4.11 The Moving Coil Galvanometer

NCERT Class 12 Physics for low vision and blind students.

Currents and voltages in circuits have been discussed extensively in Chapters 3. But how do we measure them? How do we claim that current in a circuit is 1.5 A or the voltage drop across a resistor is 1.2 V? Figure 4.24 exhibits a very useful instrument for this purpose: the moving coil galvanometer (MCG). It is a device whose principle can be understood on the basis of our discussion in Section 4.10.

[FIGURE 4.24 The moving coil galvanometer. Its elements are described in the text. Depending on the requirement, this device can be used as a current detector or for measuring the value of the current (ammeter) or voltage (voltmeter).]

The galvanometer consists of a coil, with many turns, free to rotate about a fixed axis (Fig. 4.24), in a uniform radial magnetic field. There is a cylindrical soft iron core which not only makes the field radial but also increases the strength of the magnetic field. When a current flows through the coil, a torque acts on it. This torque is given by Equation (4.26) to be

τ = N I A B

where the symbols have their usual meaning. Since the field is radial by design, we have taken sinθ = 1 in the above expression for the torque.

The magnetic torque N I A B tends to rotate the coil. A spring S provides a counter torque k φ that balances the magnetic torque N I A B; resulting in a steady angular deflection φ. In equilibrium

K φ = N I A B

where k is the torsional constant of the spring; i.e. the restoring torque per unit twist. The deflection φ is indicated on the scale by a pointer attached to the spring. We have

φ = ( N A B/k) I .. .. (4.38)

The quantity in brackets is a constant for a given galvanometer.

The galvanometer can be used in a number of ways. It can be used as a detector to check if a current is flowing in the circuit. We have come across this usage in the Wheatstone’s bridge arrangement. In this usage the neutral position of the pointer (when no current is flowing through the galvanometer) is in the middle of the scale and not at the left end as shown in Fig.4.24. Depending on the direction of the current, the pointer deflection is either to the right or the left.

The galvanometer cannot as such be used as an ammeter to measure the value of the current in a given circuit. This is for two reasons: (i) Galvanometer is a very sensitive device, it gives a full-scale deflection for a current of the order of μA. (ii) For measuring currents, the galvanometer has to be connected in series, and as it has a large resistance, this will change the value of the current in the circuit. To overcome these difficulties, one attaches a small resistance rs, called shunt resistance, in parallel with the galvanometer coil; so that most of the current passes through the shunt. The resistance of this arrangement is,

RG rs / (RG + rs) ≃ rs if RG >> rs

If rs has small value, in relation to the resistance of the rest of the circuit Rc, the effect of introducing the measuring instrument is also small and negligible. This arrangement is schematically shown in Fig. 4.25.

[FIGURE 4.25 Conversion of a galvanometer (G) to an ammeter by the introduction of a shunt resistance rs of very small value in parallel.]

The scale of this ammeter is calibrated and then graduated to read out of the value of current with ease. We define the current sensitivity of the galvanometer as the deflection per unit current. From Equation (4.38) this current sensitivity is,

φ/I = N A B/k .. .. (4.39)

A convenient way for the manufacturer to increase the sensitivity is to increase the number of turns N. We choose galvanometers having sensitivities of value, required by our experiment.

The galvanometer can also be used as a voltmeter to measure the voltage across a given section of the circuit. For this it must be connected in parallel with that section of the circuit. Further, it must draw a very small current, otherwise the voltage measurement will disturb the original set up by an amount which is very large. Usually we like to keep the disturbance due to the measuring device below one per cent. To ensure this, a large resistance R is connected in series with the galvanometer. This arrangement is schematically depicted in Fig.4.26.

[FIGURE 4.26 Conversion of a galvanometer (G) to a voltmeter by the introduction of a resistance R of large value in series.]

