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Φ21 Fall 2006
HW13 Solutions
1 Problem K32.52
Part A.
What is the magnetic eld at point P in the gure?
Solution:
The eld due to a loop of current is
~ loop = µ0 N I B̂
B
2R
where N is the number of turns and the direction of the eld B̂ is found
by the right hand rule. Divide the current into 4 parts, i.e. two straight
segments and two half loops. The straight segments do not contribute
to the magnetic eld at P because Id~` is parallel to (~r − ~r0 ) in the Biot- Figure 1: Loop of wire carrying
Savart Law. Calling the direction out of the page k̂, the total eld is
a current that causes a magnetic
eld.
¶
µ1
µ
¶
1
~ = µ0
B
2
2I
R1
k̂ −
2I
R2
k̂
=
µ0 (5 A)
4
1
1
−
0.02 m 0.01 m
k̂ = −7.85 × 10−5 T k̂
(Only the magnitude was requested, but you should know the direction!)
2 Problem SP 27.30 Crossed E~ and B~ Fields.
A particle with initial velocity ~v0 = 5.85 × 103 m/s ĵ enters a region of uniform electric and magnetic elds.
~ = 1.35 T k̂ . You can ignore the weight of the particle.
The magnetic eld in the region is B
Calculate the magnitude and direction of the electric eld in the region if the particle is
to pass through undeected, for a particle of charge +0.640 nC.
Parts A and B.
The electric eld must cause a force to exactly cancel the magnetic force. In other words, the
total force should be zero.
Solution:
F~
~
E
~ + qE
~ =0
= q~v × B
³
´
³
´ ³
´
~ = − 5.85 × 103 ĵ × 1.35 k̂ = −7.90 × 103 î
= − ~v × B
Calculate the magnitude and direction of the electric eld in the region if the particle is
to pass through undeected, for a particle of charge -0.320 nC.
Parts C and D.
Solution:
Since the total force is zero, q divides out and the same electric eld is the answer.
3 Problem SP 28.31
(Note that the numbers in this problem have been randomized.)
The gure shows, in cross section, several conductors that carry currents
through the plane of Figure 2. The currents have the magnitudes I1 =
4.10 A, I2 = 5.50 A, and I3 = 2.50 A, and the directions shown. Four
paths, labeled aHthrough d, are shown. Parts A thru D. Find the value
~ · d~` for each path, travelling in a counter-clockwise
of the integral B
direction.
1
Figure 2: Amperian paths for
problem SP 28.31.
~ , dene each path analytOne could theoretically calculate B
ically, and perform messy integrals. Instead, use Ampere's law:
Solution:
I
~ · d~` = µ0 Ienc
B
For each path, use whatever total current passes through the loop. By the right-hand-rule (ngers curl in
the direction of the path, thumb points in the positive direction of positive Ienc ), currents out of the page
are positive and into the page are negative. The integrals for each path are:
I
~ · d~` = µ0 (0) = 0 T · m
B
a
I
b
I
I
~ · d~` = µ0 (−I1 ) = −5.15 × 10−6 T · m
B
~ · d~` = µ0 (−I1 + I2 ) = 1.76 × 10−6 T · m
B
c
~ · d~` = µ0 (−I1 + I2 + I3 ) = 4.90 × 10−6 T · m
B
d
4 Problem OH 11-05
(Note the numbers in this problem have been randomized.)
An electron enters a uniform magnetic eld with magnitude 0.3 T at a
~ . Determine the radius r and pitch p (distance between
45◦ angle to B
loops) of the electron's helical path, assuming its speed is 2 × 106 m/s. Figure 3: The spiral path of a
charged particle moving in a magPart A. Give the radius of the helical path in meters.
netic eld.
~ the z direction. The motions of the
Solution: Call the direction of B
particle in the xy plane and in the z direction can be separated. With a
45◦ angle, the velocity can be split into~v = ~vk + ~v⊥ with ~vk = v cos 45◦ k̂
and v⊥ = v sin 45◦ . (The direction of ~v⊥ will vary in the xy plane.)
Considering only the x and y coordinates (because the magnetic force
~ ), the magnetic force is related to the acceleration for circular motion to nd the
must be perpendicular to B
cyclotron radius.
qv⊥ B
r
Part B.
v2
= F = ma = m ⊥
r
¢
¡
¢¡
¢ ¡√
2/2
9.11 × 10−31 kg 2 × 106 m/s
mv⊥
=
=
= 2.68 × 10−5 m
qB
(1.6 × 10−19 C) (0.3 T)
Find the pitch p of the helical path in meters.
~ . In the xy plane,
This question links the motions of the particle parallel and perpendicular to B
the particle is travelling in circular motion. One revolution takes a time
Solution:
∆t =
2π
2πr
=
v⊥
v⊥
µ
mv⊥
qB
¶
=
¡
¢
2π 9.11 × 10−32 kg
2πm
=
= 1.19 × 10−10 s
qB
(1.6 × 10−19 C) (0.3 T)
~ in that time.
And the pitch is the distance travelled along B
´¡
¡
¢ ³√
¢
p = vk ∆t = 2 × 106 m/s
2/2 1.19 × 10−10 s = 1.69 × 10−4 m
2
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