Question

A beam of protons moves in a circle of radius 0.36 m. The
protons move perpendicular to a 0.44-T magnetic field.
**(a)**What is the speed of each proton?
**(b)** Determine the magnitude of the centripetal
force that acts on each proton.

Answer #1

A beam of protons moves in a circle of radius 0.53 m. The
protons move perpendicular to a 0.21 T magnetic field. What is the
speed of each proton?

In a certain cyclotron a proton moves in a circle of radius
0.540 m. The magnitude of the magnetic field is 1.60 T. (a) What is
the oscillator frequency? (b) What is the kinetic energy of the
proton?

In a certain cyclotron a proton moves in a circle of radius
0.590 m. The magnitude of the magnetic field is 1.30 T.
(a) What is the oscillator frequency?
(b) What is the kinetic energy of the proton?

A beam of protons (proton mass is 1.67 x 10 - 27 kg) moves at 3
x 10 5 m/s through a uniform magnetic field with magnitude 2 T.
The magnetic field has exactly equal components along the positive
y and negative x axes and no component along the z axis. The
velocity of each proton lies in the xz- plane at an angle of 30
0 to the z- axis.
(a) Write the magnetic field B and the velocity...

Constants
In (Figure 1), a beam of protons moves through a uniform
magnetic field with magnitude 2.0 T , directed along the positive
z axis. The protons have a velocity of magnitude
3.0×105 m/s in the x-z plane at an
angle of 30 ∘ to the positive z axis. Find the force on a
proton. The charge of the proton is q=+1.6×10−19C.
SOLUTION
SET UP We use the right-hand rule to find the
direction of the force. The force acts...

True or False?
( ) A charge with a velocity perpendicular to a uniform B field
will move in a circular path in a plane perpendicular to the B
field.
( ) A charge with a velocity perpendicular to a uniform B field
will experience a magnetic force of F=qvB
( ) For a mass m to maintain a circular motion on an orbit of
radius r, the needed centripetal force is mv/r.
( ) For a mass m to...

A proton moves at 8.00 ✕ 107 m/s perpendicular to a
magnetic field. The field causes the proton to travel in a circular
path of radius 0.750 m. What is the field strength (in T)?

A beam of protons is moving in the +x direction with a
speed of 13 km/s through a region in which the electric field is
perpendicular to the magnetic field. The beam of protons is not
deflected in this region. The magnetic field has a magnitude of 0.8
T and points in the +y direction. Therefore, the magnitude
of the electric field is 11.570 V/m (= (13 x 10³ m/s) (0.8T)) and
the direction is along the negative z-axis.
What...

A beam of protons traveling at 1.20 km/s enters a uniform
magnetic field, traveling perpendicular to the field. The beam
exits the magnetic field, leaving the field in a direction
perpendicular to its original direction (the figure (Figure 1)).
The beam travels a distance of 1.10 cm while in the field. a. What
is the magnitude of the magnetic field?

An electron moves in a circular path perpendicular to a magnetic
field of magnitude 0.265 T. If the kinetic energy of the electron
is 4.50 ✕ 10−19 J, find the speed of the electron and
the radius of the circular path.
(a) the speed of the electron
m/s
(b) the radius of the circular path
μm

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