Problem 5.41 Determine the mutual inductance between the circular loop and the
linear current shown in Fig. P5.41.
y
a
x
d
I1
Figure P5.41: Linear conductor with current I1 next to
a circular loop of radius a at distance d (Problem 5.41).
Solution: To calculate the magnetic flux through the loop due to the current in the
conductor, we consider a thin strip of thickness dy at location y, as shown. The
magnetic field is the same at all points across the strip because they are all equidistant
(at r = d + y) from the linear conductor. The magnetic flux through the strip is
dΦ12 = B(y) · ds = ẑ
=
µ0 I
· ẑ 2(a2 − y2 )1/2 dy
2π (d + y)
µ0 I(a2 − y2 )1/2
dy
π (d + y)
1
L12 =
dΦ12
I S
Z
µ0 a (a2 − y2 )1/2 dy
=
π y=−a
(d + y)
Z
Let z = d + y → dz = dy. Hence,
p
Z
a2 − (z − d)2
µ0 d+a
L12 =
dz
π z=d−a
z
p
Z
µ0 d+a (a2 − d 2 ) + 2dz − z2
dz
=
π d−a
z
Z √
µ0
R
dz
=
π
z
where R = a0 + b0 z + c0 z2 and
a0 = a2 − d 2
b0 = 2d
c0 = −1
∆ = 4a0 c0 − b20 = −4a2 < 0
From Gradshteyn and Ryzhik, Table of Integrals, Series, and Products (Academic
Press, 1980, p. 84), we have
Z √
Z
Z
√
b0
dz
dz
R
√
√ .
+
dz = R + a0
z
z
z R
R
For
√ ¯¯d+a
R¯
z=d−a
=
¯d+a
p
¯
a2 − d 2 + 2dz − z2 ¯
z=d−a
= 0 − 0 = 0.
dz
√ , several solutions exist depending on the sign of a0 and ∆.
z R
For this problem, ∆ < 0, also let a0 < 0 (i.e., d > a). Using the table of integrals,


d+a
Z
dz
2a0 + b0 z 
1
√ = a0  √
sin−1  q
a0
−a0
z R
z b2 − 4a c
For
Z
0
0 0
z=d−a
¶¸d+a
·
µ 2
2
p
−1 a − d + dz
2
2
= − d − a sin
az
z=d−a
p
= −π d 2 − a2 .
dz
√ , different solutions exist depending on the sign of c0 and ∆.
R
In this problem, ∆ < 0 and c0 < 0. From the table of integrals,
¸
·
Z
2c0 z + b0 d+a
b
b0
dz
−1
√ = 0 √
sin−1 √
z
2
−c0
−∆ z=d−a
R
µ
·
¶¸d+a
d −z
= −d sin−1
= π d.
a
z=d−a
For
Z
Thus
i
p
µ0 h
· π d − π d 2 − a2
πh
i
p
= µ0 d − d 2 − a2 .
L12 =