Question

A proton, with a charge Q = 1.60 × 10−19 Coulombs has an initial velocity as shown below with a magnitude of v = 2.50 × 105 m/s. Take the proton mass to be M = 1.67 × 10−27 kg. The proton moves in a plane perpendicular to a 3.00 Tesla uniform magnetic field

Calculate the radius, in meters, of the circular path followed by the proton. With respect to the magnetic field lines, are the orbits in the clockwise or counterclockwise sense?

Calculate in seconds the amount of time needed for the proton to complete a circular orbit.

Calculate the frequency in Hertz of the circular orbits.

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Answer #1

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A proton with a charge qp = +e = +1.60 × 10–19 C moves with a
speed v = U × 105 m/s East through the Earth’s magnetic field,
which has a value of B = 55.0 ?T and points northward at a
particular location. What is the strength of the magnetic force F
when the proton moves eastward?

An electron and a proton are moving in circular orbits in the
earth’s magnetic field directly above the earth’s geographic north
pole, high above atmosphere, where the field’s magnitude is 40
µT.
a) Looking down on the orbits from above, what is the sense of
the orbit for each of the particles—clockwise or
counterclockwise?
b) What is the period of the orbit for the proton?
c) What is the frequency of the orbit for the electron

A proton of mass mp=1.67×10−27
kgmp=1.67×10−27 kg and a
charge of qp=1.60×10−19 Cqp=1.60×10−19
C is moving through vacuum at a constant velocity of
10,000 m/s10,000 m/s directly to the east when it
enters a region of uniform electric field that points to the south
with a magnitude of ∣E⃗∣=∣E⃗∣=2520 N/C
N/C . The region of uniform electric field is 5
mm wide in the east-west direction. How far (in meters) will the
proton have been deflected towards the south by...

11.- A proton accelerated from rest by a potential difference
acquires a speed of 5000 m / s with which it enters a region in
which there is a uniform magnetic field of 0.5 T perpendicular to
the direction in which it moves the proton.
a. Make a diagram of the forces and trajectory of the proton.
b. Determine the radius of the circular trajectory that the proton
follows within this region.
c. Determine the time it takes to complete...

An electron (m = 9.11×10?31 kg, q = 1.60×10?19 C) travels around
a 1.7 mm radius circular orbit perpendicular to a 2.8 T magnetic
field. Its speed is: (a) 0.16c, (b) 0.36c, (c) 0.94c, (d) c, (e)
2.8c.

A proton (e = 1.60 X 10-19 C and m = 1.67
X 10-27 kg) is traveling at 90.0° with respect to the
direction of a magnetic field of strength 4.50 mT experiences a
magnetic forceof 7.50 X 10-17 N. The
proton's kinetic energy is:

A proton (q = 1.6 X 10-19 C, m = 1.67 X 10-27 kg) moving with
constant velocity enters a region containing a constant magnetic
field that is directed along the z-axis at (x,y) = (0,0) as shown.
The magnetic field extends for a distance D = 0.75 m in the
x-direction. The proton leaves the field having a velocity vector
(vx, vy) = (3.9 X 105 m/s, 1.9 X 105 m/s).
1)What is v, the magnitude of the velocity...

An ionized helium atom has a mass of 6.6 × 10-27 kg
and a speed of 7.1 × 105 m/s. It moves perpendicular to
a 0.94-T magnetic field on a circular path that has a 0.016-m
radius. Determine whether the charge of the ionized atom is
+e or +2e.

A proton, with mass 1.67 × 10-27 kg and charge +1.6 × 10-19 C,
is sent with velocity 7.1 × 104 m/s in the +x direction into a
region where there is a uniform electric field of magnitude 730 V/m
in the +y direction. What are the magnitude and direction of the
uniform magnetic field in the region, if the proton is to pass
through undeflected? Assume that the magnetic field has no
x-component and neglect gravitational effects.
Draw a...

A particle with positive charge q = 4.49 10-18 C moves with a
velocity v = (3î + 5ĵ − k) m/s through a region where both a
uniform magnetic field and a uniform electric field exist.
(a) Calculate the total force on the moving particle, taking B =
(5î + 2ĵ + k) T and E = (5î − ĵ − 2k) V/m. (Give your answers in N
for each component.)
Fx = N Fy = N Fz =...

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