
as
carrying
.
It
a
no
.
placed in
of section A
magnetic field at an
area
Mf
density
:
-
a
(
of electrons in the
of electrons in the conductor Ps
number
total
conductor
BI
←
:
l
-
Aln
.
¥↑
As the force
electrons
acting
f-=eVdBsPn⊖
electron is
one
velocity
drift
where Vd is the
of
.
acting
the total force
So
on
the
conductor
is
Alnf
=
Alnlevd BsPn⊖)
=
=fAneVd)lBsPn⊖
f-
=
µnaPN¥
IIB sin ⊖
↳ direction
left
#
Force between two
Consider
two
infinite
same direction
They
Since
each
.
long straight
conductors
conductors
parallel
each other
to
at
field is produced due to
magnetic
conductor experiences
force
a
and , the
.
carrying
( ✗ and Y )
^
current
:
currents I, and Iz in
carrying
force
distance
r!
'
each conductor
through
current
,
←
.
-
-
-
-
Bi
-
-
-
,
magnetic
B,
field
=
at P due to
-
,
current I,
②
M¥j
-
←
a:
-
,
,
"
'
-
-
-
-
force
>_
-
-
-
-
Iz (d) Bi
=
Fz
Bi Iz
=
Fz
=
sin 90°
✗
(%
d
( Td =L
→
-
-
^4
^
=
✗
→
<
→
y
/ unit length))
M°{¥
MOTI
B2
③
④
-
21-18
conductor ✗ will also experience
F, due to 12 current
-
←
-
-
-
_
,
a
force
.
,
:
-
a $
-
Fi
"
(
-
,
-
i
F, -=Bz 1
F,
=
=
,
sin ⊖
Bz I , sin 90°
MoIiI_
211-8
11--1 (unit length I]
,
'
-
-
-
-
-
-
-
F,
-1-2
sin 90=1)
due to current Iz at point Q
Bz
Similarly
'
-
-
F,
Magnetic field
-
-
conductor Y dies in the
magnetic
carrying
will experience
the unit
Y
of
length
given by
-
-
-
ftp.p
-
-
As the current
field Bi , therefore
a
therefore
.
Il BSPNQ
will be
a
,
Now
the
.
held
are
parallel straight
Fleming's
can be determined by
hand rule
-
NI,
✗ <→
→
a
→
Y
_
12
We
can
Hence
observe that F, acts perpendicular to ✗ and
✗ and Y attract each other
F, Fz
So
directed towards Y
F,
=
,
⑦ Bi
⑨ Bz
but
when current next be in
opposite directions ,
the conductors will repel each other and
derived
will be
same as
above
?⃝
magnitude
Hence
↳ opposite
acting
# TORQUE
direction
current
on
current dioxin
a
→
attraction
→
repulsion
n
carrying loop/coil
current
in
uniform
I,
-1
MF
.
/
a→
¥¥¥¥!¥÷¥µ
⇐
.
fz
④ B,
⑦ Bz
Same
.
f- FEY
.
I2
rectangular) :
¥
⇐
→ i-
current
is placed in
uniform magnetic
carrying coil does
rectangular
it
not experience
It
then
experiences
When
a
a
field
a
force
Magnetic
%
on
a
torque
current
f-
but
field exerts no force
is antiparallel to I
B
Kow
,
=
on
a
.
carrying
II.b- Sino
the two
conductor
-
on
.
arms
AD and BC
arms
ABI CD
of
loop
only
because
.
The
magnetic
field
is
perpendicular
to the
AB of the loop and exerts a force fi
which is directed into the plane of the
F,
=
IIB sin 90°
on
-
c- a
→
↑F2
A¥¥
arm
it ,
loop
.
↓f
.
,
↳ (front
)
view
IIB
=
Similarly
the
,
plane
the
of the loop
Fz
Thus , the
'
field
magnetic
=
net
Il B
on
exerts
a
force Fz
on
arm CD
.
