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A proton with a kinetic energy of 4.8×10⁻¹⁶ J Part A moves perpendicular to a magnetic field of 0.34 T. What is the radius of its circular path? Express your answer using two significant figures.

User TlonXP
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Final answer:

The radius of the proton's circular path is approximately 2.85 × 10^-4 m.

Step-by-step explanation:

To find the radius of the circular path, we can use the equation for the centripetal force experienced by a charged particle moving in a magnetic field. The centripetal force is provided by the magnetic force:

F = qvB

where F is the magnetic force, q is the charge of the particle, v is its velocity, and B is the magnetic field.

Since the particle is moving in a circular path, the magnetic force is equal to the centripetal force:

F = mv^2 / r

where m is the mass of the particle and r is the radius of the circular path.

Since the charge of a proton is 1.602 × 10^-19 C and its mass is 1.67 × 10^-27 kg, we can rearrange the equations and solve for the radius:

r = mv / qB

Plugging in the given values:

r = (1.67 × 10^-27 kg) × (4.8×10^-16 J) / (1.602 × 10^-19 C) × (0.34 T)

r ≈ 2.85 × 10^-4 m

Therefore, the radius of the proton's circular path is approximately 2.85 × 10^-4 m.

User Marouane Afroukh
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