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Physics > MAGNETIC effect of Current > Flashcards

MAGNETIC effect of Current Flashcards

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1
Q

force on moving charge in a magnetic field

A

Fb = q(v X B)
q vB sinθ
θ- angle btwn v and B

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2
Q

if the particle is at rest force on it in a magnetic field

A

0

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3
Q

if the particle is moving parallel to the magnetic field then force is

particle travels in a straight line

A

θ = 0
sin0 = 0
F= 0

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4
Q

IF the particle is moving anti parallel to the magnetic field then force is

particle travels in a straigt line

A

θ = 180
sin 180 = 0
F = 0

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5
Q

IF the particle is moving perpendicular to the magnetic field then FORCE is

particle travels in a circular loop

A

θ = 90
sin 90 = 1
F= q v B
force is maximum

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6
Q

if particle travels in an arbitary value (0<θ<90) then path ?

A

hellical

pitch = 2πmvcosθ/ q B

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7
Q

the Velocity v when force is max when particle is in a uniform B

A

Fc = Fb
mv^2/r = q v B
v = qB r/ m

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8
Q

what is the work done by uniform B on a charge entering it at a velocity v

A

W= F. s
since F is perpendicular to v and displacement s is perpendicular to v
W = F s cos θ
* θ = 90 cos θ = 0*
W = 0

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9
Q

TESLA

A

the strenght of B is 1 tesla if a charg of 1C travels with a velocity 1m/s at right angle to the magnetic field which experiences a force of 1 N at that point

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10
Q

unit of MAGNETIC FORCE

1 gauss

A

Ns/Cm

10^-4 TESLA

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11
Q

LORENTZ FORCE

A

the force experienced by a charged particle moving in space where both ELECTRIC FIELD and MAGNETIC FIELD IS present

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12
Q

direction of force in UNIFORM B is given by

A

FLEMINGS LEFT HAND RULE
thumb - force
index - B
middle- velocity or current (+ve)

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13
Q

MAgnetic Force on a current Carrying wire or conductor

A

1.consider a conductor PQ of length l and area of cross section A placed at angle θ to the magnetic field
2. let I be the current flowing thru thr conductor and vd be the drift velocity and e be the charge on each e=
3. no. of e- in length l of the conductor=
n = N/V where n- no. density of e-
N = nAl where V= Al
4. force acting on each e-
F= -e (vd X B)
F = -e vd B sinθ
5. total force acting on free e-
F= nA l -e vd B isn θ
F = I l Bsinθ
F = I (lXB)

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14
Q

how does magnetic field act on current carrying conductor in a uniform B

A

due to motion of free e- inside the conductor

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15
Q

FOrce between 2 parallel current carrying conductor

A
  1. consider 2 infinite long straight conductors X1Y1 and X2Y2 of length l kept at distance r m away from each other
  2. let I1 AND I2 be the current flowing thru them in the same dirn
  3. magnetic field at P due to current I1
    B1 = μoI1/2π r
  4. force experienced by X2Y2 due to B1
    F2 = B1 I2 l
    F2 = μoI1 I2 l/2πr
  5. similarly X1Y1 also experiences a force of same amount directed toward the wire X2Y2
  6. therefore F between 2 current carrying conductor per unit length

F/l = μo 2 I1 I2/ 4π r

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16
Q

1 AMPERE in terms of FORCE

F/l = μo 2 I1 I2/ 4π r

2X 10^-7 N/m

dontmention mahnetic fireld

A

1 ampere is the current flowing thru 2 infinitely long parallel current carrying conductor placed 1 m apart from each other & which attract or repel with a force of 2X10^-7 N/m

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17
Q

Ampere’s circuital law

A

the line integral of magnetic field around a closed loop in vaccum is μo times the net current enclosed in the path

∮B.dl = μo I

μo = 4π X 10^-7

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18
Q

Magnetic field due to infinite long straight wire carrying current

A

B =μo I / 2 π r

∮dl = 2πr

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19
Q

Magnetic field due Solenoid

length is very large as compared to its diameter

A

B = μonI
B = μo N/l I

20
Q

Magnetic field due Solenoid at end points of the solenoid

length is very large as compared to its diameter

A

B = 1/2μonI
B = 1/2 μo N/l I

21
Q

relation between μ0 εo and c

A

εoμo = 4π/4π μoεo
= 4π εo μo/4π
= 1/ (9 X10^9) X 10 ^-7
= 1/ (3X10^8) ^2
1/c^2
c^2 = 1/μoεo
c= 1/ √μoεo

22
Q

Magnetic field due to infinite cylindrical wire with curren tdistributed uniformly in cross section

r>R

p lies outside

A

B = μ0 I/2πr

B is inversely proportion to r

∮dl = 2πr

23
Q

Magnetic field due to infinite cylindrical wire with current distributed uniformly in cross section

r < R

p lies inside

A

B = μ0 Ir/2πR^2

B is proportional to r

∮dl = 2πr

current throught closed loop =** I/πR^2 X πr^2**

24
Q

Magnetic field due to infinite cylindrical wire with current along the surface
B at outside will be

r>R

p lies outside

A

B = μ0 I/2πr

B is inversely proportion to r

∮dl = 2πr

25
Q

Magnetic field due to infinite cylindrical wire with current along the surface
B inside will be
r< R

curren thru closed loop is?

