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a proton has a mass of 1.67x10^-27 kg and a charge of +1.60x10^-19C it moves in a magnetic field of strength 0.800 T in a circle of radius 0.0150 m. what is the velocity of the proton

User Antoneta
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1 Answer

4 votes

Answer:

The velocity of the proton is approximately 9.28 meters per second.

Step-by-step explanation:

To find the velocity of the proton moving in a circle in a magnetic field, we can use the equation for the magnetic force on a charged particle moving perpendicular to a magnetic field:

F = q * v * B

where:

F is the magnetic force on the particle,

q is the charge of the particle (in Coulombs),

v is the velocity of the particle (in m/s), and

B is the magnetic field strength (in Tesla).

The magnetic force provides the necessary centripetal force to keep the proton moving in a circle. The centripetal force is given by:

F_c = (m * v^2) / r

where:

m is the mass of the proton (in kg), and

r is the radius of the circle (in meters).

Since the magnetic force (F) is equal to the centripetal force (F_c), we can set these two equations equal to each other and solve for the velocity (v):

q * v * B = (m * v^2) / r

Now, let's plug in the given values:

q (charge of the proton) = +1.60x10^-19 C

B (magnetic field strength) = 0.800 T

m (mass of the proton) = 1.67x10^-27 kg

r (radius of the circle) = 0.0150 m

Now, we can solve for v:

1.60x10^-19 C * v * 0.800 T = (1.67x10^-27 kg * v^2) / 0.0150 m

Let's solve for v:

1.60x10^-19 C * 0.800 T * 0.0150 m = 1.67x10^-27 kg * v^2

v^2 = (1.60x10^-19 C * 0.800 T * 0.0150 m) / 1.67x10^-27 kg

v^2 = 1.44x10^-25 C T m / 1.67x10^-27 kg

v^2 ≈ 8.632x10^1 m^2/s^2

v ≈ √(8.632x10^1 m^2/s^2)

v ≈ 9.28x10^0 m/s

v ≈ 9.28 m/s

So, the velocity of the proton is approximately 9.28 meters per second.

User Sheepgobeep
by
8.1k points
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