This document provides 30 problems related to calculating forces and torques in magnetic fields. It covers topics like calculating torque on loops and coils using the torque formula, determining required currents and magnetic field strengths, and converting galvanometers into voltmeters and ammeters by adding resistances. The problems involve setting up and solving equations for torque, current, resistance, and magnetic field based on given parameters like loop/coil area, number of turns, angle with the magnetic field, and desired deflection or measurement values.

1. Chapter 30. Forces and Torques in a Magnetic Field Physics, 6th Edition
Chapter 30. Forces and Torques in a Magnetic Field
Magnetic Torque on a Loop (α is angle that plane of loop makes with B field.)
30-1. A rectangular loop of wire has an area of 30 cm2 and is placed with its plane parallel to a
0.56-T magnetic field. What is the magnitude of the resultant torque if the loop carries a
current of 15 A ? (1 cm2 = 1 x 10-4 m2)
τ = NBIA cos α = (1)(0.56 T)(15 A)(30 x 10-4 m 2 ) cos 00 ; τ = 0.0252 N⋅m
30-2. A coil of wire has 100 turns, each of area 20 cm2. The coil is free to turn in a 4.0 T field.
What current is required to produce a maximum torque of 2.30 N⋅m?
τ 2.30 N ⋅ m
I= = ; I = 2.88 A
NBA cos α (100)(4.0 T)(20 x 10-4 m 2 )(1)
30-3. A rectangular loop of wire 6 cm wide and 10 cm long is placed with its plane parallel to
a magnetic field of 0.08 T. What is the magnitude of the resultant torque on the loop if
it carries a current of 14.0 A. [ A = (0.10 m)(0.06 m) = 6 x 10-3 m2 ]
τ = NBIA cos α = (1)(0.08 T)(14 A)(6 x 10-3 m 2 ) cos 00 ; τ = 6.72 x 10-3 N⋅m
30-4. A rectangular loop of wire has an area of 0.30 m2. The plane of the loop makes an angle
of 300 with a 0.75-T magnetic field. What is torque on the loop is the current is 7.0 A.
τ = NBIA cos α = (1)(0.75 T)(7.0 A)(0.30 m 2 ) cos 300 ; τ = 1.36 N⋅m
30-5. Calculate the B field required to give a 100-turn coil a torque of 0.5 N m when its plane
is parallel to the field. The dimensions of each turn is 84 cm2 and the current is 9.0 A.
τ 0.500 N ⋅ m
B= = ; B = 66.1 mT
NIA cos α (100)(9.0 A)(84 x 10-4 m 2 ) cos 00
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2. Chapter 30. Forces and Torques in a Magnetic Field Physics, 6th Edition
30-6. What current is required to produce a maximum torque of 0.8 N⋅m on a solenoid having
800 turns of area 0.4 m2? The flux density is 3.0 mT. What is the position of the
solenoid in the field? (For solenoid, θ is the angle that coil axis makes with B field).
The torque is a maximum when: θ = 90
τ 0.80 N ⋅ m
I= = ; I = 833 mA
NBA sin θ (800)(0.003 T)(0.4 m 2 ) sin 900
30-7. The axis of a solenoid, having 750 turns of wire, makes an angle of 340 with a 5-mT
field. What is the current if the torque is 4.0 N m at that angle. The area of each turn of
wire is 0.25 m2.
τ 4.00 N ⋅ m
I= = ; I = 7.63 A
NBA sin θ (750)(0.005 T)(0.25 m 2 )sin 340
Galvanometers, Voltmeters, and Ammeters
30-8. A galvanometer coil 50 mm x 120 mm is mounted in a constant radial 0.2-T B field. If the
coil has 600 turns, what current is required to develop a torque of 3.6 x 10-5 N m?
A = (0.05 m)(0.120 m) = 6 x 10-3 m2; θ = 900; B = 0.200 T; τ = 3 x 10-5 N m
τ 3.60 x 10-5 N ⋅ m
I= = ; I = 50.0 µA
NBA sin θ (600)(0.2 T)(6 x 10-3 m 2 ) sin 900
30-9. A galvanometer has a sensitivity of 20 µA per scale division. What current is required to
give full-scale deflection with 25 divisions on either side of the central equilibrium
position?
I = 25 div(20 µ A/div); I = 500 µA
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3. Chapter 30. Forces and Torques in a Magnetic Field Physics, 6th Edition
30-10. A galvanometer has a sensitivity of 15 µA per scale division. How many scale divisions
will be covered by the deflecting needle when the current is 60 µA?