=
the
F,
loop
is
zeoo
(as
said earlier)
,
which is
directed
out
of
But
as we
,
there will
can see
,↑E=BIl
a
torque
along
magnetic
between the
angle
be
Let the
↓f,=BIl
-
-
-
-
-
-
-
¥
-
Coil
.
angle
_
Theforce
arms
on
F,
=
⊖
M
Galvanometer
Galvanometer
field
of forces F, and Fz
of the
an
loop B not
angle with it
,
.
and the normal to the
AB and CD are F, and fz
③
Fz I Bl
=
-
( Here
KIA
=
A
Conversion of
pair
.
MAIA
#
due to the
Now , consider the case when the plane
the
field and makes
£
c-
loop
the
on
→
1<=1
for N
no
.
Ammeter
into
be converted into ammeter
s(shunt) in
a small Resistance
with the
)
of turns
:
,
Ñ=NIA→
hmˢnÉᵗ
/
stz.gg#Ig---max
can
connecting
parallel
I
As
galvanometer
through galvanometer
current
ammeter range
=
Rj
Galvanometer
=
S and
by
G
are
Resistance
connected in parallel ,
SCI Ig)
-
=
Ig Rg
s=%É
#
Conversion of Galvanometer into voltmeter
Galvanometer
can
connecting high
Ig
R
✓
=
=
=
Rg
=
be converted into voltmeter
resistance in series
,
ace
.
.
through
total resistance
Now
by
current
galvanometer
resistance
high
External potential
Galvanometer
=
resistance
Rt
to 0hm 's law ,
Rg
V=
Ig
Ig ( Rgt R)
Rt
Rg
=
R
-
Ey
-
±
Rg
④nE±
CHAPTER # 6
Notional EMF
#
Induced EMF
or
:
s
consider
a
rectangular conducting
in the plane of
conductor PQ is
to more
loop
×
×
let the rod POPS moved towards
with a constant
velocity V' assume
right
'
no
loss
,
of energy
R
×
×
⑤
×
✗
✗
✗
✗
.
=
magnetic
×
×
×
,
the
A
×
←
×
/
✗
×
×
✗
m
×
/
×
-
✗
✗
✗
✗
µ
✗
×
loop
the
×
Q
enclosed
area
✗
_
c-
linked with
✗
×e ✓
×
×
×
→
right
'
.
×
friction
due to
Pa is moved n' distance towards
Area (A) .lu
PQRS increases
flux
Therefore , the amount of
is
in
induced
the loop
An emf
Let
×
×
✗
☒
✗
±
.
✗
there is
%
4×74
PQRS
the
paper in which the
free
Induction
Electromagnetic
:
by loop
increases
.
.
then through
BTÑ
/ A)
area
∅
=
∅
=
∅
%
Induced
EMF
coil is
in the
E.
=
-
E-
force
on
the wire
-
COSO
Bla
②
-
→
/ from ⑦]
Blk
=
E.
Blk
-9ft
=
E.
=
cos ⊖
BA
Bl
@d¥)
Bev
P¥ᵈ
Hence
[
:
day
rate
means
BIL sin 90°
=
-
f-
f
131B¥) e
BIRI
F=
?
I
=
Egg
Bff ]
=
✗
Induced EMF due to rotation of Rodin
Consider
uniform
a
'
)
( external)
f-
#
of change of displacement
velocity
which is
metallic rod of
magnetic field
Area covered by the rod
=
,
✗
,
rotating
by
21T
angle
±
×
-
-
-
.
i. i
×
i
'
'd Ps placed Pna
length
as shown in the
figure
on
Hd2
Magnetic
field :-,
×
y
,
×
×
×
×
-
'
,
×
;
'
×
×
×
,
"
'
✗
-
Y
.
.
_
.
É
"
×
✗
for
%
%
I unit rotation ( area)
for
angle
⊖
rotation
To Area will be
flux
Now ,
,
A
through
=
1¥
=
1¥
=
∅
,
1 simple
unitary method
applied )
②
-
A
area
J
¥ ¥
=
B- A•
=
∅
BA cos 0°
=
∅ -13K¥ )
Induced EMF in the rod
E.