Note that the resistance of the voltmeter is now,

RG + R ≃ R : large

The scale of the voltmeter is calibrated to read off the voltage value with ease. We define the voltage sensitivity as the deflection per unit voltage. From Equation (4.38),

φ/V = (NAB/k) (I/V) = (NAB/k) (1/R) . .. (4.40)

An interesting point to note is that increasing the current sensitivity may not necessarily increase the voltage sensitivity. Let us take Equation (4.39) which provides a measure of current sensitivity. If N → 2N, i.e., we double the number of turns, then

φ/I →2(φ /I)

Thus, the current sensitivity doubles. However, the resistance of the galvanometer is also likely to double, since it is proportional to the length of the wire. In Equation (4.40), N → 2N, and R → 2R, thus the voltage sensitivity,

φ/V → φ/V

remains unchanged. So in general, the modification needed for conversion of a galvanometer to an ammeter will be different from what is needed for converting it into a voltmeter.

Example 4.13

In the circuit (Fig. 4.27) the current is to be measured. What is the value of the current if the ammeter shown (a) is a galvanometer with a resistance RG = 60.00 Ω ; (b) is a galvanometer described in (a) but converted to an ammeter by a shunt resistance rs = 0.02 Ω; (c) is an ideal ammeter with zero resistance?

Fig. 4.27

Solution:


(a) Total resistance in the circuit is,


RG + 3 = 63 Ω .


Hence, I = 3/63 = 0.048 A.

(b) Resistance of the galvanometer converted to an ammeter is,


RG rs/(RG + rs)


= 60 Ω × 0.02 Ω /( (60 + 0.02) Ω ≃ 0.02 Ω

Total resistance in the circuit is,


0.02 Ω + 3 Ω = 3.02 Ω.

Hence, I = 3/3.02 = 0.99 A.

(c) For the ideal ammeter with zero resistance, I = 3/3 = 1.00 A

SUMMARY

NCERT Class 12 Physics books for blind and visually impaired students.

1. The total force on a charge q moving with velocity v vector in the presence of magnetic and electric fields B vector and E vector, respectively is called the Lorentz force. It is given by the expression:

F vector = q (v vector ⨯ B vector + E vector)

The magnetic force q (v vector ⨯ B vector) is normal to v vector and work done by it is always zero.

2. A straight conductor of length l, carrying a steady current I, and placed in a uniform external magnetic field B vector, experiences a force F vector, given by:

F vector = I l vector ⨯ B vector

where|l vector | = l and the direction of l vector is given by the direction of the current.

3. In a uniform magnetic field B vector, a particle with a charge q executes a circular orbit in a plane normal to B vector. Its frequency of uniform circular motion is called the cyclotron frequency and is given by:

νc = qB/2 π m

This frequency is independent of the particle’s speed and radius of the circular orbit. This fact is exploited in a machine, the cyclotron, which is used to accelerate charged particles.

4. The Biot-Savart law asserts that the magnetic field dB vector due to an element dl vector carrying a steady current I at a point P at a distance of r from the current element is:

dB vector = (μ0/4π) I (dl vector ⨯ r vector)/r^3

To obtain the total field at P due to a finite conductor, we must integrate this vector expression over the entire length of the conductor.

5. The magnitude of the magnetic field due to a circular coil of radius R carrying a current I at an axial distance x from the centre is

B = μ0 I R^2/{2(x^2 + R^2)^(3/2)

At the center of the coil this reduces to

B = μ0 I/2R

6. Ampere’s Circuital Law:


Let an open surface S be bounded by a loop C. Then the Ampere’s law states that

∮ B vector ⋅dl vector = μ0 I C


where I refers to C the current passing through S. The sign of I is determined from the right hand thumb rule. We have discussed a simplified form of this law. If the magnetic field B vector is directed along the tangent to every point on the perimeter L of a closed curve and is constant in magnitude along perimeter then,

BL = μ0 I e

where Ie is the net current enclosed by the closed circuit.

7. The magnitude of the magnetic field at a distance R from a long, straight wire carrying a current I is given by:

B = μ0I/2πR

The field lines are circles concentric with the wire.

8. The magnitude of the field B inside a long solenoid carrying a current I is

B = μ0 nI


where n is the number of turns per unit length.

