Formulae Magnetic Effect of Current

7/31/2019 Formulae Magnetic Effect of Current

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3.1 Magnetic Effect of Current

Biot Savarts Law:

Value of 0 = 4 x 10-7 Tm A-1 or Wb m-1 A-1

x

P

d

r

P

I

dBI dl sin

r2

dB=0 I dl sin

4 r2

Biot Savarts Law in vector form:

dB=0 I dl x r

4 r2

dB=0 I dl x r

4 r3


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Magnetic Field due to a Straight Wire carrying current:

P

r

a

I

2

1

l

d

x B

B=

or

B B

I I

0 I (sin 1 + sin 2)

4a

B

aa 0

If the straight wire is infinitely long,then 1 = 2 = / 2

B=0 2I

4aB=

0 I

2a

Magnetic Field Lines


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I

I I

B

xx 0

ii) If the observation point is far away fromthe coil, then a


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2. Magnetic Field at the centre of the loop:

B =0 I

2a

B

a0

x

dBI

I

d

90a

O

Magnetic Field due to a Solenoid:

I I

xxxxx xx

B

TIP:

When we look at any end of the coil carrying current, if thecurrent is in anti-clockwise direction then that end of coilbehaves like North Pole and if the current is in clockwisedirection then that end of the coil behaves like South Pole.


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Lorentz Magnetic Force:

Force on a current-carrying conductor in a uniform Magnetic Field:

Fm = q (v x B)

+q B

v

F

I

-qB

v

F

Fm = (q v B sin ) nwhere is the angle between v and B

Special Cases:

If the charge is at rest, i.e. v = 0, then Fm = 0.So, a stationary charge in a magnetic fielddoes not experience any force.

If = 0or 180i.e. if the charge movesparallel or anti-parallel to the direction of themagnetic field, then Fm = 0.

If = 90i.e. if the charge movesperpendicular to the magnetic field, then theforce is maximum.

or

I

vd

d

F

I

I

B

A l-

Fm (max) = q v B

F= I (l x B)

or

F = I l B sin


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Forces between two parallel infinitely long current-carryingconductors:

Force of attraction Force or repulsion

r

F

I1

P

Q

F

I2

x

S

R

r I2

Fx

S

R

I1

F

P

Q

x

r

F21F12I1

P

Q

I2

S

R

x B1B2

F12=F21=F =0 I1 I2l

2 r

F / l =0 I1 I2

2 rN / m


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Definition of Ampere:

One ampere is that current which, if passed in each of two parallelconductors of infinite length and placed 1 m apart in vacuum causeseach conductor to experience a force of 2 x 10-7 Newton per metre oflength of the conductor.

Flemings Left Hand Rule:

Representation of Field due to Parallel Currents:I1 I2

B

I1 I2

B

N

F / l =0 I1 I2

2 rN / m

When I1 = I2 = 1 Ampere and r = 1 m, then F = 2 x 10-7 N/m.

ElectricCurrent (I)

Force F

MagneticField B


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Torque experienced by a Current Loop (Rectangular) in a uniformMagnetic Field:

| FSP | = I b B sin

| FQR | = I b B sin

FSP = I (b x B)

FQR = I (b x B)

Forces FSP and FQR are equal in magnitude but opposite indirection and they cancel out each other. Moreover they actalong the same line of action (axis) and hence do not producetorque.

| FRs | = I l B sin 90= I l B

FPQ = I (l x B)

FRS = I (l x B)

Forces FPQ and FRS being equal in magnitudebut opposite in direction cancel out each otherand do not produce any translational motion.But they act along different lines of action andhence produce torque about the axis of thecoil.

B

P

Q

R

S

x

l

b

I

I

FQR

FSP

FPQ

FRS

| FPQ | = I l B sin 90= I l B

Torque experienced by the coil is

= FPQ x PN (in magnitude)= I l B (b cos )= I lb B cos = I A B cos (A = lb)= N I A B cos (where N is the no. of turns)


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If is the angle between the normal to the coil and thedirection of the magnetic field, then + = 90 i.e. = 90- So,

= I A B cos (90 - )= N I A B sin

NOTE:One must be very careful in using the formula in termsof cos or sin since it depends on the angle takenwhether with theplane of the coil or the normal of thecoil.

Torque in Vector form:

= N I A B sin

= (N I A B sin ) n (where n is unit vector normal to the plane of the loop)

= =or

(since M = I A is the Magnetic Dipole Moment)


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PBW

P S

Q R

N Sx

T

E

T Torsion Head, TS Terminal screw, M Mirror, N,S Poles pieces of a magnet,LS Levelling Screws, PQRS Rectangular coil, PBW Phosphor Bronze Wire

LS LS

B

Torque experienced by thecoil is

= N I A B sin Restoring torque in the coilis

= k (where k isrestoring torque per unitangular twist, is the angulartwist in the wire)

At equilibrium,N I A B sin = k

I =k

N A B sin

The factor sin can beeliminated by choosing Radial

Magnetic Field.

M

Hair Spring

TS

FRS

FPQ

Lamp

Scale

Radial Magnetic Field:

N S

BP

SThe (top view PS of) plane of the coil PQRS liesalong the magnetic lines of force in whicheverposition the coil comes to rest in equilibrium.So, the angle between the plane of the coil andthe magnetic field is 0.

orthe angle between the normal to the plane ofthe coil and the magnetic field is90.

i.e. sin = sin 90= 1

I =k

N A B or I = G

k

N A Bwhere G =

is called Galvanometer constant

Mirror

2

Moving CoilorSuspended CoilorD Arsonval Type Galvanometer:

Current Sensitivity of Galvanometer:It is the defection of galvanometer per unit current.

