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PHYS 222
Worksheet 21 – Inductance
Supplemental Instruction
Iowa State University
Alek Jerauld
PHYS 222
Dr. Paula Herrera-Siklódy
3/6/12
Useful Equations
 2  M
M
di1
di
and 1  M 2
dt
dt
N2 B 2 N1 B1

i1
i2
N B
i
 self   L di
L
dt
Mutually induced emf’s
Mutual Inductance
Self inductance
Self induced emf
Related Problems
1) Two coils have mutual inductance of M= 3.25×10−4 H. The current I_1 in the first coil
increases at a uniform rate of 860 A/s. What is the magnitude of the induced emf in the
second coil? Suppose that the current described is in the second coil rather than the first.
What is the magnitude of the induced emf in the first coil? (Book 30.1)
di
  M  0.280 V
dt
2) At the instant when the current in an inductor is increasing at a rate of 6.45×10−2 A/s,
the magnitude of the self-induced emf is 1.65×10−2 V. What is the inductance of the
inductor If the inductor is a solenoid with 405 turns, what is the average magnetic flux
through each turn when the current is 0.725 A? (Book 30.7)

di
 self  L  L  self  0.256 H
dt
di / dt
N
di
,  L
i
dt
di
i
   iN   
 4.58(104 ) Wb
dt
Ndi / dt
L
3) The inductor in the figure has inductance 0.260 H and carries a current in the direction
shown that is decreasing at a uniform rate, di/dt= - 1.80×10−2 A/s. Find the self-inducted
emf. Which end of the inductor is at a higher potential? (Book 30.9)
 self   L di  4.68(103 ) V
dt
[a] will have a higher potential since the current is decreasing, going to the left.
4) A solenoid coil with 26 turns of wire is wound tightly around another coil with 330 turns.
The inner solenoid is 23.0 cm long and has a diameter of 2.30 cm. At a certain time, the
current in the inner solenoid is 0.130 A and is increasing at a rate of 1700 A/s. (Book 30.4)
(a) For this time, calculate the average magnetic flux through each turn of the inner
solenoid.
Average flux through each turn in the inner coil:
 I  di 2  2 ave
d 2 N
 2  BA  0

 0 I 2 2  9.74(108 ) Wb
di 4
turn
4 l2
(b) For this time, calculate the mutual inductance of the two solenoids;
The flux through the outer coil is the same as the inner coil (this is because we assume the
diameter of the two coils are the same):
  N d 2 
N1  0 2 2 
4
N
N
  1.95(105 ) H
M 1 1 1 2 
i2
i2
i2
(c) For this time, calculate the emf induced in the outer solenoid by the changing current in
the inner solenoid.
 self  M di1  1.95(105 )1700  3.31(102 ) V
dt