60 µ A
Number = ; Number = 15 div
15 µ A/div
30-11. A certain voltmeter draws 0.02 mA for full-scale deflection at 50 V. (a) What is the
resistance of the voltmeter? (b) What is the resistance per volt?
50 V
Rv = = 2.5 x 106 Ω; Rv = 2.5 MΩ
2 x 10-5 A
R 2.5 x 106 Ω R
= ; = 50.0 kΩ/V
V 50 V V
*30-12. For the voltmeter in Problem 30-11, what multiplier resistance must be used to convert
this voltmeter to an instrument that reads 100 mV full scale?
VB − Vg 150 V - 50 V
Rm = = ; Rm = 5.00 MΩ
If 2 x 10-5 A
*30-13. The coil of a galvanometer will burn out if a current of more than 40 mA is sent through
it. If the coil resistance is 0.5 Ω, what shunt resistance should be added to permit the
measurement of 4.00 A?
I g Rg (40 x 10-3A)(0.500 Ω)
Rs = = ; Rs = 5.05 Ω
I − Ig 4 A - 0.04 A
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4. Chapter 30. Forces and Torques in a Magnetic Field Physics, 6th Edition
*30-14. A current of only 90 µA will produce full-scale deflection of a voltmeter that is designed
to read 50 mV full scale. (a) What is the resistance of the meter? (b) What multiplier
resistance is required to permit the measurement of 100 mV full scale?
50 x 10-3 V 0.100 V - 0.050 V
(a) Rv = ; Rv = 555 Ω (b) Rm = ; Rm = 555 Ω
90 x 10-6 A 90 x 10-6 A
Adding the same resistance as voltmeter resistance doubles the range of the meter.
*30-15. An ammeter that has a resistance of 0.10 Ω is connected in a circuit and indicates a current
of 10 A at full scale. A shunt having a resistance of 0.01 Ω is then connected across the
terminals of the meter. What new circuit current is needed to produce full-scale deflection
of the ammeter? Rg
Ig Is
IsRs = IgRg; Is = I – Ig; (I – Ig) Rs = IgRg
IRs – IgRs = Ig Rg IRs = IgRs + IgRg Solve for I Rs
I
VB RL
I g ( Rg + Rs ) (10 A)(0.1 Ω + 0.01 Ω)
I= = ; I = 110 A
Rs 0.01 Ω
Challenge Problems
30-16. A wire of length 12 cm is made in to a loop and placed into a 3.0-T magnetic field. What is
the largest torque the wire can experience for a current of 6 A? C = 2πR
C 0.12 m
R= = ; R = 0.0191 m; A = π R 2 = π (0.0191 m) 2 ; A = 1.16 x 10-4 m2
2π 2π
τ = NBIA cos α = (1)(3 T)(6 A)(1.16 x 10-4 m 2 ) cos 00 ; τ = 2.09 x 10-3 N⋅m
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5. Chapter 30. Forces and Torques in a Magnetic Field Physics, 6th Edition
30-17. A solenoid has 600 turns of area 20 cm2 and carries a current of 3.8 A. The axis of the
solenoid makes an angle of 300 with an unknown magnetic field. If the resultant torque on
the solenoid is 1.25 N⋅m, what is the magnetic flux density?
τ 1.25 N ⋅ m
B= = ; B = 548 mT
NAI sin θ (600)(20 x 10-4 m 2 )(3.8 A) sin 300
30-18. A flat coil of wire has 150 turns of radius 4.0 cm, and the current is 5.0 A. What angle
does the plane of the coil make with a 1.2-T B field if the resultant torque is 0.60 N⋅m?
A = πR2 = π(0.04 m)2; A = 5.03 x 10-3 m2; I = 5 A; τ = 0.6 N⋅m
τ 0.600 N ⋅ m
cos α = = ; α = 82.40
NBIA (150)(1.2 T)(5 A)(5.03 x 10-3 m 2 )
30-19. A circular loop consisting of 500 turns carries a current of 10 A in a 0.25-T magnetic field.
The area of each turn is 0.2 m2. Calculate the torque when the plane of the loop makes the
following angles with the field: 00, 300, 450, 600, and 900.
