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Dennis_Churaev [7]
3 years ago
10

if an electron in an electron beam experiences a downward force of 2.0x10^-14N while traveling in a magnetic field of 8.3x10^-2T

west, what are the direction and the magnitude of the velocity?
Physics
1 answer:
11111nata11111 [884]3 years ago
8 0

Answer:

1.51\cdot 10^6 m/s north

Explanation:

When a charged particle moves in a magnetic field, the particle experiences a force given by the formula:

F=qvB sin \theta

where

q is the magnitude of the charge

v is its velocity

B is the magnetic field

\theta is the angle between the directions of v and B

In this problem,

q=1.6\cdot 10^{-19}C (charge of the electron)

B=8.3\cdot 10^{-2} T (strength of magnetic field)

F=2.0\cdot 10^{-14} N (force)

\theta=90^{\circ}

Therefore, the velocity is

v=\frac{F}{qB sin \theta}=\frac{2.0\cdot 10^{-14}}{(1.6\cdot 10^{-19})(8.3\cdot 10^{-2})(sin 90^{\circ})}=1.51\cdot 10^6 m/s

The direction of the force is perpendicular to both the direction of the velocity and the magnetic field, and it can be found using the right-hand rule:

. Thumb: direction of the force (downward) --> however the charge is negative, so this direction must be reversed: upward

- Middle finger: direction of the field (west)

- Index finger: direction of velocity --> north

So, the electron is travelling north.

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Answer:

comparing the anatomy of organisms

Explanation:

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It would mean that the change in motion or the acceleration of an object depends on both the size of the force and the mass of the object. Hence, the correct option is (c).

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3 0
2 years ago
Read 2 more answers
as an aid in understanding this problem. The drawing shows a positively charged particle entering a 0.61-T magnetic field. The p
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Answer:

E = 420.9 N/C

Explanation:

According to the given condition:

Net\ Force = 2(Magnetic\ Force)\\Electric\ Force - Magnetic\ Force = 2(Magnetic\ Force)\\Electric\ Force = 3(Magnetic\ Force)\\qE = 3qvBSin\theta\\E = 3vBSin\theta

where,

E = Magnitude of Electric Field = ?

v = speed of charge = 230 m/s

B = Magnitude of Magnetic Field = 0.61 T

θ = Angle between speed and magnetic field = 90°

Therefore,

E = (3)(230\ m/s)(0.61\ T)Sin90^o

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3 years ago
As a summer intern assigned to the Olympics Kayaking committee, you need to determine the speed of water flowing through a rapid
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Answer:

11.56521 m/s

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V_2 = Velocity in second section

From the continuity equation we get

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The speed of the river through the section of rapids is 11.56521 m/s

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