A proton (charge q = 1.6 10^-19 C, mass m = 1.67 10^-27 kg) enters a uniform magnetic field B = 0.5 T perpendicular to the field lines with speed v = 2.0 * 10^6 m/s. What is the radius of its circular trajectory?
Answer and explanation
Correct answer: B. 4.18 cm
- Step 1: The magnetic Lorentz force provides the centripetal force: q v B = m * v^2 / r.
- Step 2: Radius formula: r = m v / (q B).
- Step 3: Substitute values: r = (1.67 10^-27 2.0 10^6) / (1.6 10^-19 * 0.5).
- Step 4: r = (3.34 10^-21) / (0.8 10^-19) = 0.04175 m = 4.18 cm.
Answer
4.18 cm.
Why not the others: 2.09 cm uses B = 1.0 T; 8.35 cm uses v = 4.0 * 10^6 m/s; 1.05 cm divides by 4.
Practise more Electromagnetism, Wave Optics and Modern Physics.
Free questions with an explanation for every one, and full timed Bagrut (Secondary School Matriculation) mock tests.
More Electromagnetism, Wave Optics and Modern Physics questions
- A 1.0 kg block attached to a horizontal spring oscillates with simple harmonic motion. When its displacement is x = 0.05 m, its speed is v =…
- Two identical spheres A and B undergo an elastic oblique collision in two dimensions. Sphere A is initially moving at speed v and sphere B i…
- What is the period of a simple pendulum of length L = 0.98 m in a location where g = 9.8 m/s^2 for small amplitude oscillations?
- A 0.1 kg bullet traveling horizontally at 400 m/s embeds into a 3.9 kg wooden block initially at rest on a frictionless table. What fraction…
- In a damped harmonic oscillator, the mechanical energy decreases by 10% in each full cycle. After 3 complete cycles, what percentage of the…
- A mass m suspended from a vertical spring oscillates with frequency f. If the spring is cut in half and the same mass m is suspended from on…