MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 14: MAGNETISM AND E/M INDUCTION
WORKSHEET SOLUTIONS
1. A circular loop carrying a current of 1.42 A is oriented in a magnetic field of 2.74 T. The loop has an
area of 0.17 m2 and is mounted on an axis, perpendicular to the magnetic field, which allows the loop to
rotate. What is the torque on the loop when its plane is oriented at an 18o angle to the field?
1A.
(1)  = BIA(sin
(2)  = (2.74 T)(1.42 A)(0.17 m2)(sin(90-18))o
(3)  = 0.63 N.m
2. A current-carrying wire of length 0.41 m is positioned perpendicular to a uniform magnetic field. If
the current is 12.5 A and it is determined that there is a resultant force of 4.05 N on the wire due to the
interaction of the current and field, what is the magnetic field strength?
2A.
(1) F = BIL(sin
(2) F = BIL
(3) B = F / IL
(4) B = (4.05 N) / (12.5 A)(0.41 m)
(5) B = 0.79 T
3. A charged particle, mass 3.51 x 10-25 kg and charge 3.24 x 10-19 C, moves in a circular orbit
perpendicular to a uniform magnetic field of 0.75 T. Find the time for the proton to make one complete
circular orbit.
3A.
(1) T = (2m) / (qB)
(2) T = (2)(3.51 x 10-25 kg) / (3.24 x 10-11 C)(0.75 T)
(3) T = 9.08 x 10-14 s
4. An electron moves through a region of crossed electric and magnetic fields. The electric field E =
1225 V/m and is directed straight down. The magnetic field B = 1.15 T and is directed to the left. For
what velocity v of the electron into the paper (perpendicular to the plane) will the electric force exactly
cancel the magnetic force?
Example 4. Diagram
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
4A.
(1) FE – FB = 0
(2) FE = FB
(3) FE = Eq
(4) FB = qvB(sin
(5) FB = qvB
(6) Eq = qvB
v = E / B
(8) v = (1225 V/m) / (1.15 T)
(9) v = 1065 m/s
5. A proton is released such that its initial velocity is from left to right across this page. The proton’s
path, however, is deflected in a direction toward the bottom edge of the page due to the presence of a
uniform magnetic field. What is the direction of this field?
(A) out of the page
6. Two parallel cables of a high voltage transmission line carry equal and opposite currents of 2215 A.
The distance between the cables is 1.17 m. What is the magnetic force acting on a 21 meter length of
each cable?
6A.
(1) F / L = oI1I2 / 2r
(2) F = oI1I2L / 2r
(3) F = (4 x 10-7 T.m/A)(2215 A)(2215 A)(21 m)/(2) (1.17m)
(4) F = 55.3 N
7. A square loop (L = 0.200 m) consists of 50 closely-wrapped turns which each carry a current of
0.500 A. The loop is oriented as shown in a uniform magnetic field of 0.40 T directed in the positive y
direction. What is the magnitude of the torque on the loop?
Example 7. Diagram
7A.
(1)  = BnIA(sin
(2)  = (0.40 T)(50)(0.500 A)(0.04m2)(sin 60o)
(3)  = 0.35 N.m
PHYSICS
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
8. A superconducting solenoid is to be designed to generate a magnetic field of 14.7 T. If the solenoid
winding has 1715 turns/meter, what is the required current?
8A.
(1) B = onI
(2) I = B / on
(3) I = (14.7 T) / (4 x 10-7 A.m/T)(1715)
(4) I = 6821 A
9. A positive charge and a negative charge enter a region where the magnetic field is uniform and into
the page (between the poles of a magnet with a square cross section). Describe the deflections of each
charge:
Example 9. Diagram
9A. The positive charge deflects toward the bottom of the page and the negative charge towards the top
of the page. While in the field, their paths are circular arcs.
10. The two charges now travel antiparallel to a uniform magnetic field, as shown below. Describe the
deflections of each charge:
10A. There is no deflection since the magnetic force is zero. The charges continue in straight line
motion.
11. A long straight wire passes through a region where the magnetic field is uniform and directed out of
the page. The current is directed to the top of the page.