,
=
-
e-
-
_
E
BIG dd¥
Btw
=
E.
dd¥
¥1B %-)
=
where
,
w
angular velocity ( d
=
Hence PIED
#
Self
-
Induction
Consider
a
of Solenoid
turns with length l and cross-section area A
having
it So there will be
magnetic field at
flowing
through
B
it
the solenoid
solenoid
I is the current
point in
'
Mo¥
magnetic flue
∅
:O
we
.
also know
from ② &④
,
=
∅
given
MONI
LI
Lf
=
↳
[ This
by
to
a
.
product of
B and area of each
turn
by product of fun present
the
no
N
✗
M_◦N÷
=
equal
a
A
✗
will be
=
,
represent
per turn will be
∅
,
'
,
=
And
N
.
given
Now,The magnetic flush
% Total
:
-
②
④
-
Monet
M¥ µµeP¥ᵈ
is
self inductance of
-
a
solenoid
.
in each turn
.
of
turns
.
and
.
# Mutual inductance
of
solenoids :
two
-
s
,→Niwmˢ
llltllllllllldmd.TN
solenoids stands≥ each of
consider two
long
length
and Nz are the no of turns in the solenoid
,
.
respectively
52
s , and
.
'
sz
52 Ps wound
are considered
section A'
over S ,
I , is the current
flowing through
closely
to have the
,
so
↳
both the solenoids
of
same area
'
N2 tums
-
cross
l
-
.
Now ,
I,
the
Ps
magnetic field
B,
produced
B,
MoNe
=
Si
.
at any
-
point inside solenoids
,
⑦
magnetic flux linked with each turn of sa ie equal to B
magnetic flux linked with solenoid sa having Na turns is
And , the
Total
∅
∅z=(µoN¥)
∅
=
,
but
∅z
from ④ d④
MI ,
%
And , if the core is
M
=
filled
M
=
AN≥
( WWII) A
④
MI ,
=
,
A
.
B , A N2
=
,
i.
due to current
-
=(M°%N
MoNyN#
with
a
,
-
④
where m is the coefficient
between S, and S2
of mutual induction
A
µmeeP¥ᵈ
magnetic
MNiN2A_
/ from ②)
material
of
permeability
M
CHAPTER
# AC
voltage applied
to a
#7
Alternating
:
Resistance
Current
:
Alternating
%
EMF
current
Eosin wt
→
through
I
the
§
=
WAVE
FORM
✗
⑦
C-
5-
④
Tosin wt
tlencepnred
-
=
,
DIAGRAM
phase difference
:
fosinwt
Eo
-
.
-
-
Iosinwt
-
-
Alternating
-
-
-
-
-
EMF
i. e.
;
-
-
-
i
-
*
Io
,
I
,
!
i
L
to an Inductor :
Eosin wt
→
e=
-
the
-
ummmm
'
⑦
F.
c-
C-
=
=
=
µ→ -1
-
£
inductor due to the current
I
=
↳ spnwt
.
Ldtˢd
Lenz law, the induced emf
say ,
→
.
An EMF will induce in
can
-
↑wᵗ
'
'
we
-
fi
Consider an inductor of inductance L
connected in series with a circuit containing
to
and EMF
→
voltage applied
According
between current
^
Phaser diagram
AC
no
^
-
É
✓
#
Eosin wt
=
EosP¥wt
say that there is
we can
-
circuit,
I
comparing ② &④
-
mm
1-② -1
_
Consider a resistor tester of resistance R is
connected in series with a circuit containing
-
e
-
L
dI=
dI=
(-1%-1)
¥t
Edt
Eosinwtdt
will oppose the
alternating
EMF
.
for
current
total
both side
integrating
,
C- sinwtdt
fd 2=1 I
¥ tioswwt)
◦
EI
/
coswt
=
,
(
when sin wt
④
on
I
comparing
i. e.
