Physics 2220 – Module 07 Homework
01.
A proton moves in the magnetic field B = 0.50 i T with a speed of 1.0 × 10 7 m/s in the directions shown in
the figure. For each, what is the magnetic force F on the proton? Give your answers in component
form.
The force on a charged particle moving in an external magnetic field will follow:
⃗
⃗
F = q ⃗v × B
(a)
The angle between the velocity vector and magnetic field vector is 45 degrees.
−19
F = (1.6 × 10
F = qvB sin (θ)
7
o
−13
C) (1.0 × 10 m/s) (0.50 T) sin (45 ) = 5.7 × 10 N
The velocity and magnetic field vectors form a plane on the x-z plane. Using the right hand rule,
the force is pointing in the positive y-direction.
F = (5.7 × 10−13 ̂j) N
(b)
The angle between the velocity vector and magnetic field vector is 90 degrees.
−19
F = (1.6 × 10
F = qvB sin (θ)
7
o
−13
C) (1.0 × 10 m/s) ( 0.50 T) sin (90 ) = 8.0 × 10 N
The velocity and magnetic field vectors form a plane on the x-y plain. Using the right hand rule,
the force is pointing in the positive z-direction.
F = (8.0 × 10−13 k̂ ) N
02.
An electron moves in the magnetic field B = 0.50 i T with a speed of 1.0 × 10 7 m/s in the directions
shown in the figure. For each, what is the magnetic force F on the proton? Give your answers in
component form.
(a)
The angle between the velocity vector and magnetic field vector is 90 degrees.
−19
F = (1.6 × 10
F = qvB sin (θ)
7
o
−13
C) (1.0 × 10 m/s) ( 0.50 T) sin (90 ) = 8.0 × 10 N
The velocity and magnetic field vectors form a plane on the x-z plane. Using the right hand rule,
the force is pointing in the negative y-direction.
̂ N
F = (−8.0 × 10−13 k)
(b)
This one is a bit more tricky to eyeball the angle between the two vectors.
The velocity vector can be written like:
⃗v = v x ̂i + v y ̂j + v z k̂
v x = 0 m/s
v y = −v cos (45o )
v z = v sin (45o )
Now solve for the force vector:
⃗
F = q ⃗v × ⃗
B
̂ × (0.50 ̂i T)
⃗ = qv (−cos (45 ) ̂j + sin (45o ) k)
F
⃗ = (−1.6 × 10−19 C) (1.0 × 10 7 m/s) (cos (45o ) (0.50 T) k̂ + sin (45 o) (0.50 T) ̂j)
F
⃗
F = (−5.7 × 10−13) k̂ + (−5.7 × 10−13 ̂j) N
̂
⃗ = (−5.7 × 10−13 N) ( ̂j + k)
F
o
03.
An electron travels with v = 5.0 × 106 i through a point in space where the electric field is E = (2.0 × 105 i
– 2.0 × 105 j) V / m and the magnetic field is B = 0.10 k T. What is the force on the electron?
Use the Lorentz Force Equation:
⃗
⃗)
F = q (⃗
E + ⃗v × B
Electric field part:
F⃗E = q ⃗
E
F⃗E = (−1.60 × 10 C) (2.0 × 105 ̂i − 2.0 × 10−5 ̂j) N/C
F⃗E = (−3.2 × 10−14 ̂i + 3.2 × 10−14 ̂j) N
−19
Magnetic field part:
F⃗B = q ⃗v × ⃗
B
−19
6
F⃗B = (−1.60 × 10 C) ((5.0 × 10 ̂i m/s) × (−0.10 k̂ T))
̂i × k̂ = − ̂j
−19
F⃗B = (−1.60 × 10 C) (5.0 × 10 6 m/s) (−0.10 T) (−̂j)
F⃗B = −8.0 × 10−14 ̂j N
Combined:
−14
−14
−14 ̂
⃗
F = (−3.2 × 10 ̂i + 3.2 × 10 ̂j) N − 8.0 × 10
j N
−14 ̂
−14 ̂
⃗ = (−3.2 × 10
F
i − 4.8 × 10
j) N
04.
