Torque on a coil

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1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 6.6.1 Formula of torque Consider a rectangular coil (loop) of wire with side lengths a and b that it can turn about axis PQ. The coil is in a magnetic field of flux density B and the plane of the coil makes an angle θ with the direction of the magnetic field. A current I is flowing round the coil as shown in Figure 6.31. 6.6 Torque on a coil

Transcript of Torque on a coil

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6.6.1 Formula of torque Consider a rectangular coil (loop) of wire with side lengths a and

b that it can turn about axis PQ. The coil is in a magnetic field of

flux density B and the plane of the coil makes an angle θ with

the direction of the magnetic field. A current I is flowing round the coil as shown in Figure 6.31.

6.6 Torque on a coil

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Q

P

b

a

Figure 6.31aFigure 6.31a

θ

BB

B

B

BF

F

1F

I I

I

I1F

A

φ

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From the Figure 6.31b, the magnitude of the force F1 is given by

From the Figure 6.31a, the forces F lie along the axis PQ.

φsin2

b

φsin2

b

θB

B

B

1F

1F

A

φ

Q

φ

φ2

brotationrotationrotationrotation

Figure 6.31b: side viewFigure 6.31b: side view

90sin1 IlBF = al =and

IaBF =1

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From the Figure 6.31a, the forces F lie along the axis PQ. The resultant forceresultant force on the coil is zerozero but the nett torquenett torque is

not zeronot zero because the forces F1 are perpendicular to the axis PQ

as shown in Figure 6.31a.

The forces F1 cause the coil to rotatecoil to rotate in the clockwiseclockwise

direction about the axis PQdirection about the axis PQ as shown in Figure 6.31b. The magnitude of the nett torque about the axis PQ (refer to

Figure 6.31b) is given by

−= φφ sin

2sin

2 11

bF

bFτ

IaBF =1

−= φsin

22 1

bF and

( )

−= φsin

22

bIaB

φsinIabB−= coil) of area(Aab =and

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since thus

For a coil of N turns, the magnitude of the torque is given by

φsinIABτ =θ−=φ 90

( )θIAB −=τ 90sin

θIABcos=τ

φsinNIABτ =

OR

θNIABτ cos=

(6.14)(6.14)

(6.15)(6.15)

where coil on the torque:τdensityflux magnetic :B

coil in the flowscurrent :IBA

and areactor between ve angle :φBθ

and coil theof plane ebetween th angle :(coils) turnsofnumber :N

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From the eq. (6.14), thus the formula of torque in the vector form is given by

The torque is zerotorque is zero when θθ = 90= 90°° oror φφ = 0= 0°° and is maximummaximum

when θθ = 0= 0°° or or φφ = 90= 90°° as shown in Figures 6.32a and 6.32b.

( )BANIτ ×= (6.16)(6.16)

0=φ

90=θB A

0sinNIABτ =

90cosNIABτ =OR

0=τ

Figure 6.32aFigure 6.32a

B

A

0=θ

90=φ

Figure 6.32bFigure 6.32b90sinNIABτ =

0cosNIABτ =OR

NIABτ =max

plane of the coil

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In a radial fieldradial field, the plane of the coilplane of the coil is always parallelalways parallel to the magnetic fieldmagnetic field for any orientation of the coil about the vertical axis as shown in Figure 6.33.

Hence the torquetorque on the coil in a radial fieldradial field is always constantconstant and maximummaximum given by

Radial field is used in moving coil galvanometer.

0=θ

90=φORSSNN

coilfixed soft iron cylinder

radial field

Figure 6.33: Plan view of moving coil meterFigure 6.33: Plan view of moving coil meter

90sinNIABτ = 0cosNIABτ =OR

NIABτ = maximummaximum

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A 50 turns rectangular coil with sides 10 cm × 20 cm is placed vertically in a uniform horizontal magnetic field of magnitude 2.5 T. If the current flows in the coil is 7.3 A, determine the torque acting on the coil when the plane of the coil is

a. perpendicular to the field,

b. parallel to the field,

c. at 75° to the field.

Solution :Solution :

The area of the coil is given by

a.

Example 13 :

A 7.3 T; 5.2 turns;50 === IBN

( )( ) 2222 m 100.210201010 −−− ×=××=A

From the figure, θ = 90° and φ = 0° , thus the torque on the coil is

φsinNIABτ =θNIABτ cos= OR

B

A90=θ

90cosNIAB= 0sinNIAB=0=τ

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Solution :Solution :

b.

c.