Voltage Sensitivity of Galvanometer:It is the defection of galvanometer per unit voltage.

k

N A B

I

=

kR

N A B

V

=


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S.No.Ammeter Voltmeter

1 It is a low resistance instrument. It is a high resistance instrument.

2 Resistance is GS / (G + S) Resistance is G + R

3

Shunt Resistance is(GIg) / (I Ig) and is very small.

Series Resistance is(V / Ig) - G and is very high.

4It is always connected in series. It is always connected in parallel.

5Resistance of an ideal ammeter iszero.

Resistance of an ideal voltmeter isinfinity.

6Its resistance is less than that of thegalvanometer.

Its resistance is greater than that of thevoltmeter.

7

It is not possible to decrease therange of the given ammeter.

It is possible to decrease the range of thegiven voltmeter.

Conversion of Galvanometer to Ammeter:

Galvanometer can be converted into ammeterby shunting it with a very small resistance.Potential difference across the galvanometerand shunt resistance are equal.

(I Ig ) S = Ig G S =

Ig

I Ig

Conversion of Galvanometer to Voltmeter:

Galvanometer can be converted intovoltmeter by connecting it with a very highresistance.Potential difference across the given loadresistance is the sum of p.d acrossgalvanometer and p.d. across the highresistance.

V = Ig (G + R)

GI Ig

Is = I - Ig

S

or R =V

Ig

- G

GIg R

V

or

Difference between Ammeter and Voltmeter:


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N

S

D1 D2 +

W

B

D1, D2 Dees N, S Magnetic Pole PiecesW Window B - Magnetic Field

H FOscillator

D2D1

W

B

Theory:

The magnetic force experienced by the charge provides centripetalforce required to describe circular path.

mv2 / r = qvB sin 90 (where m mass of the charged particle,q charge, v velocity on the path ofradius r, B is magnetic field and 90is theangle b/n v and B)

v =B q r

m

If t is the time taken by the charge to describe the semi-circular pathinside the dee, then

t = r

vor t =

m

B q

Time taken inside the dee depends only onthe magnetic field and m/q ratio and not onthe speed of the charge or the radius of thepath.

If T is the time period of the high frequency oscillator, then for resonance,

T = 2 t or T =2mB q

If fis the frequency of the high frequency oscillator (Cyclotron

Frequency), then

f =2m

B q

Working: Imagining D1 is positive and D2 is negative, the + vely charged particle keptat the centre and in the gap between the dees get accelerated towards D2. Due to

perpendicular magnetic field and according to Flemings Left Hand Rule the chargegets deflected and describes semi-circular path.

When it is about to leave D2, D2 becomes + ve and D1 becomes ve. Therefore theparticle is again accelerated into D1 where it continues to describe the semi-circularpath. The process continues till the charge traverses through the whole space in thedees and finally it comes out with very high speed through the window.

Cyclotron:


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Maximum Energy of the Particle:

Kinetic Energy of the charged particle is

K.E. = m v2= m (B q r

m)2 =

B2 q2 r2

m

Maximum Kinetic Energy of the charged particle is when r = R (radius of the Ds).

= B2 q2 R2

mK.E. max

The expressions for Time period and Cyclotron frequency only whenm remains constant. (Other quantities are already constant.)

m =m0

[1 (v2 / c2)]

If frequency is varied in synchronisation with the variation of mass of thecharged particle (by maintaining B as constant) to have resonance, then thecyclotron is called synchro cyclotron.

If magnetic field is varied in synchronisation with the variation of mass of thecharged particle (by maintaining f as constant) to have resonance, then thecyclotron is called isochronous cyclotron.

But m varies with v according toEinsteins Relativistic Principle as per

Amperes Circuital Law:

The lineintegral B . dlfor a closed curve is equal to0times the netcurrentIthreading through the area bounded by the curve.

B . dl =0 I

B . dl =B . dl cos 0

B . dl = B= d= B (2 r) = ( 0 I / 2 r)x 2 r

I

B

B

r

O

dl

I

Current is emergingout and the magneticfield is anticlockwise.

Proof:

B . dl =0 I

NOTE: Cyclotron can not be used for accelerating neutral particles. Electrons cannot be accelerated because they gain speed very quickly due to their lighter massand go out of phase with alternating e.m.f. and get lost within the dees.


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Magnetic Field at the centre of a Straight Solenoid:

I I

xxxxx xx

P Q

RS

B . dl = 0 I0(where I0 is the net currentthreading through the solenoid)

B . dl =B . dl +PQ

B . dl +QR

B . dl +RS

B .SP

B

B . dl cos 0+ B . dl cos 90+ 0 . dl cos 0+ B . dl cos 90== B dl = B.a and 0I0 =0n a I

(where n is no. of turns per unit length, a is the length of the pathand I is the current passing through the lead of the solenoid)

a

a

B = 0n I

Magnetic Field due to Toroidal Solenoid (Toroid):

I

d

BP

O QB = 0

B = 0

B . dl = 0 I0

B . dl cos 0

= B dl = B (2 r)r

And 0 I0=0 n (2 r) I

B = 0n I

B . dl =

NOTE:The magnetic field exists only in the tubulararea bound by the coil and it does not exist

in the area inside and outside the toroid.i.e. B is zero at O and Q and non-zero at P.

B 0

Formulae Magnetic Effect of Current

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