Question

An electron at point AAA in (Figure 1) has a speed v0v0 of
1.44×10 1 of 1 |
Part A Part complete Find the direction of the magnetic field that will cause the electron to follow the semicircular path from AAA to BBB. The magnetic field must go into the page. SubmitPrevious Answers Correct Part B Find the magnitude of the magnetic field that will cause the electron to follow the semicircular path from AAA to BBB. Express your answer in teslas.
SubmitRequest Answer Part C Find the time required for the electron to move from AAA to BBB. Express your answer in seconds.
SubmitRequest Answer Part D What magnetic field would be needed if the particle were a proton instead of an electron? Express your answer in teslas.
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Answer #1

An electron at point A in the figure has a speed
vo of 1.41 x 106 m/s. Find (a) the
magnitude and direction of the magnetic field that will cause the
electron to follow the semicircular path from A to B the diaimeter
of the semicircular is 10cm and (b)the time required for the
electron to move from A to B. (c) What magnetic field would be
needed if the particle were a proton instead of an electron?
what is...

An electron at point A in the below figure has a speed,
v0, of 2.5 X 106 m/s.
15cm from point A to B
Find the magnitude and direction of the magnetic field,
B, that will cause the electron to follow the
semi-circular path from point A to point B
Calculate the time required for the electron to move from point
A to point B
Repeat part (a) replacing the electron with a Helium nucleus
(He2+, 2 protons and 2...

A straight vertical wire carries a current of 1.60 AA downward
in a region between the poles of a large electromagnet where the
field strength is 0.550 TT and is horizontal.
Part A
Part complete
What are the direction of the magnetic force on a 1.20 cmcm
section of this wire if the magnetic-field direction is toward the
east?
The direction is southward.
SubmitPrevious Answers
Correct
Part B
What are the magnitude of the magnetic force on a 1.20 cmcm...

14 of 18 Constants A proton is traveling horizontally to the
right at 4.80×106 m/s . Part A Find (a)the magnitude and (b)
direction of the weakest electric field that can bring the proton
uniformly to rest over a distance of 3.50 cm . E =
1.317•10−81.317{\cdot}10^{-8} N/C SubmitPrevious AnswersRequest
Answer Incorrect; Try Again; 2 attempts remaining Part B θ = ∘
counterclockwise from the left direction SubmitRequest Answer Part
C How much time does it take the proton to...

An electron with a speed of 6.00×106 m/sm/s, collides
with an atom. The collision excites the atom from its ground state
(0 eVeV) to a state with an energy of 3.40 eVeV.
What is the speed of the electron after the collision?
Express your answer in meters per second.

A conducting disk with radius aaa, thickness h , and
resistivity ρ is inside a solenoid of circular cross section. The
disk axis coincides with the solenoid axis. The magnetic field in
the solenoid is given by B=bt, with b a constant.
Part A
Find the expression for the current density in the disk as a
function of the distance r from the disk center.
Express your answer in terms of the variables a, b, h, ρ, and
t.
j...

A straight, 2.50 mm wire carries a typical household current of
1.50 AA (in one direction) at a location where the earth's magnetic
field is 0.550 gauss from south to north.
Figure
of 0
Part A
Part complete
Find the direction of the force that our planet's magnetic field
exerts on this wire if is oriented so that the current in it is
running from west to east.
--
The force is directed upward.
SubmitPrevious AnswersRequest Answer
Correct
Part B...

An electron that has a velocity with x component 1.9 × 106 m/s
and y component 3.4 × 106 m/s moves through a uniform magnetic
field with x component 0.033 T and y component -0.15 T. (a) Find
the magnitude of the magnetic force on the electron. (b) Repeat
your calculation for a proton having the same velocity.

An electron that has a velocity with x component 2.0 × 106 m/s
and y component 3.9 × 106 m/s moves through a uniform magnetic
field with x component 0.035 T and y component -0.21 T. (a) Find
the magnitude of the magnetic force on the electron. (b) Repeat
your calculation for a proton having the same velocity.

The net force on a current loop in a uniform magnetic
field is zero. But what if B? is not
uniform? The figure (Figure 1) shows a square loop of wire that
lies in the xy-plane. The loop has corners at
(0,0),(0,L),(L,0), and (L,L)
and carries a constant current I in the clockwise
direction. The magnetic field has no x-component but has
both y- and z-components: B?
=(B0z/L)j^+(B0y/L)k^,
where B0 is a positive constant.
Part G
Find the magnitude of the...

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