Download Name: Magnetic Field and Lorentz Force

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Magnetic Field and Lorentz Force - Practice
1. A proton traveling at 23.0° with respect to the direction of a magnetic field of strength 2.60 mT
experiences a magnetic force of 6.50 × 10-17 N. Calculate the proton's speed. [e = 1.602 × 10-19 C]
2.

m
An electron that has a velocity of v  2.0 x 106 î  3.0 x 106 ĵ
moves through the uniform
s

magnetic field B  0.030 î  0.15 ĵ T . Find the force on the electron due to the magnetic field.
[me = 9.11 × 10-31 kg]
3. An electron accelerated from rest through potential difference V1 = 1.00 kV enters the gap
between two parallel plates having separation d = 20.0 mm and potential difference V2 = 100 V. The
lower plate is at the lower potential. Neglect fringing and assume that the electron's velocity
vector is perpendicular to the electric field vector between the plates. In unit-vector notation,
what uniform magnetic field allows the electron to travel in a straight line in the gap?
4. A particle moves along a counterclockwise circular path in a region of
uniform magnetic field of magnitude B = 4.00 mT. The particle is either a
proton or an electron (you must decide which). It experiences a magnetic
force of magnitude 3.20 × 10-15 N. [mp = 1.67 × 10-27 kg]
A. What is the particle?
B. What is the particle's speed?
C. What is the radius of the circle?
D. What is the period of the motion?
5. An electron of kinetic energy 1.20 keV circles in a plane perpendicular to a uniform magnetic field.
The orbit radius is 25.0 cm. [1 eV = 1.602×10−19 J]
A. Find the electron's speed.
B. Find the magnitude of the magnetic field.
C. Find the circling frequency.
D. Find the period of the motion.
Solutions:
1. 3.99 x 105 m/s
4. A. electron
2. 6.2 x 10-14 N k̂
B. 4.99 x 106 m/s
3. -2.67 x 10-4 T k̂
C. 7.10 x 10-3 m
D. 8.93 x 10-9 s
5. A. 2.05 x 107 m/s
B. 4.67 x 10-4 T
C. 1.31 x 107 Hz
D. 7.65 x 10-8 s