A

0
current through the closed loop is 0

26
Q

magnetic moment

A

m = N I A

27
Q

TORQUE ON CURRENT LOOP IN B FEILD

A

𝜏 =N I A B sinθ
𝜏 = m B sin θ
𝜏 = m X B
unit: N/m

28
Q

MOVING COIL GALVANOMETER

A
  1. device - detect presence of electric current
  2. construction: consist a coil with many turns free to rotate in anout a fixed axis
    also a cylindrical soft iron core
  3. working:
    * suppose the coil PQRS is suspended freely in B field
    * let l be lenght PQ and b the the breadth QR
    * N be no. of turns and A be area where A = l X b
    * Let B be the strength of B field and I be the current passing thru it
    * let at any instance α be the angle which normal drawn to the plane of the coil with the B field
    * the coil expresses torque 𝜏 = NIABsinα
    * due to deflecting 𝜏, the coil rotates and suspension wire gets twisted
    * a restoring torque is set up
    * let θ be the twist produced due to rotation of coil and k be the restoring torque per twist
    * in EQUILIBRIUM position: deflecting 𝜏 = restoring 𝜏
    * NIAB = kθ
    * I = (k/ NIAB) θ
    * I = G θ
    * i/e/, I is proportional to θ

it means that deflection produced is proportional to the current flowing thru GAL

29
Q

2 purpose of soft iron core in GAL

A
  1. to make B field radial
  2. to increase strenght of B field
30
Q

current sensitivity

unit

A

deflection per unit current flowing thru it
θ/I = NAB / k

unit rad / A

31
Q

voltage sensitivity

unit

A

defection per unit voltage flowing thru it
θ/ V = NAB/kR

unit rad/ v

32
Q

figure of merit

currentproducedeflection

A

amount of current which produces one scale deflection in GAL

33
Q

REASON why galvanometre cant be used to measure a current in crk

A
  1. galvanometer is sensitive. it shows full deflection for μA of the current
  2. galvanometer has high resistance and it will affect the original current
34
Q

why increasing the current sensitivity may not increase the voltage sensitivity

A

when CS increases no. of turns also increases but RESISTANCE also increases which adversly affect VS

35
Q

how to convert gal to amm

FOR IDEAL AMM: R = 0

A

connect a shunt resistance parallel with the GAL

36
Q

How to convert GAL to VOLT

FOR IDEAL VOLT: R=infinity

A

connect a high resistance in series with GAL

37
Q

BIOT - SAVART LAW

A

magnitude of B field is
1. directly prop to current
2. directly prop to length of current element
3. directly prop to sinθ
4. inversely prop to square of distance from current element

B = μo/4π I dl sinθ/ r^2

38
Q

BIOT - SAVART LAW vector form

A

B = μo/4π I dl Xr/ r^3

39
Q

μ value
SI UNIT
DIMENSION

A

4π X 10^-7
Tm/A
[MLA-2T-2]

[T]= [M L A-1 T-2]

40
Q

COMPARE COULOMBS AND BIOT SAVART LAW

A

SIMILARITY:
1. inverse square dependence on distance
2. obeys superposition principle

dissimilarity
1. B field produced due to vector source and E field due to sclar source
2. angle dependence and no angle dependnce

41
Q

magnetc field at centre of current carrying loop
for N turns

A

B= μoNI/2R
R- RADIUS
I-CURRENT

42
Q

magnetic field due to semi-circular arc

A

B=μoI/4R

μoI/4πR θ/2π

θ= π

43
Q

magnetic field due to quadrant

A

B= μoI/8R

μoI/4πR θ/2π

θ= π/2

44
Q

magnetic field on axis of current carrying loop
for n turns

A

B= μo n I a^2/2(a^2 +x^2)^3/2

45
Q

torque experienced by a current carrying loop in uniform magnetic field
when the loop is placed such that uniform B is in th eplane of loop

A

𝜏 = N I A B

46
Q

torque experienced by a current carrying loop in uniform magnetic field
WHEN the plane of the loop i snot along B but makes an angle with it

A

𝜏=m X B
= mBsinθ

Nm

Am2

Physics (8 decks)

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