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disa [49]
3 years ago
9

You have a horizontal cathode-ray tube (CRT) for which the controls have been adjusted such that the electron beam should make a

single spot of light exactly in the center of the screen. You observe, however, that the spot is deflected to the right. It is possible that the CRT is broken. But as a clever scientist, you realize that your laboratory might be in either an electric or a magnetic field. Assuming that you do not have a compass, any magnets, or any charged rods, how can you use the CRT itself to determine whether the CRT is broken, is in an electric field, or is in a magnetic field
Physics
1 answer:
Mama L [17]3 years ago
4 0

Answer:

Please look in the explanation section

Explanation:

If the beam is in a magnetic field, a simple check is to rotate the CRT a quarter of a turn. If the beam is to the right, a magnetic field should be produced. This is because a magnetic field would be up or down that will allow the beam to be pushed to the right or left. In conclusion, if turning the CRT a quarter of a turn, nothing should happen because the CRT is perpendicular to the magnetic field.

A check to see if the beam is in an E field, you should rotate the CRT a quarter of a turn and see if the beam is centered. If you turn the CRT again another quarter of a turn and see if the beam is bent in the opposite direction. If both tests are true, then the CRT must be in an E field. Otherwise, the CRT will be broken.

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If two micro coulomb of charge is flowing in a circuit for 5 minutes. What is the amount of current in the circuit?
Murrr4er [49]

Answer:

6.67×10¯⁹ A

Explanation:

From the question given above, the following data were obtained:

Quantity of electricity (Q) = 2 μC

Time (t) = 5 mins

Current (I) =?

Next, we shall convert 2 μC to C. This can be obtained as follow:

1 μC = 1×10¯⁶ C

Therefore,

2 μC = 2 μC × 1×10¯⁶ C / 1 μC

2 μC = 2×10¯⁶ C

Next, we shall convert 5 mins to seconds. This can be obtained as follow:

1 min = 60 secs

Therefore,

5 min = 5 min × 60 sec / 1 min

5 mins = 300 s

Finally, we shall determine the current in the circuit. This can be obtained as follow:

Quantity of electricity (Q) = 2×10¯⁶ C

Time (t) = 300 s

Current (I) =?

Q = It

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I = 2×10¯⁶ / 300

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2 years ago
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