For a toroid one obtains,

B = μ0 N I/2πr


where N is the total number of turns and r is the average radius.

9. Parallel currents attract and anti-parallel currents repel each other.

10. A planar loop carrying a current I, having N closely wound turns, and an area A possesses a magnetic moment m vector where,

m vector = N I A vector

and the direction of m vector is given by the right hand thumb rule :


curl the palm of your right hand along the loop with the fingers pointing in the direction of the current. The thumb sticking out gives the direction of m vector (and A vector)

When this loop is placed in a uniform magnetic field B vector,


the force F vector on it is:


F vector = 0

And the torque on it is,


τ vector = m vector ⨯ B vector

In a moving coil galvanometer, this torque is balanced by a counter-torque due to a spring, yielding

K φ = N I A B


where φ is the equilibrium deflection and k the torsion constant of the spring.

11. An electron moving around the central nucleus has a magnetic moment μl given by:

μl = (e/2m) l

where l is the magnitude of the angular momentum of the circulating electron about the central nucleus. The smallest value of μl is called the Bohr magneton μB and it is μB = 9.27 × 10^−24 J/T

12. A moving coil galvanometer can be converted into a ammeter by introducing a shunt resistance rs, of small value in parallel. It can be converted into a voltmeter by introducing a resistance of a large value in series.

TABLE

Physical Quantity	Symbol	Nature	Dimensions	Units	Remarks
Permeability of free space	μ0	Scalar	[MLT^−2A^−2]	T m A^−1	4 π × 10^−7 T m A^−1
Magnetic Field	B vector	Vector	[M T^−2A^−1]	T (tesla)
Magnetic Moment	M vector	Vector	[L^2A]	A m^2 or J/T
Torsion Constant	k	Scalar	[M L^2T^−2]	N m rad^−1	Appears in MCG

POINTS TO PONDER

1. Electrostatic field lines originate at a positive charge and terminate at a negative charge or fade at infinity. On the other hand, Magnetic field lines always form closed loops.

2. The discussion in this Chapter holds only for steady currents which do not vary with time.

When currents vary with time Newton’s third law is valid only if momentum carried by the electromagnetic field is taken into account.

3. Recall the expression for the Lorentz force,

F vector = q (v vector ⨯ B vector + E vector)

This velocity dependent force has occupied the attention of some of the greatest scientific thinkers. If one switches to a frame with instantaneous velocity v vector, the magnetic part of the force vanishes. The motion of the charged particle is then explained by arguing that there exists an appropriate electric field in the new frame. We shall not discuss the details of this mechanism. However, we stress that the resolution of this paradox implies that electricity and magnetism are linked phenomena (electromagnetism) and that the Lorentz force expression does not imply a universal preferred frame of reference in nature.

4. Ampere’s Circuital law is not independent of the Biot-Savart law. It can be derived from the Biot-Savart law. Its relationship to the Biot-Savart law is similar to the relationship between Gauss’s law and Coulomb’s law.

EXERCISES

Q4.1 A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B vector at the centre of the coil?

A4.1 π × 10^−4 T ≃ 3.1 × 10^−4 T

Q4.2 A long straight wire carries a current of 35 A. What is the magnitude of the field B vector at a point 20 cm from the wire?

A4.2 3.5 × 10^−5 T

Q4.3 A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B vector at a point 2.5 m east of the wire.

A4.3 4 × 10^−6 T, vertical up

Q4.4 A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

A4.4 1.2 × 10^−5 T, towards south

Q4.5 What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T?

A4.5 0.6 N m^−1

Q4.6 A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?

A4.6 8.1 × 10^−2 N; direction of force given by Fleming’s left-hand rule

Q4.7 Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

A4.7 2 × 10^−5 N; attractive force normal to A towards B

Q4.8 A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B vector inside the solenoid near its centre.

A4.8 8 π × 10^−3 T ≃ 2.5 × 10^−2 T

Q4.9 A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

A4.9 0.96 N m

Q4.10 Two moving coil meters, M1 and M2 have the following particulars:


R1 = 10 Ω, N1 = 30,


A1 = 3.6 × 10^−3 m^2, B1 = 0.25 T

R2 = 14 Ω, N2 = 42,


A2 = 1.8 × 10^−3 m^2, B2 = 0.50 T


(The spring constants are identical for the two meters).

