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ANEK [815]
3 years ago
13

13. A proton moves at 7.50×107 m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path o

f radius 0.800 m. What is the field strength
Physics
1 answer:
diamong [38]3 years ago
5 0

The magnetic force acting on a charged particle moving perpendicular to the field is:

F_{b} = qvB

F_{b} is the magnetic force, q is the particle charge, v is the particle velocity, and B is the magnetic field strength.

The centripetal force acting on a particle moving in a circular path is:

F_{c} = mv²/r

F_{c} is the centripetal force, m is the mass, v is the particle velocity, and r is the radius of the circular path.

If the magnetic force is acting as the centripetal force, set F_{b} equal to F_{c} and solve for B:

qvB = mv²/r

B = mv/(qr)

Given values:

m = 1.67×10⁻²⁷kg (proton mass)

v = 7.50×10⁷m/s

q = 1.60×10⁻¹⁹C (proton charge)

r = 0.800m

Plug these values in and solve for B:

B = (1.67×10⁻²⁷)(7.50×10⁷)/(1.60×10⁻¹⁹×0.800)

B = 0.979T

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Explanation:

The given data is as follows.

    Mass of small bucket (m) = 4 kg

    Mass of big bucket (M) = 12 kg

    Initial velocity (v_{o}) = 0 m/s

    Final velocity (v_{f}) = ?

  Height H_{o} = h_{f} = 2 m

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Thus, we can conclude that the speed with which this bucket strikes the floor is 4.43 m/s.

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