P2B_P4_part_1_MOVING_CHARGES_AND_MAGNETISM[1] - Read-Only-1.pdf

• Christian oersted discovered magnetic field surrounding a current
carrying wire
• The direction of magnetic field depends on direction of current
Oersted’s Experiment
⑭

Magnetic Field
• Space surrounding a magnet, where its influence is felt.
• Strength of magnetic field is represented by magnetic field
intensity 𝐵
• SI Unit is Tesla(T) or Wb/m2
--
-
T - D
↳
Sn ①
-
rected I =
B
.
N
-
-
-
webn =
B
-
- -
-
webd m
-
m2
-
-

Magnetic Lorentz Force
• It is the force acting on a moving charged particle in an external
magnetic field
F = q v B sinϴ
⃗
𝐅 = q(𝐯 x 𝐁)
- --
- -
- -
-
⑬ v
E
-
↑
& B =
0
- ---
V -
In -
- angle
b
-

• It is the force acting on a moving charged particle moving in an
combined magnetic field(B) and electric field(E)
F = Felectric + Fmagnetic
⃗
𝐅 = q𝑬 +q(𝐯 x 𝐁)
Lorentz force
Fe = 9E
- Fi =
q (UXB)
--
- -

Write the expression for Lorentz force acting on a moving charge.
[SAY – 2020]
-
- - - -
-
F =
qE + q(vXx)
-
-
-

An electron is placed in a magnetic field of 3 × 10 T. find the force
acting on the electron
-
E
-
-
* X X X F =
q(v-xi)
i -
>v =
X X X X
XXXX

Motion of a charged particle in a uniform magnetic field
Charged particle entering parallel to external magnetic field
- -
- - -
O
D = O
=
B
q -v
i F=
gubsin o
F = 0
-
S
-
Constant reclocity
-
-

The path of a charged particle entering parallel to uniform magnetic
field will be
a)Circular
b)Helical
c)Straight line
d)None of these (1)[March 2020]
-
~
-

Flemings Left Hand Rule
Direction of Magnetic Field
Fore finger
Direction of Current
Middle finger
Direction of Force on
positively charged particle
Thumb
-
-
-
-
-

Charged particle entering perpendicular to
external magnetic field
• When a charged particle of mass(m) and charge(q) moves
perpendicular to Mag. Field of strength(𝐵) with a velocity(v),
constant force acts right angles to the motion of particle
This provides necessary
centripetal force and path of
particle will be a CIRCLE
- -
-
-
-
&

f i
-v
av
S
↓ &
0= 90 9 F= qu BSin90
f qui
-
-

3. Charged particle entering at an angle 𝜃 to
external magnetic field
n men i
·
10
----v cose
&
S

A charged particle enters a uniform magnetic field at an angle of
40∘
. It's path becomes_____________
[march 2019]
~
backcal -
-
-

Two charged particles 𝑞 and 𝑞 are moving through a uniform
magnetic field (𝐵) as shown in figure:
(a) What is the shape of path of 𝑞 and 𝑞 .
[march 2018]
- -
-
9
a
heckcal
92 -
q
92
/
-
-
-

Biot – Savart’s Law
• Magnetic field at a point due to current carrying element is
directly proportional to
Strength of current(I)
Length of element(dl)
Sine of angle between element and
line joining element and the point
Inversely proportional to square of
distance between element and point
- - -
=
-
dis & I
%
Eu
dis
add 20
e
dis sind
M
di

dis 2 I del sind
-
22
* dot paiabichty
&B I do Idd Sind offar
- -
space
&
GTT 22 -
Gi
X107
Mo =
-

Biot - Savarts Law
d𝑩 =
μ𝟎
𝟒𝝅
𝑰𝒅𝒍 𝒔𝒊𝒏ϴ
𝒓𝟐
μ𝟎 = Permeability of
free space
𝟒𝝅 x 10-7 Tm/A
d𝑩 =
μ𝟎
𝟒𝝅
𝑰(𝒅𝒍 𝐱 𝒓)
𝒓𝟑
𝑽𝒆𝒄𝒕𝒐𝒓 𝑭𝒐𝒓𝒎
=
=
-
-
~

