MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics: 8.02
Group Problem Solution: Self Inductance Toroid
Calculate the self-inductance L of a toroid with a square cross section with inner radius
a , outer radius b = a + h , (height h ) and N square windings.
Solution: We first use Ampere’s Law to find the direction and magnitude of the magnetic
field inside the toroid, as a function of distance r from the central axis, ( a < r < b ). The
magnetic field is tangential so we choose a circle of radius r centered around the
symmetry axis of the toroid with a < r < b , for the Amperian loop. We calculate the line
integral in the counterclockwise direction.
Then the value of the line integral is
!
!
" B ! ds = " Bds = B " ds = Bl = B2# r
All the N turns of current at r = a flow out of the plane of the figure, and hence cut
through the disk defined by our Amperian loop. By our choice of circulation direction,
this current contribute a positive amount and the enclosed current is
I enc =
""
! !
J ! dA = NI .
surface
Note that the current at r = b passes into the plane but is outside the Amperian loop. We
can apply Ampere’s Law to find the magnetic field as a function of r
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics: 8.02
! !
! !
" B ! d s = µ0 "" J ! dA .
closed
loop
surface
B2! r = µ0 NI .
Because the value for the magnetic field is positive, the magnetic field points in the same
counterclockwise direction as we chose for calculating the line integral around the circle.
So the magnitude of the magnetic field for the region a < r < b is
B=
µ0 NI
.
2! r
Outside this region the magnetic field is zero.
We next calculate the magnetic flux through one turn
!turn
b
! !
µ NI b
= # B " dA = h # Bdr = h 0 ln
2$
a
turn
a
The total magnetic flux is therefore
!total = N!turn = h
µ0 N 2 I b
ln .
2"
a
The self-inductance is
L=
!total
µ N2 b
= h 0 ln .
I
2"
a
As a check we can calculate the total energy stored in the magnetic field by integrating
the magnetic field density over the volume of the toroid. Since the magnetic field is nonuniform, in order to calculate the total magnetic energy, we need to integrate over the
volume of the toroid. We use as a volume element a cylindrical shell of radius r ,
thickness dr , and height h . The integral is now only over the variable r .
U mag
1
=
2 µ0
b
!
2
B dVvol
all space
2
1
=
B 2 h2" rdr
!
2 µ0 a
b
hµ0 N 2 I 2
h" # µ0 NI &
=
rdr =
µ0 !a %$ 2" r ('
4"
2 2
dr hµ0 N I
b
! r = 4" ln a
a
b
MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics: 8.02
The total energy stored in the magnetic field U mag =
1 2
LI so we can determine the self2
induction
L=
2U mag
I2
=
2hµ0 N 2 I 2
4! I 2
in agreement with our previous calculation.
ln
2
b hµ0 N
b
=
ln
a
2!
a