MAGNETIC effect of Current Flashcards by Unknown Unknown

force on moving charge in a magnetic field

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

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

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

particle travels in a straight line

θ = 0
sin0 = 0
F= 0

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

particle travels in a straigt line

θ = 180
sin 180 = 0
F = 0

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

particle travels in a circular loop

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

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

hellical

pitch = 2πmvcosθ/ q B

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

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

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

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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TESLA

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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unit of MAGNETIC FORCE

1 gauss

Ns/Cm

10^-4 TESLA

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LORENTZ FORCE

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

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direction of force in UNIFORM B is given by

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

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MAgnetic Force on a current Carrying wire or conductor

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

due to motion of free e- inside the conductor

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FOrce between 2 parallel current carrying conductor

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

F/l = μo 2 I1 I2/ 4π r

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1 AMPERE in terms of FORCE

F/l = μo 2 I1 I2/ 4π r

2X 10^-7 N/m

dontmention mahnetic fireld

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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Ampere’s circuital law

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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Magnetic field due to infinite long straight wire carrying current

B =μo I / 2 π r

∮dl = 2πr

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Magnetic field due Solenoid

length is very large as compared to its diameter

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B = μonI
B = μo N/l I

Magnetic field due Solenoid at end points of the solenoid

length is very large as compared to its diameter

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B = 1/2μonI
B = 1/2 μo N/l I

relation between μ0 εo and c

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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

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

r>R

p lies outside

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B = μ0 I/2πr

B is inversely proportion to r

∮dl = 2πr

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

r < R

p lies inside

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B = μ0 Ir/2πR^2

B is proportional to r

∮dl = 2πr

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

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

r>R

p lies outside

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B = μ0 I/2πr

B is inversely proportion to r

∮dl = 2πr

Magnetic field due to infinite cylindrical wire with current along the surface B inside will be **r< R** | curren thru closed loop is?

0 current through the closed loop is 0

magnetic moment

m = N I A

TORQUE ON CURRENT LOOP IN B FEILD

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

MOVING COIL GALVANOMETER

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

2 purpose of soft iron core in GAL

1. to make B field radial 2. to increase strenght of B field

current sensitivity | unit

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

voltage sensitivity | unit

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

figure of merit ## Footnote currentproducedeflection

amount of current which produces one scale deflection in GAL

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

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

why increasing the current sensitivity may not increase the voltage sensitivity

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

how to convert gal to amm | FOR IDEAL AMM: R = 0

connect a shunt resistance parallel with the GAL

How to convert GAL to VOLT | FOR IDEAL VOLT: R=infinity

connect a high resistance in series with GAL

BIOT - SAVART LAW

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

BIOT - SAVART LAW vector form

B = μo/4π I dl Xr/ r^3

μ value SI UNIT DIMENSION

4π X 10^-7 Tm/A [MLA-2T-2] | [T]= [M L A-1 T-2]

COMPARE COULOMBS AND BIOT SAVART LAW

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

magnetc field at centre of current carrying loop for N turns

B= μoNI/2R R- RADIUS I-CURRENT

magnetic field due to semi-circular arc

B=μoI/4R | μoI/4πR θ/2π ## Footnote θ= π

magnetic field due to quadrant

**B= μoI/8R** | μoI/4πR θ/2π ## Footnote θ= π/2

magnetic field on axis of current carrying loop for n turns

B= μo n I a^2/2(a^2 +x^2)^3/2

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

𝜏 = N I A **B**

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

𝜏=m X **B** = m**B**sinθ | Nm ## Footnote Am2

MAGNETIC effect of Current Flashcards

(46 cards)