MASSCHUSETTS INSTITUTE OF TECHNOLOGY
ESG Physics
8.02 with Kai
Spring 2003
Problem Set 8 Solution
Problem 1: 30-12
Determine the magnetic field (in terms of I, a and b) at the origin due to the current loop
shown below.
Solution:
The magnetic field at point P due to a finite length is
µI
B = 0 ( cos θ1 − cos θ 2 )
4π r
where θ1 and θ 2 are the angles shown above.
(1.1)
So, to find the magnetic field at O, we can use the above formula. We let point P to be
the origin, and θ 2 be the angle nearer to the x-axis.
The problem can be divided into 3 parts:
Part I
Consider the left part of the wire. When the current is infinitely far away from the x-axis,
θ1 → 0 , so
cos θ1 = 1
(1.2)
and
d
cos θ 2 =
(1.3)
2
d + a2
Therefore

µI
µI
d
(1.4)
B1 = 0 ( cosθ1 − cos θ 2 ) = 0 1 −

4π a
4π a 
d 2 + a2 
And B1 is out of page
Part II
Consider the part of the wire which is parallel to the x-axis. Denote θ1 to be the angle on
the left. We have r = d and
a
cos θ1 =
(1.5)
2
a + d2
a
cos θ 2 = cos (π − θ1 ) = − cos θ1 = −
(1.6)
a2 + d 2
Therefore,
B2 =
µ0 I 
a
a
+
 2
4π d  a + d 2
a2 + d 2

µ 0 Ia
=
2
2
 2π d a + d
(1.7)
B2 is into the page
Part III
Consider the part of the wire on the right. It is not difficult to observe that
B3 = B1
(1.8)
As a result, the total magnitude of the magnetic field is
G
G G
G
G G
B tot = B1 + B 2 + B3 = 2B1 + B 2
(1.9)
=
=
?
µ0 I 
µ 0 Ia
d
k
1 − 2
k −
2
2π a 
a +d 
2π d a 2 + d 2
µ0 I
µ 0 Ia
2
2
?
)
(
k
d +a −d k−
2π a a 2 + d 2
2π d a 2 + d 2
µ0 I
d a 2 + d 2 − d 2 − a 2 kˆ
=
2
2
2π ad a + d
(
)
(1.10)
(1.11)
(1.12)
Problem 2: 30-28
In the figure below, both currents are in the negative x direction.
(a) Sketch the magnetic field pattern in the yz plane.
(b) At what distance d along the z axis is the magnetic field a maximum?
Solution
(a)
Currents are into the paper
(b)
Combining the contribution of the magnetic field from each wire,
the z components cancel each other on the z axis. Therefore
G
(2.1)
B = Bˆj
Denote B as the magnitude of the contribution of magnetic field
1
by one of the wire, then
2 µ 0 Iz
 µ I  z 
(2.2)
B = 2 B1 sin θ = 2  0    =
2
2
π
r
r
2

   2π ( a + z )
where r is the distance from the point on the z axis to one of the
wire.
Therefore
B=
µ 0 Iz
π ( a2 + z2 )
(2.3)
dB
= 0 , so we get
dz
dB µ 0 I  1
2z2 
=
−
 2
=0
dz
π  a + z2 a2 + z 2 
To find the maximum of B, we set
(2.4)
µ 0 I  a 2 − z 2 
=0
π  ( a 2 + z 2 )2 
(2.5)
z=a
(2.6)


which gives
Problem 3 30.34
A solenoid 2.50 cm in diameter and 30.0 cm long has 300 turns and carries 12.0 A.
(a) Calculate the flux through the surface of a disk of radius 5.00 cm that is positioned
perpendicular to and centered on the axis of the solenoid.
(b) Figure P30.34b shows an enlarged end view of the same solenoid. Calculate the flux
through the blue area, which is defined by an annulus that has an inner radius of 0.400 cm
and outer radius of 0.800 cm.
Solution:
(a)
G G
Φ B = B ⋅ A = BA
where A is the cross sectional area of the solenoid.. Then
 µ NI 
Φ B =  o  (π r 2 ) = 7.40 µ Wb
 1 
(b)
 µ NI 
Φ B = B ⋅ A = BA =  0  π ( r22 − r12 )
 1 
(
(3.1)
(3.2)
)
(3.3)
which gives
Φ B = 2.27 µ Wb
(3.4)
Problem 4: 30-35
Consider the hemispherical closed surface in Figure P30.35. If the hemisphere is in a
uniform magnetic field that makes an angle θ with the vertical, calculate the magnetic
flux
(a) through the flat surface S1 and
(b) through the hemispherical surface S2
Solution:
(a)
G G
Φ B , flat = B ⋅ A = Bπ R 2 cos (180° − θ ) = − Bπ R 2 cos θ
(b) The net flux out of the closed surface is zero, therefore
Φ B ,curved = −Φ B , flat = Bπ R 2 cos θ
(4.1)
(4.2)
Problem 5: 30.61
Solution:
(a)
B=
µ0 I
= 2.74 × 10−4 T
2π r
(b) At point C, conductor AB produces a field
(5.1)
G
1
2.74 × 10−4 T ) − j , conductor DE
(
2
( )
G
1
2.74 × 10−4 T ) − j , BD produces no field, and AE produces
(
2
G
negligible field. The totle field at C is 2.74 ×10−4 T − j
( )
produces a field of
( )
(c)
FB = IL × B = 1.15 × 10−3 N
(d)
a=
∑ F = ( 0.384ms ) i
m
-2
(5.2)
(5.3)
(e) The bar is already so far from AE that it moves through nearly constant magnetic
field. The force acting on the bar is constant, and therefore the bar’s acceleration
is constant
(f)
this gives
v 2f = vi2 + 2ax
(5.4)
v f = 0.999m/s ( i )
(5.5)
Problem 6 : 30-67
A wire is bent into the shape shown below and the magnetic field is measures at P1 when
the current in the wire is I.
The same wire is then formed into the shape shown below, and the magnetic field is
measured at P2 when the current is again I.
If the total length of wire is the same in each case, what is the ratio of
B1
?
B2
Solution:
For the first figure, we can divide the wire into four equal segments with length l. Then,
we can apply equation (2.1) with θ1 = 90° and θ 2 = 135° for the four segments, so the
total magnetic field at P1 is
1  µ0 I
µ I
 µ I
B1 = 4  0 ( cosθ1 − cos θ 2 )  = 0  0 +
=
2
2π l
 4π l
 πl 
For the second figure, since the length of wire is the same, we have π R = 4l , so
4l
R=
π
For a circular loop of wire, the magnitude of magnetic field at the center is
µ Iθ
B2 = 0
4π R
Since in this case, θ = π , thus
µ I (π ) µ 0 I π
B2 = 0
=
 4l  16l
4π  
π 
Therefore
(7.1)
(7.2)
(7.3)
(7.4)
 µ0 I 
B1  2π l  8 2
=
=
≈ 1.15
B2  µ 0 Iπ  π 2


 16l 
(7.5)
Problem 7: 31-20
Example 2
Consider the arrangement shown below.
Assume that R = 6.00 Ω, l = 1.20 m , and a uniform 2.50-T magnetic field is directed into
the page. At what speed should the bar be moved to produce a current of 0.500 A in the
resistor?
Solution:
Since we know that
ε = Blv
so we can express the current by
ε Blv
I= =
R
R
Substituting the given values, we can find that
v = 1.00 m/s
(9.1)
(9.2)
(9.3)