(a) τ = NBIA cos α = (500)(0.25 T)(10.0 A)(0.20 m 2 ) cos 00 ; τ = 250 N⋅m
(b) τ = (250 N⋅m) cos 300; τ = 217 N⋅m; (c) τ = (250 N⋅m) cos 450; τ = 177 N⋅m
(d) τ = (250 N⋅m) cos 300; τ = 125 N⋅m; (e) τ = (250 N⋅m) cos 00; τ = 0 N⋅m
30-20. A solenoid of 100 turns has a cross-sectional area of 0.25 m2 and carries a current of 10 A.
What torque is required to hold the solenoid at an angle of 300 with a 40-mT magnetic
field? ( The angle of the solenoid axis is θ = 300 )
τ = NBIA sin θ = (100)(0.04 T)(10.0 A)(0.25 m 2 ) sin 300 ; τ = 5.00 N⋅m
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6. Chapter 30. Forces and Torques in a Magnetic Field Physics, 6th Edition
30-21. A solenoid consists of 400 turns of wire, each of radius 60 mm. What angle does the axis
of the solenoid make with the magnetic flux if the current through the wire is 6 A, the flux
density is 46 mT, and the resulting torque is 0.80 N m?
A = πR2 = π(0.06 m)2 = 0.0113 m2; τ = 0.80 N⋅m; B = 46 mT; I = 6 A; θ = ?
τ 0.800 N ⋅ m
sin θ = = ; θ = 39.80
NBIA (400)(0.046 T)(6 A)(0.0113 m 2 )
*30-22. A galvanometer having an internal resistance of 35 Ω requires 1.0 mA for full-scale
deflection for a current of 10 mA. What multiplier resistance is needed to convert this
device to a voltmeter that reads a maximum of 30 V?
VR 30 V
Rm = − Rg = − 35 Ω; Rm = 29,965 Ω
Ig 0.001 A
*30-23. The internal resistance of a galvanometer is 20 Ω and it reads full scale for a current of
10 mA. Calculate the multiplier resistance required to convert this galvanometer into a
voltmeter whose range is 50 V. What is the total resistance of the resulting meter?
VR 50 V
Rm = − Rg = − 20 Ω; Rm = 4980 Ω
Ig 0.01 A
Rv = Rm + Rg = 4980 Ω + 20 Ω ; Rv = 5000 Ω
*30-24. What shunt resistance is needed to convert the galvanometer of Problem 30-22 to an
ammeter reading 10 mA full scale?
I g Rg (0.001A)(35 Ω)
Rs = = ; Rs = 3.89 Ω
I − Ig 0.01 A - 0.001 A
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7. Chapter 30. Forces and Torques in a Magnetic Field Physics, 6th Edition
*30-25. A certain voltmeter reads 150 V full scale. The galvanometer coil has a resistance of 50
Ω and produces a full-scale deflection on 20 mV. Find the multiplier resistance of the
voltmeter.
VR Vg 0.020 V
Rm = − Rg ; Ig = = = 4 x 10-4 A
Ig Rg 50.0 Ω
VR 150 V
Rm = − Rg = − 50 Ω; Rm = 374,950 Ω
Ig 4 x 10-4 A
*30-26. A galvanometer has a coil resistance of 50 Ω and a current sensitivity of 1 mA (full
scale). What shunt resistance is needed to convert this galvanometer to an ammeter
reading 2.0 A full scale?
I g Rg (0.001A)(50 Ω)
Rs = = ; Rs = 0.025 Ω
I − Ig 2 A - 0.001 A
*30-27. A laboratory ammeter has a resistance of 0.01 Ω and reads 5.0 A full scale. What shunt
resistance is needed to increase the range of the ammeter tenfold?
I I R
= 10; Rs = g g ; IRs − I g Rs = I g Rg ; IRs = I g ( Rs + Rg );
Ig I − Ig
I R + Rg
= s = 10; Rs + Rg = 10 Rs ; Rs (10 − 1) = Rg
Ig Rs
Rg 0.01 Ω
Rs = = ; Rs = 1.11 mΩ
9 9
Thus, by adding 1.11 µΩ in parallel, the meter
range will extend to 50 A full scale.
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8. Chapter 30. Forces and Torques in a Magnetic Field Physics, 6th Edition
*30-28. A commercial 3-V voltmeter requires a current of 0.02 mA to produce full-scale
deflection. How can it be converted to an instrument with a range of 150 V?
VB − Vg 150 V - 3 V
Rm = = ; Rm = 7.35 Ω
Ig 0.02 x 10-3A
Adding Rm = 7.35 Ω in series will increase the range to 150 V.