=
I
=
I
=
E)
-
-
will be
-
Io sin
② &④
¥ ( sing wt
¥ sin / wt TE)
-
,
(
wt
we
Ig
the
E)
peak
phase difference
=
-
value
sin ⊖
I◦=E?⃝
i. e.
.
-
that I and E have
see
to )
sin
④
I will be
tlencethored
-
-
]
SIME -0-1=0so
%
different phase
between I and E.
∅ aft aft ¥
=
+
-
∅ ¥
=
.
:
voltage
leads current
.
÷
"
;
waveform diagram for
1- and E
¥
i
◦
'
¥
-
-
-
-
-
-
-
-
-
-
-
-
-
_qfo
i
Phasor
diagram
for
I and C-
§
'
i
s
;
÷
'
↑
¥
:
iii.
±
wt
;
d-
-
tot)
fwt
"
-
-
-
-
-
-
-
'
¥
E)
-
-
-
-
# AC
voltage applied
Consider
in
a
series
to
→ 1-
capacitor :
-
a
capacitor of capacitance C' is connected
contain AC of EMF of Eosinwt
g
-
The maximum voltage of the capacitor will
,
charge
Instantaneous
on
capacitor q=cv
,
g- Ceos
I
I
=
=
I
-
I
=
I
Now,
I will be
Max
To
So
④
,
I
=
◦
:
is
Cfo wcoswt
=
CEO
W
(sing
(wt E)
+
(sin
+
leading
+
)
a
behind
and E
Z and E
-
⑤
I.
_
phase difference between
the
,
∅
∅
diagram for
)
wt
will become
wt
PID
that there is
phase difference between I
Waveform
E)
,
Hence
we see
.
CEO
Eow
To sin
the AC
c.
I◦
c
of
day ( Eosinwt)
ddtlsinwt)
when
=
v=
EMF
to
dd¥
/ peak)
② and 1⑤
Comparing
due to which current
,
=
equal
be
f:
in wt
in the circuit
current
E- Eosin wt
②
E- Eosin wot
Also
V=E
'
voltage
=
=
aft
.
+
E
¥
÷:
I
wt
-
2
and E
%
phaser
diagram
for
,
%
7- and E
:
☒
-
-
-
-
-
;
wt
!
Impedance
Consider
a
in series
LCR circuit connected to
voltage drop
across
vi. 1%
4--1
.
×
,
"
!
-1¥
'
I
AC
an
R
Mmm
Ya
-1
T
in series
source
②
resistance , capacitor and
} [ E.
②
✗a.
=
e-vi.
WL
±
)
"
ka
-
-
I
-
'
Phaser
diagram
Consider ,
for t.GR circuit
Eo Ps the total
circuit
In the above
E.
:-,
%
Now ,
Kc
voltage
=
diagram
( Vik )
across
,
-
all the
let K > Vc
"
⑤
,
!
[
.
components
-
%i
¥
voltage supplied in the
.
phasor
k=c
t
/_mmm_t
-
→
L
""
-
Ps
-1
1-
E
inductor
VR= I R
-
-
-
,
-
LCR circuit :
.
Here
-
M↑wt
'
#
-
→
Vc
V=J2t
V=t(2X+IR
V=tIÉ+R}
✓
=IfÉ+R
F. É
=
=/ (✗É
2
is
called
impedance
pfoeePn¥
Resonating frequency
#
Resonance
reactance
when inductive
occurs
circuit
in series LCR
:
-
reactance becomes
¥
=
WZ
w
=
¥
=
211-0
¥
=
V
=
↳
Proved
2T¥ Hertel
± V , ✗<
to Xc and resonance will
and the frequency is known as
will become
occur
,
equal
resonating frequency
Power in
current in the
power
p
=
by
the
Now,
Eosin wt
EoI÷ /
=
average power over a
of the above equation
R H S
But we can see that
-
-
be
.
zero
%
( % positive
Cos
=
cycle
cos
is
✗
∅
Érms Isms
&
=
=
tan
a
"
/ ×c-¥- )
% sin /wt ∅)
-
cos
-
(2 wt
given by
+
)
∅)
the
-
②
average of
the two terms in
.
time
the
dependent % Its average
negative second half)
∅
P=k÷) ( E.) '
P
%
where
;
∅
the second term is
only
half of the cosine cancels
Eo÷
P=
RLC circuit drives
the source is :
C- I
P
series
a
a
-
? instantaneous
.
