Download Homework 10 Solution

Homework 10
Exercises:
27.1 Magnetism
27.2 Magnetic Field
27.3 Magnetic Field Lines and Magnetic Flux
27.4 Motion of Charged Particles in a Magnetic Field
27.5 Applications of Motion ion Charged Particles
27.6 Magnetic Force on a Current-Carrying Conductor
27.7 Force and Torque on a Current Loop
Question 1
y
In a region of space, there is a constant magnetic field that is given by
the following vector field.
!
ˆ
B = (10 mT )(iˆ) + (20 mT )( j)
B
At one moment in time, a charge of 5 nC is traveling with the following
velocity.
!
ˆ
v = (8 m/s)(k)
x
a. What is the force vector on the charge at this moment?
z
v
b. What is the angle between the force vector and the x axis?
c. What is the angle between the force vector and the y axis?
d. What is the angle between the force vector and the z axis?
Solution
a. The magnetic force on the charge is
!
! !
ˆ ( (10 mT )(iˆ) + (20 mT )( j)
ˆ )⎤
FB = q(v × B) = (5 nC ) ⎡⎣ (8 m/s)(k)×
⎦
⎡ (iˆ) ( j)
ˆ ⎤
ˆ (k)
⎢
⎥
⎢
⎥
ˆ
ˆ ) = det ⎢ 0 0
ˆ
8(k)× ( (10)(iˆ) + (20)( j)
8 ⎥ = (−160)(iˆ) − (−80)( j)
⎢ 10 20 0 ⎥
⎢
⎥
⎣
⎦
!
y
ˆ⎤
FB = (5 nC ) ⎡⎣ (−160 m/s ⋅ mT )(iˆ) − (−80 m/s ⋅ mT )( j)
⎦
!
ˆ FB = (−800 ×10−12 N )(iˆ) + (400 ×10−12 N )( j)
B
F
x
z
v
page 1
Question 2
A current-carrying swing has a length of L and a width of w. The strings are massless and the
cross bar has a mass of m. The current in the bar is I. A constant magnetic field B is turned on
pointing upward.
I
θ
L
m
w
B
B
a. At what angle will the swing be resting?
Say the magnetic field is angled 30° counter-clockwise.
θ
B
B
b. At what angle will the swing be resting now?
Solution
a. The magnetic field on the current will produce a magnetic force and it will be constant since the
current, the length, and the magnetic field are constant. The force will be this.
!
FB = ILB sin φL&B (iˆ) = IwB sin 90°(iˆ) = IwB(iˆ)
Applying Newton's second law to the bar in the horizontal direction,
!
!
FB + Tx = 0 ⇒ IwB = T sin θ
Applying Newton's second law to the bar in the vertical direction,
!
!
Fg + Ty = 0 ⇒ mg = T cos θ
Together,
tan θ =
IwB
mg
⎛ IwB ⎞⎟
⇒ θ = tan−1 ⎜⎜
⎟
⎜⎝ mg ⎟⎟⎠
page 2
B. The magnetic force is now
!
!
!
!
FB = ILB sin φL&B (30°) = IwB sin 90°(30°) = IwB(30°)
Applying Newton's second law to the bar in the horizontal direction,
!
!
FB,x + Tx = 0 ⇒ IwB cos 30° = T sin θ
Applying Newton's second law to the bar in the vertical direction,
!
!
!
FB,y + Fg + Ty = 0 ⇒ mg = T cos θ + IwB sin 30°
Together,
tan θ =
IwB( 3/ 2)
mg − IwB(1/ 2)
⎛ IwB( 3/ 2) ⎞⎟
⎟⎟
⇒ θ = tan−1 ⎜⎜⎜
⎜⎝ mg − IwB(1/ 2) ⎟⎠
page 3
Question 3
A machine is made of a velocity selector followed by a mass spectrometer. The velocity selector
uses an electric field of 5 N/C and a magnetic field of 1 mT. The mass spectrometer uses a
constant magnetic field of 2.5 mT. A singly charges atom is accelerated to some speed passes
through the velocity selector undeflected, enters the mass spectrometer, and lands a distance of
1.62 meters from the entry point.
+e
velocity selector
v
mass spectrometer
v
Es & Bs
2R & Bm
What kind of atom is it? See section 27.5 for more information.
Solution
The equation that describes the velocity of a charged particle that passes through a velocity
selector is this.
!
! ! !
Fs = q(E − v × Bs ) = 0 ⇒ E = vB
The equation that describes the diameter of the arc that a charged particle will have in the mass
spectrometer is this. In short, the magnetic force on a moving charged particle is the only
contribution to the centripetal force.
mv 2
= qvBm
r
The question about the type of atom is answered by the mass of the particle.
m=
qBmr
qB B r
(1.6 ×10−19 C )(2.5 mT )(1 mT )(0.81 m)
= m s =
v
E
5 N/C
m=
(1.6 ×10−19 C )(2.5 mT )(1 mT )(0.81 m)
= 6.48 ×10−26 kg
5 N/C
This is the mass of an atom so it must be a certain integer number of amu’s (atomic mass units).
m = 6.48 ×10−26 kg = N(amu) ⇒ N =
6.48 ×10−26 kg
= 39
1.66 ×10−27 kg
The atom has an atomic weight of 39 which is potassium.
page 4