Exercise 29.1 Hints: Getting Started | I'm Stuck A proton moves with a speed of 2.0 × 106 m/s horizontally, at a right angle to a magnetic field. What magnetic field strength is required to just balance the weight of the proton and keep it moving horizontally? B =

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Exercise 29.1
Hints: Getting Started | I'm Stuck
A proton moves with a speed of 2.0 × 106 m/s horizontally, at a right angle to a
magnetic field. What magnetic field strength is required to just balance the weight
of the proton and keep it moving horizontally?
B =
Transcribed Image Text:Exercise 29.1 Hints: Getting Started | I'm Stuck A proton moves with a speed of 2.0 × 106 m/s horizontally, at a right angle to a magnetic field. What magnetic field strength is required to just balance the weight of the proton and keep it moving horizontally? B =
Problem An electron in a television picture tube moves toward
the front of the tube with a speed of 8.8 x 106 m/s along the x
axis (Fig. 29.5). The neck of the tube is surrounded by a coil of
wire that creates a magnetic field of magnitude 0.050 T, directed
at an angle of 60° to the x axis and lying in the xy plane.
Calculate the magnetic force on and acceleration of the electron.
Strategy Use Equation 29.2.
Figure 29.5 The magnetic force F, on
the electron is in the negative z direction
when v and B lie in the xy plane.
Solution
Using Equation 29.2, we find the magnitude of the magnetic force.
FB = 1g|vB sin e = (1.60 x 10-19 C)(8.8 x 106 m/s)(0.050 T)(sin 60°)
FB =
N
Because v x B is in the positive z direction (from the right-hand rule) and the charge is negative, Fg is in the
negative z direction.
Once we have determined the magnetic force, we have a problem where the electron is a particle under a net
force and the acceleration is determined from Newton's second law. The mass of the electron is m, = 9.1 x 10-31
kg, and so the acceleration is in the negative z direction and is calculated as follows.
Fв
a =
me
Fв
9.1 x 10-31 kg
m/s?
Transcribed Image Text:Problem An electron in a television picture tube moves toward the front of the tube with a speed of 8.8 x 106 m/s along the x axis (Fig. 29.5). The neck of the tube is surrounded by a coil of wire that creates a magnetic field of magnitude 0.050 T, directed at an angle of 60° to the x axis and lying in the xy plane. Calculate the magnetic force on and acceleration of the electron. Strategy Use Equation 29.2. Figure 29.5 The magnetic force F, on the electron is in the negative z direction when v and B lie in the xy plane. Solution Using Equation 29.2, we find the magnitude of the magnetic force. FB = 1g|vB sin e = (1.60 x 10-19 C)(8.8 x 106 m/s)(0.050 T)(sin 60°) FB = N Because v x B is in the positive z direction (from the right-hand rule) and the charge is negative, Fg is in the negative z direction. Once we have determined the magnetic force, we have a problem where the electron is a particle under a net force and the acceleration is determined from Newton's second law. The mass of the electron is m, = 9.1 x 10-31 kg, and so the acceleration is in the negative z direction and is calculated as follows. Fв a = me Fв 9.1 x 10-31 kg m/s?
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