BA

90=φ

From the figure, θ = 0° and φ = 90° , thus the torque on the coil is

θNIABτ cos=( ) ( ) ( )( ) 0cos5.2100.23.750 2−×=

m N 3.18=τ

A 7.3 T; 5.2 turns;50 === IBN

B

A

15=φ75=θ

From the figure, θ = 75° and φ = 15°,thus the torque on the coil is

θNIABτ cos=( ) ( ) ( )( ) 75cos5.2100.23.750 2−×=

m N 72.4=τ

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A galvanometer consists of a coil of wire suspended in the magnetic field of a permanent magnet. The coil is rectangular shape and consists of many turns of fine wire as shown in Figure 6.34.

6.6.2 Moving-coil galvanometer

Figure 6.34Figure 6.34

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When the current II flows through the coil flows through the coil, the magnetic field magnetic field exerts a torque on the coilexerts a torque on the coil as given by

This torque is opposed by a spring which exerts a torque, τs

given by

The coil and pointer will rotatecoil and pointer will rotate only to the point where the spring torque balances the torque due to magnetic fieldspring torque balances the torque due to magnetic field, thus

NIABτ =

kθτ s =where constant torsional:k

radianin coil theof anglerotation :θ

sττ =kθNIAB =

NAB

kθI =

(6.17)(6.17)

(6.18)(6.18)

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A rectangular coil of 10 cm × 4.0 cm in a galvanometer has 50 turns and a magnetic flux density of 5.0 × 10−2 T. The resistance of the coil is 40 Ω and a potential difference of 12 V is applied across the galvanometer, calculate the maximum torque on the coil.

Solution :Solution :

The area of the coil is given by

The current through the galvanometer is

Therefore the maximum torque on the coil is

Example 14 :

( )( ) 2322 m 100.4100.41010 −−− ×=××=A

( )4012 I=IRV =A 3.0=I

NIAB=maxτ

m N 100.3 3max

−×=τ

; 04 T; 100.5 turns;50 2 Ω=×== − RBNV 12=V

( ) ( ) ( )( )23 100.5100.43.050 −− ××=

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OhmmeterOhmmeter It is used to measure the unknown resistance of the resistormeasure the unknown resistance of the resistor. Figure 6.35 shows the internal connection of an Ohmmeter.

6.6.3 Electrical instruments

QP

∞ 0

MR SR

ε

XR

Ω

where

resistance (coil)meter :MRresistance variable:SR

resistanceunknown :XR

Figure 6.35Figure 6.35

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When nothing is connected to terminals P and Qnothing is connected to terminals P and Q, so that the circuit is open (that is, when RR → → ∞∞ ), there is no current and no current and no deflectionno deflection.

When terminals P and Q are short circuitedP and Q are short circuited (that is when RR = 0 = 0), the ohmmeter deflects full-scaledeflects full-scale.

For any value of RX the meter deflection depends on the value of

RX.

AmmeterAmmeter It is used to measure a current flows in the circuitmeasure a current flows in the circuit. Ammeter is connected in seriesconnected in series with other elements in the

circuit because the currentcurrent to be measured must pass directlymust pass directly through the ammeterthrough the ammeter.

An ammeter should have low internal resistancelow internal resistance (RM) so that the currentcurrent in the circuit would not affectedwould not affected.

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The maximum readingmaximum reading from the ammeter is known as full scale full scale deflectiondeflection (fs).

If the full scale current passing through the ammeter then the potential difference (p.d.) across that ammeter is given by

If the meter is used to measure currents that are larger than its full scale deflection (I >II >Ifsfs), some modification has to be done.

A resistor has to be connected in parallel with the meter resistor has to be connected in parallel with the meter (coil) resistance (coil) resistance RRMM so that some of the current will current will

bypasses the meter (coil) resistancebypasses the meter (coil) resistance. This parallel resistor is called a shuntshunt denoted as RS.

Figure 6.36 shows the internal connection of an ammeter with a shunt in parallel.

Mfsfs RIV =where resistance )meter(coil :MR

current scale full :fsI(p.d.) difference potential scale full :fsV

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Since shunt is connected in parallel with the meter (coil) resistance then

0 max

A

MR

SR

I fsI

SI

I

Figure 6.36Figure 6.36

SM RR VV =and fsS III −=SSMfs RIRI =

( ) SfsMfs RIIRI −= Mfs

fsS R

II

IR

= (6.19)(6.19)

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VoltmeterVoltmeter It is used to measure a potential difference (p.d.) across measure a potential difference (p.d.) across

electrical elements in the circuitelectrical elements in the circuit. Voltmeter is connected in parallelconnected in parallel with other elements in the

circuit therefore its resistance must be largerresistance must be larger than the than the resistance of the element so that a very small amount ofresistance of the element so that a very small amount of current only can flows through itcurrent only can flows through it. An ideal voltmeterideal voltmeter has infinity resistanceinfinity resistance so that no current exist in it.