Example 11. Diagram
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
11a. What is the direction of the magnetic force on the wire?
A. To the right.
11b. Calculate the magnitude of the force.
A.
(1) F = BIL(sin
(2) F = BIL
(3) F = (0.40 T)(10 A)(0.20 m)
(4) F = 0.80 N
12. A square loop 1.50 m on a side is placed in a magnetic field of 0.150 T. If the field makes an angle
of 12.0o with the normal to the plane of the loop, determine the magnetic flux through the loop.
12A.
(1)  = BA(cos
(2) A = s2
(3) A = (1.50 m)2
(4) A = 2.25 m2
(5)  = (0.150 T)(2.25 m2)(cos 12o)
(6)  = 3.30 x 10-1 T.m2
13. A 125-turn solenoid (r = 0.35 m) is placed in a magnetic field of 0.225 T. The field makes a 75o
angle with the normal to the plane of the loop. The magnetic field then decreases to zero in 0.1 s.
Determine the average voltage induced during this time interval.
13A.
(1)  = BA(cos
(2) A = r2
(3) A = (0.35 m)2
(4) A = 0.385 m2
(5)  = (B)Acos
(6) Bo = 0.225 T
(7) B = 0 T
(8) B = 0.225 T – 0 T
(9) B = 0.225 T
(10)  = (0.225 T)(0.385 m2)(cos 75o)
(11)  = 2.24 x 10-2 T.m2
(12)  = - N / t
(13)  = (125)(2.24 x 10-2 T.m2) / (0.1 s)
(14)  = 28.0 V
14. A wire loop (r = 0.4 m) is placed in a magnetic field of 1.20 T. The field makes a 45o angle to the
plane of the loop. The wire loop is shrunk to a radius of 0.05 m in 0.07 s. Determine the average
voltage induced during this time interval.
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
14A.
(1)  = BA(cos
(2) Ao = r2
(3) Ao = (0.4 m)2
(4) Ao = 5.03 x 10-1 m2
(5) A = r2
(6) A = (0.05 m)2
(7) A = 7.85 x 10-3 m2
(8)  = B(A)cos
(9) A = 5.03 x 10-1 m2 – 7.85 x 10-3 m2
(10) A = 4.95 x 10-1 m2
(11)  = (1.20 T)(4.95 x 10-1 m2)(cos 45o)
(12)  = 4.20 x 10-1 T.m2
(13)  = / t
(14)  = (4.20 x 10-1 T.m2) / (0.07 s)
(15)  = 6.00 V
15. A 0.41 m wire is moved perpendicular to a 0.58 T magnetic field at a speed of 2.50 m/s. What emf
is induced across the ends of the wire?
15A.
(1)  = BLv
(2)  = (0.58 T)(0.41 m)(2.50 m/s)
(3)  = 0.59 V
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTERS 13 - 14: ELECTRIC DC AND MAGNETISM
QUIZ SOLUTIONS
1. A color television set draws about 3.25 A when connected to 120 V. What is the cost (with electrical
energy at 8 cents/kWh) of running the color TV for 14 hours?
1A.
(1) Determine joules used: P = IV
(2) P = (3.25 A)(120 V)
(3) P = 390 W = 390 J/s
(4) (14 hours) (3600 sec / 1 hour) (390 J / 1 sec) = 1.97 x 107 J
(5) 8.64 x 107 J
(1 kWh) = 5.47 kWh
-------------3.6 x 106 J
(6) Determine cost: (5.47 kWh) (8 cents)
(7) Cost = 43.7 cents
2. A wire is utilized to determine the melting point of a metal. The resistance of the wire is 4.00  at
20.0oC and increases to 7.034  as the metal starts to melt.  = 4.78 x 10-3/oC. What is the melting
temperature (in oC)?
2A.
