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?
(e = 1.60 × 10^-19 C, m_proton = 1.67 × 10^-27 kg)

Question

Quizplus · Accepted Answer

The magnetic force provides the centripetal force for circular motion: $ qvB = \frac{mv^2}{r} $. Solving for B gives $ B = \frac{mv}{qr} $. Substituting the given values: $ B = \frac{(1.67 \times 10^{-27} \, \text{kg})(3.0 \times 10^6 \, \text{m/s})}{(1.60 \times 10^{-19} \, \text{C})(0.20 \, \text{m})} = 0.16 \, \text{T} $.

A Proton Having a Speed of 3