To measure a potential difference that are larger than its full scale deflection (V > VV > Vfsfs), the voltmeter has to be modified.

A resistor has to be connected in seriesA resistor has to be connected in series with the meter with the meter (coil) resistance (coil) resistance RRMM so that only a fraction of the total p.d.

appears across the RM and the remainder appears across the serial resistor.

This serial resistorserial resistor is called a multipliermultiplier OR bobbinbobbin denoted as RB.

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1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Figure 6.37 shows the internal connection of a voltmeter with a multiplier

in series.

Since the multiplier is connected in series with the meter (coil) resistance then the current through them are the same, current through them are the same, IIfsfs.

0 max

V

MRBR

Electrical Electrical elementelement

V1I

fsI

I

Figure 6.37Figure 6.37

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The p.d. across the electrical element is given by

Hence the multiplier resistance is MB RR VVV +=

MfsBfs RIRIV +=

−=fs

MfsB I

RIVR (6.20)(6.20)

Note:Note: To convert a galvanometer to ammetergalvanometer to ammeter, a shunt shunt

(parallel resistor)(parallel resistor) is used. To convert a galvanometer to voltmetergalvanometer to voltmeter, a multiplier multiplier

(serial resistor)(serial resistor) is used.

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A milliammeter with a full scale deflection of 20 mA and an internal resistance of 40 Ω is to be used as an ammeter with a full scale deflection of 500 mA. Calculate the resistance of the shunt required.

Solution :Solution :

By applying the formula of shunt resistor, thus

Example 15 :

A 10 500 ; 40A; 1020 3M

3fs

−− ×=Ω=×= IRI

Mfs

fsS R

II

IR

=

Ω= 67.1SR

( )40102010500

102033

3

×−×

×= −−

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A galvanometer has an internal resistance of 30 Ω and deflects full scale for a 50 µA current. Describe how to use this galvanometer to make

a. an ammeter to read currents up to 30 A.

b. a voltmeter to give a full scale deflection of 1000 V.

Solution :Solution :

a. We make an ammeter by putting a resistor in parallel (RS) with

the internal resistance, RM of the galvanometer as shown in

figure below.

Example 16 :

Ω=×= − 30A; 1050 M6

fs RI

SI

I MRfsI

GG

SR

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Solution :Solution :

a. Given

Since RS in parallel with RM therefore

b. We make a voltmeter by putting a resistor in series (RB) with the

internal resistance, RM of the galvanometer as shown in figure

below.

Ω=×= − 30A; 1050 M6

fs RIA 30=I

SM RR VV =and fsS III −=

( ) SfsMfs RIIRI −=

in parallel.in parallel.

SSMfs RIRI =

( )( ) ( ) S66 105030301050 R−− ×−=×

Ω×= − 100.5 5SR

VfsI

MR

GG

BR

fsI

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Solution :Solution :

b. Given

Since RB in series with RM therefore

Ω=×= − 30A; 1050 M6

fs RIV 1000=V

MB RR VVV +=

MfsBfs RIRIV +=

( ) ( )( )30105010501000 6B

6 −− ×+×= R

Ω×= 100.2 7BR in series.in series.

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Exercise 6.5 :1. A moving coil meter has a 50 turns coil measuring 1.0 cm by

2.0 cm. It is held in a radial magnetic field of flux density 0.15 T and its suspension has a torsional constant of 3.0×10−6 N m rad−1. Determine the current is required to give a deflection of 0.5 rad.

ANS. :ANS. : 1.0 1.0×× 1010−− 33 A A

2. A milliammeter of negligible resistance produces a full scale deflection when the current is 1 mA. How would you convert the milliammeter to a voltmeter with full scale deflection of 10 V?

ANS. :ANS. : 1.0 1.0×× 101044 ΩΩ in series in series3. A moving-coil meter has a resistance of 5.0 Ω and full scale

deflection is produced by a current of 1.0 mA. How can this meter be adapted for use as :

a. a voltmeter reading up to 10 V,

b. a ammeter reading up to 2?

ANS. :ANS. : 9995 9995 ΩΩ in series; 2.5 in series; 2.5×× 1010−− 33 ΩΩ in parallel in parallel