Critical Thinking Problems
30-29. Consider the rectangular, 4 cm by 6 cm loop of 60 turns of wire in Fig. 30-11. The plane
of the loop makes an angle of 400 with a 1.6-T magnetic field B directed along the x-axis.
The loop is free to rotate about the y-axis and the clockwise current in the coil is 6.0 A.
What is the torque and will it turn toward the x axis or toward the z axis?
y
A = (0.06 m)(0.04 m) = 2.4 x 10-3 m2; α = 400
4 cm
τ = NBIA cos α = (60)(1.6 T)(6 A)(0.0024 m2)cos 400 60 turns
τ = 1.06 N⋅m, The downward direction of the current 6 cm
B
I
in the near segment means that the force producing the
400 x
torque will cause the loop to rotate AWAY from x axis.
τ = 1.06 N⋅m, toward z-axis z
*30-30. A voltmeter of range 150 V and total resistance 15,000 Ω is connected in series with
another voltmeter of range 100 V and total resistance 20,000 Ω. What will each meter
read when they are connected across a 120-V battery of negligible internal resistance?
To determine what the readings will be we need to know the galvanometer currents Ig
required for full-scale deflection of the galvanometer in each voltmeter. These are found
by dividing the full-scale voltages by their galvanometer resistances (Ohm’s law).
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9. Chapter 30. Forces and Torques in a Magnetic Field Physics, 6th Edition
150 V 100 V
*30-30. (Cont.) For A: I gf = = 0.010 A; For B: I gf = = 0.005 A
15,000 Ω 20,000 Ω
Since voltmeters are in series: Iv = Ig 150-V meter A 100-V meter
RgA RgB
GA GB
120 V
Iv = = 3.43 mA ; Ig = 3.43 mA
15 kΩ + 20 kΩ RmA RmB
Iv
Consider reading of Voltmeter A (150-V full):
VA = Iv(RmA + RgA) = (3.43 mA)(15 kΩ)
120-V source C
VA = Iv(RmA + RgA) = (3.43 mA)(20 kΩ)
VA = 51.5 V; VB = 68.6 Ω
*30-31. The magnetic moment µ is a vector quantity whose magnitude for a coil of N turns of
area A is given by NIA, where I is the current in the coil. The direction of the magnetic
moment is perpendicular to the plane of the coil in the direction given by the right hand
thumb rule (see Fig. 29-19). If such a coil is placed into a uniform B field, show that
the torque has a magnitude given by:
τ = µB sin θ
This is sometimes written as a vector (cross) product τ = µ x B.
For a coil or loop: τ = NBIA cos α [ cos α = sin θ ]
Recall that the angle θ is the compliment of the angle α µ
that the plane of the loop makes with the B field, thus the θ B
α
magnetic moment is perpendicular to the plane of the loop.
Simply substitute µ = NIA into τ = (NIA)B sin θ, then write:
τ = µB sin θ
.
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10. Chapter 30. Forces and Torques in a Magnetic Field Physics, 6th Edition
*30-32. Verify the answer obtained for Problem 30-19 by applying the formula derived in the
previous problem. Do not confuse the angle θ with the angle α that the plane of the loop
makes with the B field. [ Recall that θ + α = 900 and cos α = sin θ ]
Thus angles change from: α = 00, 300, 450, 600, and 900 to θ = 900, 600, 450, 300, and 00
The maximum torque is still 250 N⋅m, from τ = NBIA cos α or from τ = NBIA sin θ:
µ = NIA = (500)(10.0 A)(0.20 m2) = 1000 A⋅m2 directed at angle θ
Applying τ = µB sinθ = (1000 A⋅m2)(0.25 T) sin θ for each angle gives:
(a) 250 N⋅m; (b) 217 N⋅m; (c) 177 N⋅m; (d) 125 N⋅m; and (e) 0 N⋅m
*30-33. The internal wiring of a three-scale voltmeter is shown in Fig. 30-12. The
galvanometer has an internal resistance of 40 Ω and a current of 1.00 mA will produce
full-scale deflection. Find the resistances R1, R2, and R3 for the use of the voltmeter for
10, 50, and 100 V. (Multiplier resistances Rm)
R2 R3
Rm =
VR
− Rg =
10 V
− 40 Ω; R1
Ig 0.001A
Rm = 9960 Ω Rg 50 V
V 50 V G 10 V 100 V
Rm = R − Rg = − 40 Ω; Rm = 49,960 Ω
Ig 0.001A
VR 100 V
Rm = − Rg = − 40 Ω; Rm = 99,960 Ω
Ig 0.001A
172