LCR Circuit :
voltage E- Eosin wt applied to
circuit given
by is
pospnlwt ∅ )
we know that
◦
capacitive
Xc
=
WL
Average
to
.
✗[
#
equal
I :&
∅
•
cos
Proved
Hence
∅
.
-
-
¥ ¥)
✗
.
will
#
Energy
consider
Voltage
As
we
stored in
an
Inductor
an
:
-
inductor of inductance L connected to
source E
know ,
shown in
as
figure
P= EI
P
[
Liddy
=
% C-
Mmm
a
.
=dd¥)
¥ 12¥ 1° P=dd÷)
:
=
dw
=
LI DI
②
-
Io
Integrating
both sides
(
fdw / LIDI
W=
LJIDI
=
,
0
Io
Io
=
max current
in the
circuit
w=% !
w=L
1¥ ]
°
-
W=
121102
[ This
To
-
work is
0=1-21202
stored in the circuit
µµeP¥ᵈ
as
magnetic potential
energy
.
CHAPTER
Relation between critical
#
consider
to
rarer
a
(1)
#9
refractive
and
angle
Ray Optics
:
index of
medium :
a
medium (M)
light ray travelling from denser
According
µ since
:
µ spnc
µ
(1)
=
fit
sin 90°
,
I
>
1-
=
90°
'
I
=
m -1
;
.
to Snell 's law
air
i
Ési=c
=
Ill
I
proved
since
µ
Hank
#
Retraction
figure shows
surface
at
spherical surface :
a
refraction
by convex refracting
M'
.
-
-
made by
and 8 be the
incident ray , normal and refracted
ray with the principle axis
let
'
angle
xp
g-
-
¥
-
-
-
-
I¥_É
;
me
-8-1
±
-
#
-
-
-
-
✓
.
The normal
drawn
from
centre of curvature (c)
direction of incident
,
ray
In AOMC ,
Now ,
Now
.
surface
refracting
are measured
the convex
All distances
is taken + re
0-1--4+13
73=-02
0-2=73
Bysnetsaw :
as ⊖ , &
µ, sin ⊖ ,
0-2
are
µ,
0-1
µ , / ✗ + B)
Here ,
-
:
Mz Sin 0-2
%
very small ,
=
=
tan
i.
②
Mi
=
=
µ -02
Mz / B- 8)
h_
y
-
th E)
sin -0 , ≈ ⊖ ,
-
②
very small
≈
✗
U
¥
8-
+
-
2
-
tan B
Y
+
=
9,1348 are
tant
,
.
.
In ACMI ,
%
passes through the
from pole and the
¥
=
≈
≈
B
V
Molk ¥)
-
and sin 0-2
≈
-02
-
-
µ
,
/ ¥ 1)
-
¥
¥
-
¥
→
-
Mfs
=
¥
-
Malta E)
=
¥
-
MTµµP¥ᵈ
=
A
# Lens Maker Formula
:
Consider a convex lens ( thick) , let an object
Ps placed on the principle axis at O'
'
.
formed
image
I.
The
Ps at
convex thick
by the
im
I, l
first surface
through
A-
.
i.
¥
,
←ñÉ
ni
E)
%
na
further ,
}
If surface ADC is not present then
be
°
• .
formed
at I ,
According
to
as
shown in
image
the
figure
m_#,=%-
¥
-
-
ABC is not
figure
According
.
refraction formula
to
n¥= -7
② d④
:
-
-
¥
-
na-r÷+- MI
④
=
nz-r.rs
+
-
m-rn÷
=
¥
-
A
%
-
""
-
-
-
,
•
☐
{
c
¥ %
-
,
R2
n¥ˢ
/
-
-
±
,
--=vi
,
-
.