(1) R = Ro(1 + (T – To))
(2) T = [(R – Ro) / (Ro)] + To
(3) T = [(7.034  – 4.00 ) /
(4.78 x 10-3/oC)(4.00 )] + 293 K
(4) T = 158.7 K + 293 K
(5) T = 451.7 K
(6) T = Tc + 273.15
(7) Tc = T – 273.15
(8) Tc = 451.7 – 273.15 = 178.6oC
3. An electric car is designed to run off a bank of 18.0 V batteries with total energy storage of 3.15 x
107 J. If the electric motor draws 9200 W in moving the car at a steady speed of 22 m/s, how far will the
car go (in km) before it runs out of fuel?
3A.
(1) Determine time car can move:
Time = (3.15 x 107 J) (9200 W)
(2) Time = (3.15 x 107 J) (1 sec / 9200 J)
(3) Time = 3424 s
(4) Determine distance car can move: (3424 s) (22 m/s) = 7.53 x 104 m
(5) 7.53 x 104 m (1 km / 1000 m) = 75.3 km
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
4. The heating coil of a hot water heater has a resistance of 27.0 ohm and operates at 180 V. What is its
power rating?
4A.
(1) P = V2 / R
(2) P = (180 V)2 / 27.0 
(3) P = 1200 W
5. A certain material is in a room at 42.0oC. If the absolute temperature K of the material is doubled,
its resistance also doubles. What is the value for , the temperature coefficient of resistivity?
5A.
(1) R = Ro(1 + (T – To))
(2) (T – To) = (R/Ro) – 1
(3) R = 2Ro
(4) (T – To) = (2Ro/Ro) – 1
(5) (T – To) = 1
(6)  = 1 / (T – To)
(7)  = 1 / (630 K – 315 K)
(8)  = 1 / 315 K
(9)  = 3.17 x 10-3 / oC
6. Three resistors, with values of 4.00, 6.00, and 18.0 ohms, respectively, are connected in parallel.
What is the overall resistance of this combination?
6A.
(1) 1 / Req = 1 / R1 + 1 / R2 + 1 / R3
(2) 1 / Req = 1 / 4 + 1 / 6 + 1 / 18
(3) 1 / Req = 9 / 36
(4) 1 / Req = 17 / 36 
(5) Req = (36/ 17) 
Questions 7 - 8. A circuit obeys Ohm’s law. It contains a 12 volt battery and a 4 resistor.
7. The current in this circuit is:
7A.
(1) V = IR
(2) V = IReq
(3) I = V / Req
(4) I = 12V / 4
(5) I = 3.0 A
8. The total power dissipated by the circuit is:
8A.
(1) P = IV
(2) P = (3A)(12V)
(3) P = 36 W
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
9. The following three appliances are connected to a 120 volt house circuit: (i) toaster, 1325 W (ii)
coffee pot, 950 W; and (iii) microwave, 675 W. If all were operated at the same time, what total current
would they draw?
9A.
(1) P = IV
(2) I = P / V
(3) I = 2950 W / 120 V
(4) I = (2950 J/s) / (120 C/J)
(5) I = 24.58 C/s = 24.6 A
10. How long does it take for 1.5 C to pass through a frictionless wire in a 12 amp circuit?
10A.
(1) I = Q / t
(2) t = Q / I
(3) t = 1.5 C / 12 A
(4) t = 0.125 s
11. A metal wire has a radius of 1.45 mm, is 1.85 m long, and has a resistance of 15.0 ohms. What must
be the length of a second wire made of the same type of metal if its radius is 0.10 mm and also has a
resistance of 15.0 ohms?
11A.
(1) R1 = L1 / A
(2)  = R1A / L1
(3) A = r2
(4)  = R1r2 / L1
(5)  = (15 ) (1.45 x 10-3 m)2 / 1.85 m
(6)  = 5.356 x 10-5 .m
(7) R2 = L2 / A
(8) L2 = R2A / 
(9) L2 = R2(r2) / 
(10) L2 = (15 )  (1 x 10-4 m)2 / 5.356 x 10-5 .m
(11) L2 = 8.80 x 10-3 m
(12) L2 = 0.9 cm
12. A circular loop carrying a current of 3.25 A is oriented in a magnetic field of 1.50 T. The loop has
an area of 0.17 m2 and is mounted on an axis, perpendicular to the magnetic field, which allows the loop
to rotate. What is the torque on the loop when its plane is oriented at a 42.0o angle to the field?