Magnetic Field due to current carrying circular
conductor
• Consider a circular coil of radius, a carrying a current I.
• Let P be point on the axis of coil at a distance, r from the centre - O
- - -
di
⑬
discost
WhdSe
#
-
z
I *
dissing
- W
&
-----
dissing
~ x
idi
n dicosd
ad

Magnetic Field due to current carrying circular
conductor
*
Sin 90 =
1
dB I
do since -
-
-
git xh
&B Mo T
dd
-
-
I
G
= -
Git x2
-
Total magnetic feel
dB = dasind + desind
= 2 dis sind
F
F

dis' = 2 do Id sind
Sind =
wit xh
dB' = 2) Mo I di a B = do I a Ita
-
-
- - Zit x3
2 pit x3
--
↑
a
-
dis' = do I de a
2x3
2π x3
I
I
B =
do I de

Right Hand Grip Rule
• If the current carrying element is grasped
in the right hand with the thumb pointing
in the direction of current.
• Then the outer fingers lying in
perpendicular direction gives the
direction of magnetic field
-

Ampere’s Circuital Law
• The line integral of magnetic field around any closed loop is equal
to μ0 times the total current enclosed by the loop
𝐵. 𝑑𝑙 = μ0 𝐼
𝐵 𝑑𝑙𝑐𝑜𝑠𝛳 = μ0 𝐼
I
- -
- =
- =
- del
- -
-
> 11
I I -
& M
-
I 11 -
I

B At Point Due To An Infinitely Long Straight
Wire Carrying Current
P
=
-
SB.
di = do I
NE
SBdd Cosc = do I
[
0 = 8
B
M
↑
Sidd =
clot
B/dd = do I
-
-

B At Point Due To An Infinitely Long Straight
Wire Carrying Current
BLITZ = do I
do I
= -
B -
2π y
-

Magnetic Field Intensity Due to Solenoid
D
- C
-
-
-
-
i
-
·
BE
i----
-
A B
- L -
GB. del = Mo I
A
A
C
Side
+ Bidd
+fide
+
I
de -lot
B
S

t t
Ba
"
op
·
self p
Si
-
!
O
O
6
= MOI I B L = Mow I
B
Mol MONI
&
Bac =
N =
n
IB =
-
L
B Sdd =
Mol B = MonI
-

SAY
EXAM PYQ(2021)
State Ampere's circuital law. Use this law to obtain the expression for
the magnetic field inside a current carrying solenoid at an axial point
near the centre.
3 Marks
BOARD
EXAM PYQ(2021)

Force acting on a current carrying conductor placed in
magnetic field
Consider a rod of uniform cross-section ‘A’ and length ‘l’ . Keep
this in a magnetic field of ‘B’
n number of electrons per unit volume ( number density )
Total charge in the conductor = nAlq
𝑉 average drift velocity of all the charges
F =

Moving Coil Galvanometer
• Principle : A current carrying rectangular coil placed in a magnetic
field experiences a torque
• This torque is proportional to the strength of the current passing
through it
-
-
D

Conversion of galvanometer into voltmeter
• By putting high resistance R in series with the galvanometer
R =
- - -
VG
-
-
& um F O
Ig high
on

Conversion of galvanometer into ammeter
• By putting low resistance S in parallel with the galvanometer
S =
S = Shunt Resistance
Ig 9
6
-
=
--Ig O
T
dow
reach
mu

A galvanometer with coil resistance 12Ω shows full-scale deflection
for a current of 2.5 mA. How will you convert it in to an ammeter of
range 0 – 7.5 A ?

P2B_P4_part_1_MOVING_CHARGES_AND_MAGNETISM[1] - Read-Only-1.pdf

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