②
-
as
will
refraction formula :
,
Adding
{
,
(ABC) :
then
present
behave like object and the
image I will
second surface will be formed at
by
image
shown in
I
,
{
☐
i
refraction through second surface ( ADC) :
If the surface
Now
ri
'
'
B
Iz
>
i
c
"
8
I
>
g
Ne
dens
1) Refraction
'
i
Ni
¥
cnzz.nl#--ni( ± ±)
-
+
F-
n¥
in ni
-
1¥
-
nil ¥ )
=
,
( tr E.)
n
-
hi
¥-11k
f#
E) ±
=
-
,
In 1)
-
=
(¥
µuP¥ᵈ
-
,
A
Prism :
through
Refraction
¥
=
,
consider a triangular prism , let a ray of
PQ strikes on the face AB of the \
refracted
.
and then refracted by the
of the
t.a.ee AB towards the base
prism BC and again OR is
the false AC away from the normal
prism
by
light
,
¥
if
8,
=
≥
A
=
f-
angle
angle
In AQNR
In
of
angle
and
AB
Lr,
quadrilateral
+
by face
respectively
LK
AQNR
+
,
LQNR
+
+
Lk
+
Also
,
A
8=81+82
f- (T a) + ( e
-
8
=
(ite)
-
-82)
Coin)
=
=
360°
360° -180°
=
=
180°
LA
=
+
-
L∅NR
LA
=
8
③
LQNR +90°
+
LOHR
+
-
LON R
Lr , -1282
or
180°
LQNR
LA
Lr ,
=
LA +90°
LA
from ② I
s
C
of prism
of deviation
,
"
s
B
refraction
AC
É
IF
incidence
of
-
-
P
andAngle
r
,
,
.
>
F-
•↑É
,
"
"
,
+
82
-
④
⑤
f- ( ite)
Pte
or
when
,
=
-
( from eg④)
A
8+A
f- 8min
,
-
⑤
F- e
then
81--82--8
I.
becomes
eg
,
rtr
28
=
=
A
A
a-
And ,
it i
becomes
eg
-
2?
i
Now
,
According
µ
/sm¥ )
1¥
s# =
sin
-
=
8m
+
A
8m-¥
-
④
to Snell 's law,
gn÷=
µ.
8Mt A
=
=
④
(
where
his retractive
material
present
( from ④ & v10)
;¥;¥t
si
proved
Her
't
)
index of the
in the
prism
CHAPTER # 10
# Position and width of the
distance between any
The
fringe
wave
:
interference :
in
consecutive
two
width of
Optics
bright
dark
fringe
and the distance between any two consecutive
fringe
equal
is
to the
a
bright fringed↑
Consider light from two slit stand { superimposed
dark
at point P on the screen
fringes 1
bright and
dark
fringe
equal to the
is
width of
•
si
-
-
-
-
-
-
-
-
-
P
A
a
◦
↑
y
!
.
"
the distance between two slits stands,
and D be the distance between slit and screen
let d be
S2
-
-
-
-
-
-
-
☐
-
-
-
B
→
.
Now at point P
path difference of
the
,
DR
In DSIAP
Pytha
:
9s , p 2=5
.④
er
,
Szp 2-
⇐P
S,
P
P
Assuming
2+
A
+
( Sap
S , P)
-
very
^
-
.
S, P
-
APZ
( y ¥)
-
-
②
In
2-④
+
D2 +
-
Sip)
close to 0
↳
=
+
y
✗
( y ¥12
-
dz
2yd
=
such that S , P
,
④ D) DX
+
=
2yd
XD DX -12yd
4¥
on =
☐
x=nd
Y;:¥#
-
when
gn=0
n=l
i
n
=
n
,
y
y
g
g
=
Y
¥
=
.
