Question

An electron at point AAA in (Figure 1) has a speed v0v0 of 1.44×106 m/sm/s ....

An electron at point AAA in (Figure 1) has a speed v0v0 of 1.44×106 m/sm/s .

Figure

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.

nothing

TT

SubmitRequest Answer

Part C

Find the time required for the electron to move from AAA to BBB.

Express your answer in seconds.

nothing

ss

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.

nothing

TT

SubmitRequest Answer

Provide Feedback

Next

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
An electron at point A in the figure has a speed vo of 1.41 x 106...
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...
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...
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 ....
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...
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...
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)...
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...
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...
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...
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...