A circular area with a radius of 8.00 cm lies in the xy-plane. Assume the direction of the area vector is in
the +z direction. Calculate the magnitude of the magnetic flux through the circle due to a uniform
magnetic field of magnitude B = 0.450 T if the field is directed:
Recall the equation for the special case of a uniform magnetic field:
Φ B = BA cos (θ)
(a)
In the +z direction
The angle between the magnetic field vector and area vector will be 0º
Φ B = BA cos (θ) = (0.450 T) (π (0.08 m)2 ) cos (0ο ) = 9.05 × 10−3 Wb
(b)
At an angle 42.1º from the positive +z direction
The angle between the magnetic field vector and area vector will be 42.1º
Φ B = BA cos (θ) = (0.450 T) (π (0.08 m)2 ) cos (42.1 ο ) = 6.71 × 10−3 Wb
(c)
In the +y direction
The angle between the magnetic field vector and area vector will be 90º
2
ο
Φ B = BA cos (θ) = (0.450 T) (π (0.08 m ) ) cos (90 ) = 0 Wb
(d)
In the -x direction
The angle between the magnetic field vector and area vector will be 90º
Φ B = BA cos (θ) = (0.450 T) (π (0.08 m)2 ) cos (90ο ) = 0 Wb
05.
An alpha particle (a helium nucleus containing two protons, two neutrons and mass 6.64 × 10-27 kg) is
traveling horizontally at 35.6 km /s and enters a uniform, vertical magnetic field with magnitude of 1.20
T. What is the radius of the path followed by the alpha particle?
The particle will follow a circular path, so we can apply Newton's 2nd Law to the circular motion and
equate it to the magnetic force.
mv 2
r
−27
mv (6.64 × 10 kg) (35000 m/s)
r=
=
= 6.05 × 10−4 m
−19
qB
(2 (1.60 × 10 C)) (1.20 T)
F = qvB = mar =
06.
Radio astronomers detect electromagnetic radiation at 45 MHz from an interstellar gas cloud. They
suspect this radiation is emitted by electrons spiraling in a magnetic field. What is the magnetic field
strength inside the gas cloud?
Use the cyclotron frequency:
B=
2 π mf cycl
q
qB
f cycl = ω =
2π 2πm
2 π (9.11 × 10−31 kg) (45 × 10 6 Hz)
=
1.6 × 10−19 C
B = 1.6 × 10−3 T
07.
What magnetic field strength and direction will levitate the 2.0 g wire in the figure?
The magnetic force (upward) on the wire needs to
cancel out the force of gravity (downward).
F⃗B = I ⃗
L×⃗
B
F⃗G = m ⃗g
Using the right hand rule, for the magnetic force to be
upward, the magnetic field must be pointing 90
degrees from the direction of the current. Therefore,
the direction of the magnetic field must be out of
the page.
Since both forces are on the same axis, the vector notation can be dropped and the magnetic field
strength can be solved.
F⃗B = F⃗G
o
ILB sin (90 ) = mg
(0.002 kg) (9.8 m/s 2)
mg
B=
=
= 0.131 T
−2
IL
(1.5 A) (10 × 10 m)
08.
A square current loop of 5.0 cm on each side carries a 500 mA current. The loop is in a 1.2 T uniform
magnetic field. The axis of the loop, perpendicular to the plane of the loop, is 30 o away from the field
direction. What is the magnitude of the torque on the current loop?
Use the definition of torque in terms of the magnetic moment and magnetic field.
−3
τ = IABsin (θ) = (500 × 10
⃗
⃗τ = ⃗
μ×⃗
B=I ⃗
A×B
−2
2
o
−4
A) (5.0 × 10 m) (1.2 T) sin (30 ) = 7.5 × 10 N m