Determine the ratio of


(a) current sensitivity and


(b) voltage sensitivity of M2 and M1.

A4.10 (a) 1.4,


(b) 1

Q4.11 In a chamber, a uniform magnetic field of 6.5 G (1 G = 10^−4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 10^6 m s^−1 normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.5 × 10^−19 C, me = 9.1×10^−31 kg)

A4.11 4.2 cm

Q4.12 In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.

A4.12 18 MHz

Q4.13


(a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.


(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area? (All other particulars are also unaltered.)

A4.13


(a) 3.1 Nm,


(b) No, the answer is unchanged because the formula τ = N I A vector × B vector is true for a planar loop of any shape.

ADDITIONAL EXERCISES

Q4.14 Two concentric circular coils X and Y of radii 16 cm and 10 cm, respectively, lie in the same vertical plane containing the north to south direction. Coil X has 20 turns and carries a current of 16 A; coil Y has 25 turns and carries a current of 18 A. The sense of the current in X is anticlockwise, and clockwise in Y, for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.

A4.14 5π × 10^−4 T = 1.6 × 10^−3 T towards west.

Q4.15 A magnetic field of 100 G (1 G = 10^−4 T) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about 10^−3 m^2. The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns m^−1. Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.

A4.15 Length about 50 cm, radius about 4 cm, number of turns about 400, current about 10 A. These particulars are not unique. Some adjustment with limits is possible.

Q4.16 For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,

B = μ0 IR^2 N/{2 (x^2 + R^2)^(3/2)}

(a) Show that this reduces to the familiar result for field at the centre of the coil.

(b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by,

B = 0.72 × μ0 N I/R, approximately.

[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

A4.16 (b) In a small region of length 2d about the mid-point between the coils,

B = (μ0 IR^2N/2) ×[{(R/2 + d)^2 + R^2}^(−3/2) + {(R/2 − d)^2 + R^2}^(−3/2)

≃ (μ0 IR^2 N/2) × (5R^2/4)^(−3/2) × [(1 + 4d/5R)^(−3/2) + (1 − 4d/5R)^(−3/2)]

≃ (μ0 IR^2 N/2R^3) × (4/5)^(3/2) × [1 − 6d/5R + 1 + 6d/5R]

where in the second and third steps above, terms containing d^2/R^2 and higher powers of d/R are neglected since d/R << 1. The terms linear in d/R cancel giving a uniform field B in a small region:

B = (4/5)^(3/2) × (μ0 IN/R) ≃ 0.72 × (μ0 IN/R)

Q4.17 A toroid has a core (non-ferromagnetic) of inner radius 25 cm and outer radius 26 cm, around which 3500 turns of a wire are wound. If the current in the wire is 11 A, what is the magnetic field


(a) outside the toroid,


(b) inside the core of the toroid, and


(c) in the empty space surrounded by the toroid.

A4.17 Hint: B for a toroid is given by the same formula as for a solenoid:


B = μ0 nI , where n in this case is given by n = N/2πr. The field is non-zero only inside the core


surrounded by the windings.


(a) Zero,


(b) 3.0 × 10^−2 T,


(c) zero.


Note, the field varies slightly across the cross-section of the toroid as r varies from the inner to outer radius. Answer (b) corresponds to the mean radius r = 25.5 cm.

Q4.18 Answer the following questions:

(a) A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?

(b) A charged particle enters an environment of a strong and non-uniform magnetic field varying from point to point both in magnitude and direction, and comes out of it following a complicated trajectory. Would its final speed equal the initial speed if it suffered no collisions with the environment?

(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction. Specify the direction in which a uniform magnetic field should be set up to prevent the electron from deflecting from its straight line path.

A4.18


(a) Initial v vector is either parallel or anti-parallel to B vector.


(b) Yes, because magnetic force can change the direction of v vector, but not its magnitude.