12A.
(1)  = BIA(sin
(2)  = (1.50 T)(3.25 A)(0.17 m2)(sin(90- 42))o
(3)  = (1.50 T)(3.25 A)(0.17 m2)(sin(48))o
(4)  = 0.62 N.m
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
13. A current-carrying wire of length 0.472 m is positioned perpendicular to a uniform magnetic field.
If the current is 12.75 A and it is determined that there is a resultant force of 7.21 N on the wire due to
the interaction of the current and field, what is the magnetic field strength?
13A.
(1) F = BIL
(2) B = F / IL
(3) B = (7.21 N) / (12.75 A)(0.472 m)
(4) B = 1.20 T
14. A proton, mass 1.67 x 10-27 kg and charge 1.60 x 10-19 C, moves in a circular orbit perpendicular to
a uniform magnetic field of 3.15 T. Find the time for the proton to make one complete circular orbit.
14A.
(1) T = (2m) / (qB)
(2) T = (2)(1.67 x 10-27 kg) / (1.60 x 10-19 C) (3.15T)
(3) T = 2.08 x 10-8 s
(4) T = 20.8 ns
15. An electron moves through a region of crossed electric and magnetic fields. The electric field E =
1375 V/m and is directed straight down. The magnetic field B = 0.72 T and is directed to the left. For
what velocity v of the electron into the paper (perpendicular to the plane) will the electric force exactly
cancel the magnetic force?
15A.
(1) FE – FB = 0
(2) FE = FB
(3) FE = Eq
(4) FB = qvB(sin
(5) FB = qvB
(6) Eq = qvB
v = E / B
(8) v = (1375 V/m) / (0.72 T)
(9) v = 1910 m/s
16. Two parallel cables of a high voltage transmission line carry equal and opposite currents of 2300 A.
The distance between the cables is 2.15 m. What is the magnetic force acting on a 47.0 meter length of
each cable?
16A.
(1) F / L = oI1I2 / 2d
(2) F = oI1I2L / 2d
(3) F = (4 x 10-7 T.m/A)(2300 A)(2300 A)
(47.0 m) / (2)(2.15 m)
(4) F = 23.1 N
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MR. SURRETTE
VAN NUYS HIGH SCHOOL
17. A square loop 1.35 m on a side is placed in a magnetic field of 0.250 T. If the field makes an angle
of 18.0o with the normal to the plane of the loop, determine the magnetic flux through the loop.
17A.
(1)  = BA(cos
(2) A = s2
(3) A = (1.35 m)2
(4) A = 1.83 m2
(5)  = (0.250 T)(1.83 m2)(cos18o)
(6)  = 4.33 x 10-1 T.m2
18. A wire loop (r = 0.5 m) is placed in a magnetic field of 1.45 T. The field makes a 45o angle to the
plane of the loop. The wire loop is shrunk to a radius of 0.08 m in 0.07 s. Determine the average
voltage induced during this time interval.
18A.
(1) F = BA(cos
(2) Ao = r2
(3) Ao = (0.5 m)2
(4) Ao = 7.85 x 10-1 m2
(5) A = r2
(6) A = (0.08 m)2
(7) A = 2.01 x 10-2 m2
(8)  = B(A)cos
(9) A = 7.85 x 10-1 m2 – 2.01 x 10-2 m2
(10) A = 7.65 x 10-1 m2
(11)  = (1.45 T)(7.65 x 10-1 m2)(cos 45o)
(12)  = 7.85 x 10-1 T.m2
(13)  = /t
(14)  = (7.85 x 10-1 T.m2) / (0.07 s)
(15)  = 11.2 V
19. A 0.52 m wire is moved perpendicular to a 1.72 T magnetic field at a speed of 3.15 m/s.
What emf is induced across the ends of the wire?
19A.
(1)  = BLv
(2)  = (1.72 T)(0.52 m)(3.15 m/s)
(3)  = 2.82 V
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