BITSAT2021PhysicsMagnetic Effects of CurrentActual
A proton moving with a velocity 3 10⁵ ~m / s enters a magnetic field of 0.3 tesla at an angle of 30^ with the field. The radius of curvature of its path will be ( e / m for proton =10⁸ C / kg )
Options
- A2 ~cm
- B0.5 ~cm
- C0.02 ~cm
- D1.25 ~cm
Correct answer
B. 0.5 ~cm
Step-by-step solution
r= m v B e = 3 10⁵ 30^ 0.3 10⁸ 3 10⁵ 1 2 3 10⁷ =0.5 10⁻²m=0.5 ~cm