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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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Solution:

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