(c) B vector should be in a vertically downward direction.

Q4.19 An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field


(a) is transverse to its initial velocity,


(b) makes an angle of 30° with the initial velocity.

A4.19


(a) Circular trajectory of radius 1.0 mm normal to B vector.


(b) Helical trajectory of radius 0.5 mm with velocity component 2.3 × 10^7 m s^−1 along B vector.

Q4.20 A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. In the same region, a uniform electrostatic field is maintained in a direction normal to the common axis of the coils. A narrow beam of (single species) charged particles all accelerated through 15 kV enters this region in a direction perpendicular to both the axis of the coils and the electrostatic field. If the beam remains undeflected when the electrostatic field is 9.0 × 10^−5 V m^−1, make a simple guess as to what the beam contains. Why is the answer not unique?

A4.20 Deuterium ions or deuterons; the answer is not unique because only the ratio of charge to mass is determined. Other possible answers are He(++), Li(+++) , etc.

Q4.21 A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A is set up in the rod through the wires.


(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?


(b) What will be the total tension in the wires if the direction of current is reversed keeping the magnetic field same as before? (Ignore the mass of the wires.) g = 9.8 m s^−2.

A4.21


(a) A horizontal magnetic field of magnitude 0.26 T normal to the conductor in such a direction that Fleming’s left-hand rule gives a magnetic force upward.


(b) 1.176 N.

Q4.22 The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?

A4.22 1.2 N m^−1; repulsive. Note, obtaining total force on the wire as 1.2 × 0.7 = 0.84 N, is only approximately correct because the formula F = (μ0/2πr) I1 I2 for force per unit length is strictly valid for infinitely long conductors.

Q4.23 A uniform magnetic field of 1.5 T exists in a cylindrical region of radius10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,


(a) the wire intersects the axis,


(b) the wire is turned from N-S to northeast-northwest direction,


(c) the wire in the N-S direction is lowered from the axis by a distance of 6.0 cm?

A4.23
(a) 2.1 N vertically downwards
(b) 2.1 N vertically downwards (true for any angle between current direction and B since l sin θ remains fixed, equal to 20 cm)
(c) 1.68 N vertically downwards

Q4.24 A uniform magnetic field of 3000 G is established along the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. 4.28? What is the force on each case? Which case corresponds to stable equilibrium?

Fig. 4.28

A4.24 Use τ vector = I A vector ⨯ B vector and F = I l vector ⨯ B vector
(a) 1.8 × 10^−2 N m along -y direction
(b) same as in (a)
(c) 1.8 × 10^−2 N m along -x direction
(d) 1.8 × 10^−2 N m at an angle of 240° with the +x direction
(e) zero
(f) zero
Force is zero in each case. Case (e) corresponds to stable, and case (f) corresponds to unstable equilibrium.

Q4.25 A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5.0 A, what is the
(a) total torque on the coil,
(b) total force on the coil,
(c) average force on each electron in the coil due to the magnetic field?
(The coil is made of copper wire of cross-sectional area 10^−5 m^2, and the free electron density in copper is given to be about 10^29 m^−3.)

A4.25
(a) Zero,
(b) zero,
(c) force on each electron is evB = IB/(nA) = 5 × 10^−25 N.
Note: Answer (c) denotes only the magnetic force.

Q4.26 A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g lies inside the solenoid (near its centre) normal to its axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of 6.0 A in the wire. What value of current (with appropriate sense of circulation) in the windings of the solenoid can support the weight of the wire? g = 9.8 m s^−2.

A4.26 108 A

Q4.27 A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?

A4.27 Resistance in series = 5988 Ω

Q4.28 A galvanometer coil has a resistance of 15 Ω and the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?

A4.28 Shunt resistance = 10 m Ω

Congratulations! You have completed this chapter. I hope you enjoyed studying this chapter. In case you found any difficulties in this chapter or have any suggestions to improve it, please write to us at ‘blind2Visionary@gmail.com’.

End of Chapter 4 MOVING CHARGES AND MAGNETISM