AP EAMCET202018 Sep 2020Morning ShiftPhysicsMagnetic Effects of CurrentActual
A proton moving with a velocity (2.5 10^7 ~m / s ), enters a magnetic field of intensity (2.5 ~T ) making an angle (30^ ) with the magnetic field. The force on the proton is
Options
- A(3 10⁻¹² ~N )
- B(5 10⁻¹² ~N )
- C(6 10⁻¹² ~N )
- D(9 10⁻¹² ~N )
Correct answer
B. (5 10⁻¹² ~N )
Step-by-step solution
Velocity of proton, (v=25 10^7 ~m / s ) ( Magnetic field, aligned B & =2.5 ~T & =30^ aligned ) Magnetic force on proton in magnetic field is given as ( aligned F & =B q v & =2.5 1.6 10⁻¹⁹ 2.5 10^7 30^ & =6.25 1.6 10⁻¹² 1 2 =5 10⁻¹² ~N aligned )