Central bright
0
=
D£BBszp2= 5,132
=
/ Ytdz )2
( SP SP)
Casey) for Maximal
:
eq⑤
-
P2 =D 2-1
S , P)
-
,
D2
=
Now
Sz P
=
two waves is
( 1st B. F)
n
( nth
B.
f.)
fringe)
≈
Szp =D
D2
+
BPZ
( y f) 2-④
+
+
Castle)
for minima
DK
:
=
(2n 1) 412
1) %
-
y☐d=(2n
when
n= I
,
n
-
2
g
g
i
y=IdD_ [ 1st DF )
Y 3- [ 2ⁿᵈ F)
=
☐
i
n- n
,
y
(2n¥
=
To Alternate Dark &
Now ,
for
Expression
-
( nth
D-
Bright fringes
fringe
F)
appear
width :
-
difference between 2 consecutive
of dark fringes & bright fringes
The
bright fringes gives
.
☆dark Yn+
-
=
=
A-
Similarly
,
for bright
.
→
,
Yn
In
-117¥ n¥
-
¥
73
=
1¥
the
fringe
width
CHAPTER # LL
de
#
-
For
Broglie Equation
:
Dual Nature of Radiation and Matter
:
radiation of frequency G)
the energy of one photon Ps :
a
wavelength A) propagating in
&
E- hv
to
According
Einstein
mass
me
=
2
-
hit me
,
1m=k
Now , momentum of each
photon
-
is
,
,
②
energy equivalence
-
C-
Comparing ② & ④
-
raccoon
,
⑤
④
P
=
MC
P=h¥ ¢
✗
P
hip
=
=
¥
,
P
it
Let
us
K e.
.
take
can
an
example
be written
e⊖
,
as :
accelerated
KE
.
of
e
-0
|p=m_
&
multiplying
square
Now
,
root
'
'
m
K
both sides :
=
mk
Izmit
=
2m K
both side
from de
-
:
broglie
eg
Izmit
=
m2 v2
☒
=
1m27
my
=
12mi
"
:
-
=
-
¥
¥
through
1:
→
a
this is
A-
E)
de€9 ?
potential diff
'
.
v
'
,
then
1k=
&
The linear momentum
=
D=
d-
-
¥
¥
moving
with
velocity
'
r
'
are
:
-
its
1ᵗ-¥
substituting
h=
6-63×10-3
'
"
Js
-9.1×10-3 kg
e- 1.6×10-19C
M
-
}
1=-1
}¥-
CHAPTER
#
Using
# 12
Atoms
:
atoms derive the expression for total
Hydrogen
Energy of e⊖ in stationary states of the atom
Consider an electron of mass m and
e
charge
with
v around a nucleus
velocity
revolving
Then the centripetal
atomic number
↑
having
is
electron
the
force required by
provided by
⊕
⑤
attraction between nucleus
electrostatic
Bohr 's
theory of
,
-
_
,
'
2
and
force of
according
electron
Fe
=
k9i9÷
to
,
'
'
e
.
equation
×
:
'
,
"
"
"
"
Fc
'
--
m¥
=
⊕É
k¥e=mjˢ
ke¥e
Acc
.
to
Bohr 's
Now
,
=
Postulates
±
:
-
m v2
:
-10
mvr
n¥-
=
-
④
miff n¥÷×¥zez
=
Mr
n÷✗
=
41T€
-2 @ 2
n2h2I
✗=
mazes
↳ Radius
Now ,
Velocity
of
e-
in
=
MIT Ze
Nz¥
=
2
3¥
.
↳
nth orbit
rift
m✓n2h2I
"
-
orbits :
stationary
mvr
of
tlenllʳᵈ
Velocity of
pouted
e⊖ in nth
energy
level
④
Now,
f)
Energy
Kif
.
-0in
of
e
stationary orbits :
-
f- MV
=
E) PE
=
Hi,
=
i. e.
=
=
kteh.ae?e)
F. 2
%nEn%→
Tf
1<9,91
r
=Im(ZnnI%)2
KE
=
PE
KE + PE
-F¥n→
.
=
-85%7%-2