A proton having a speed of 3.0 × 10⁶ m/s in a direction perpendicular to a uniform magnetic field moves in a circle of radius 0.20 m within the field.What is the magnitude of the magnetic field?
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Question

A proton having a speed of 3.0 × 10⁶ m/s in a direction perpendicular to a uniform magnetic field moves in a circle of radius 0.20 m within the field.What is the magnitude of the magnetic field?

A proton having a speed of 3.0 × 10<sup>6</sup> m/s in a direction perpendicular to a uniform magnetic field moves in a circle of radius 0.20 m within the field.What is the magnitude of the magnetic field? , A) 0.080 T B) 0.16 T C) 0.24 T D) 0.32 T E) 0.36 T

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Quizplus · Accepted Answer

The radius of the circular path can be determined using the formula:
r = mv/qB
where r is the radius, m is the mass of the proton, v is its velocity, q is the charge of the proton, and B is the magnetic field.
Substituting the given values, we get:
0.20 m = (1.67 × 10^-27 kg)(3.0 × 10^7 m/s)/(1.60 × 10^-19 C)B
Solving for B, we get:
B = (1.67 × 10^-27 kg)(3.0 × 10^7 m/s)/(1.60 × 10^-19 C)(0.20 m)
B = 0.16 T
Therefore, the magnitude of the magnetic field is 0.16 T, which is option B.

A Proton Having a Speed of 3