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Object 1 in the diagram at the right represents a proton moving perpendicularly to a magnetic field of magnitude 0.55T at a speed of 5.6 x 10^3 m/s.

a. Draw and label a vector showing the direction of the force on the proton.
b. Calculate the magnitude of the force.

Object 1 in the diagram at the right represents a proton moving perpendicularly to-example-1

1 Answer

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Answer:

Step-by-step explanation:

a. The direction of the force on the proton can be determined using the right-hand rule for magnetic fields. If the right hand is oriented such that the fingers point in the direction of the magnetic field (from north to south in this case) and the thumb points in the direction of the proton's velocity (to the right in this case), then the palm of the hand will point in the direction of the force on the proton. Using this rule, we can see that the force on the proton points downward, as shown in the diagram below:

N

|

<----|---->

| | |

| | |

|____|____|

|

S

b. The magnitude of the force on the proton can be calculated using the formula:

F = qvB

where F is the magnitude of the force, q is the charge of the particle, v is its velocity, and B is the magnetic field strength. For a proton, which has a charge of +1.6 x 10^-19 coulombs, the magnitude of the force is:

F = (1.6 x 10^-19 C)(5.6 x 10^3 m/s)(0.55 T) = 4.5 x 10^-14 N

Therefore, the magnitude of the force on the proton is 4.5 x 10^-14 N.

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