Online Homework 7 Solution
Video Tutor: Current-Carrying Wire in Magnetic Field
Part A The figure shows a wire that is connected to a power supply and
suspended between the poles of a magnet. When the switch is closed, the wire
deflects in the direction shown. Which of the dashed boxes A–D represents the
position of the north magnetic pole?
Sol:)
本題電流方向為流出螢幕，再由右手定則即可判斷磁場方向。
So the answer is “C”.
1
Video Tutor: Magnet and Electron Beam
Part A The figure shows the path of a charged particle moving in a magnetic
field directed into the screen.What is the particle’s charge?
Sol:)
v
v v
FB = qv × B
q<0
q 小於 0 使得 FB 成為圓周運動之向心力。
So the answer is “negative”.
2
Torque on a Current Loop in a Magnetic Field
Part A A current I flows in a plane rectangular current loop with height ω and
v
horizontal sides b. The loop is placed into a uniform magnetic field B in such
v
a way that the sides of length ω are perpendicular to B , and there is an angle
v
θ between the sides of length b and B . Calculate τ , the magnitude of the
torque about the vertical axis of the current loop due to the interaction of the
current through the loop with the magnetic field. Express the magnitude of the
torque in terms of the given variables. You will need a trigonomeric function
[e.g., sin (θ ) or cos(θ ) ]. Use B for the magnitude of the magnetic field.
Sol:)
磁場在長度為 ω 的導線上所造成的力 F1 如下圖。
F1 = IωB
b
由 F1 造成的力矩 τ 1 = 2 IωB cos θ
2
磁場在長度為 b 的導線上所造成的力大小相等方向相反，互相抵消。
因此 τ = IbωB cos(θ )
3
Part B Give a more general expression for the magnitude of the torque τ .
Rewrite the answer found in Part A in terms of the magnitude of the magnetic
dipole moment of the current loop m. Define the angle between the vector
perpendicular to the plane of the coil and the magnetic field to be ψ, noting
that this angle is the complement of angle θ in Part A. Give your answer in
terms of the magnetic moment m, magnetic field B, and ψ.
Sol:)
v v v
The more general vector form of this expression is τ = m × B
v
v
其中， m = IA
So τ = mB sin (φ )
Part C A current I flows around a plane circular loop of radius r, giving the
loop a magnetic dipole moment of magnitude m. The loop is placed in a
v
uniform magnetic field B , with an angle ψ between the direction of the field
lines and the magnetic dipole moment as shown
in the figure. Find an expression for the
magnitude of the torque τ on the current loop.
Express the torque explicitly in terms of r, I, π,
ψ, and B (where τ and B are the magnitudes of
the respective vector quantities). Do not use m.
You will need a trigonomeric function [e.g.,
sin (θ ) or cos(θ ) ].
Sol:)
Area of circle A = r 2π
v
v
v
τ = m× B
τ = mB sin (φ ) = Ir 2πB sin (φ )
4
Problem 27.72
Zeeman effect. In the Bohr model of the hydrogen atom, the electron is
held in its circular orbit of radius τ about its proton nucleus by
electrostatic attraction. If the atoms are placed in a weak magnetic field
v
B , the rotation frequency of electrons rotating in a plane perpendicular
v
to B is changed.
v
Part A Find ∆f , assuming the force due to B is much less than that due to
electrostatic attraction of the nucleus. Express your answer in terms of the
variables B, m, e, and appropriate constants. If there is more than one answer,
separate them by a comma.
Sol:)
未加磁場前，電子繞原子核做圓周運動之向心加速度來自於電子與原子核間的吸
引力。
(r ⋅ 2πf 0 )
ke 2
v2
=
m
=
m
r
r
r2
2
ke
⇒ f0 = 2 3
4π mr
2
v = rω
ω = 2 πf
加磁場後
ke 2
(r ⋅ 2πf ) ⇒ f 2 m qB f - f 2 = 0
± q(r 2πf )B = m
0
2
r
2πm
r
2
f = f 0 + ∆f
( f 0 + ∆ f )2 m
qB
( f 0 + ∆f ) - f 02 = 0
2πm
qB
qB
⇒ f 02 + 2 f 0 ∆f + ∆f 2 m
f0 m
∆f - f 02 = 0
2πm
2πm
neglect
neglect
qB
⇒ ∆f = ±
4πm
5
( ∆f
<< f 0 )
Problem 27.69
A sort of "projectile launcher" is shown in the figure. A large current
moves in a closed loop composed of fixed rails, a power supply, and a
very light, almost frictionless bar
touching the rails. A 0.80 T magnetic
field is perpendicular to the plane of
the circuit.
Part A If the rails are a distance d = 25 cm apart, and the bar has a mass of
1.4 g, what constant current flow is needed to accelerate the bar from rest to
20 m/s in a distance of 1.4 m? Express your answer to two significant figures
and include the appropriate units.
Sol:)
F = IdB = ma
在 1.4 公尺內加速至 20 m/s，a = 142.86 m/s2
由上式可得 I = 1.0 A
Part B In what direction must the field point?
Sol:)
由右手定則可判斷
磁場方向為 ”downward”
6