Set 6. Chpt. 29 – Due at 11:00 pm Thurs., Apr. 22, 2010
Page 783. HWR 7th edition (2005). 29: 6, 11, 15, 31, 41 and 43.
Chapter 29.
6.
In Fig. 29-37, a wire forms a semicircle of radius R = 9.26 cm and two
(radial) straight segments each of length L = 13.1 cm. The wire
carries current i = 34.8 mA. What are the (a) magnitude and (b)
direction (into or out of the page) of the net magnetic field at the
semicircle’s center of curvature C?
(a) Recalling the straight sections discussion in Sample Problem 29-1, we see that the
current in the straight segments collinear with C do not contribute to the field at that
point.
Eq. 29-9 (with φ = π) indicates that the current in the semicircular arc contributes µ 0i 4 R
to the field at C. Thus, the magnitude of the magnetic field is
B=
µ 0i
4R
=
(4π ×10 −7 T ⋅ m A)(0.0348A)
= 1.18 ×10 −7 T.
4(0.0926m)
(b) The right-hand rule shows that this field is into the page.
11.
Figure 29-41 show a proton moving at velocity v = (-200 m/s) j
toward a long straight wire with current i = 350 mA. At the instant
shown, the proton’s distance from the wire is d = 2.89 cm. In unitvector notation, what is the magnetic force on the proton due to the
current?
We assume the current flows in the +x direction and the particle is at some distance d in
the +y direction (away from the wire). Then, the magnetic field at the location of a proton
r µi
$ Thus,
with charge q is B = 0 k.
2πd
r
r r µ iq r
F = qv × B = 0 v × k$ .
2πd
e
j
e j
r
In this situation, v = v − $j (where v is the speed and is a positive value), and q > 0.
Thus,
r µ iqv
F= 0
2πd
µ 0iqv ˆ
(4π ×10 −7 T ⋅ m A)(0.350A)(1.60 ×10−19 C)(200m/s) ˆ
ˆ
ˆ
−j ×k = −
i =−
i
2πd
2π (0.0289 m)
ˆ
= (−7.75 ×10 −23 N)i.
15.
(( ) )
In Fig. 29-45, four long straight wire are perpendicular to the page,
and their cross sections form a square of edge length a = 20 cm. The
currents are out of the page in wires 1 and 4 and into the page in wires
2 and 3, and each wire carries 20 A. In unit-vector notation, what is
the net magnetic field at the square’s center?
Each wire produces a field with magnitude given by B = µ0i/2πr,
where r is the distance from the corner of the square to the center.
According to the Pythagorean theorem, the diagonal of the square
has length 2a , so r = a 2 and B = µ 0i 2πa . The fields due to
the wires at the upper left and lower right corners both point toward
the upper right corner of the square. The fields due to the wires at
the upper right and lower left corners both point toward the upper
left corner. The horizontal components cancel and the vertical
components sum to
Btotal
−7
2 µ 0 i 2 ( 4π × 10 T ⋅ m A ) ( 20 A )
=4
=
= 8.0 × 10−5 T.
cos 45° =
πa
π ( 0.20 m )
2 πa
µ 0i
In the calculation cos 45° was replaced with 1 2 . The total field
r
ˆ
points upward, or in the +y direction. Thus, Btotal = (8.0 ×10−5 T)j.
31.
In Fig. 29-45, four long straight wires are perpendicular to the page,
and their cross sections form a square of edge length a = 13.5 cm.
Each wire carries 7.50 A, and the current are out of the page in wires
1 and 4 and into the page in wires 2 and 3. In unit-vector notation,
what is the net magnetic force per meter of wire length on the wire 4?
r
r
r
r
We use Eq. 29-13 and the superposition of forces: F4 = F14 + F24 + F34 . With θ = 45°, the
situation is as shown next:
r
The components of F4 are given by
µ 0i 2
F4 x = − F43 − F42 cos θ = −
2πa
−
µ 0i 2 cos 45°
−
µ 0i 2 sin 45° µ 0i 2
2 2πa
3µ 0i 2
=−
4πa
and
F4 y = F41 − F42 sin θ =
µ 0i 2
2πa
2 2πa
=
4πa
.
Thus,
12
2 12
4y
F4 = ( F + F
2
4x
)
  3µ i 2  2  µ i 2  2 
=  − 0  +  0  
 4πa   4πa  
10 ( 4π ×10 −7 T ⋅ m A ) ( 7.50A )
10 µ 0i 2
=
=
4 πa
4π ( 0.135m )
2
= 1.32 ×10−4 N/m.
r
and F4 makes an angle φ with the positive x axis, where
φ = tan −1
FG F IJ = tan FG − 1IJ = 162° .
H 3K
HF K
4y
−1
4x
In unit-vector notation, we have
r
F1 = (1.32 ×10 −4 N/m)[cos162°ˆi + sin162°ˆj] = (−1.25 ×10 −4 N/m)iˆ + (4.17 ×10 −5 N/m)jˆ
41.
A 200-turn solenoid having a length of 25 cm and a diameter of 10 cm
carries a current of 0.29 A. Calculate the magnitude of the magnetic
field B inside the solenoid.
It is possible (though tedious) to use Eq. 29-26 and evaluate the contributions (with the
intent to sum them) of all 200 loops to the field at, say, the center of the solenoid. This
would make use of all the information given in the problem statement, but this is not the
method that the student is expected to use here. Instead, Eq. 29-23 for the ideal solenoid
(which does not make use of the coil diameter) is the preferred method:
B = µ 0in = µ 0i
FG N IJ
H lK
where i = 0.30 A, l = 0.25 m and N = 200. This yields B = 0.0030 T.
43.
A toroid having a square cross section, 5.00 cm on a side, and an inner
radius of 15.0 cm has 500 turns and carries a current of 0.800 A. (It is
made up of a square solenoid – instead of a round one as in Fig. 29-17
– bent into a doughnut shape.) What is the magnetic field inside the
toroid at (a) the inner radius and (b) the outer radius?
(a) We use Eq. 29-24. The inner radius is r = 15.0 cm, so the field there is
B=
µ 0iN
2 πr
c4π × 10
=
−7
hb
gb g = 5.33 × 10
T ⋅ m / A 0.800 A 500
b
g
2 π 0150
.
m
(b) The outer radius is r = 20.0 cm. The field there is
B=
µ 0iN
2 πr
c4π × 10
=
−7
hb
gb g = 4.00 × 10
T ⋅ m / A 0.800 A 500
b
g
2 π 0.200m
−4
T.
−4
T.