NEET2004PhysicsMagnetic Effects of CurrentActual
A proton enters a magnetic field of intensity 1.5 ~Wb / m ^2 with a velocity 2 10^7 ~m / s in a direction at an angle 30^ with the field. The force on the proton will be (charge on proton is 1.6 10⁻¹⁹ C )
Options
- A2.4 10⁻¹² ~N
- B4.8 10⁻¹² ~N
- C1.2 10⁻¹² ~N
- D7.2 10⁻¹² ~N
Correct answer
A. 2.4 10⁻¹² ~N
Step-by-step solution
Here: q=1.6 10⁻¹⁹ C , B=1.5 ~Wb / m ^2 , v=2 10^7 ~m / s , =30^ or 30^ = 1 2 Force on proton is given by F=q v B aligned & =1.6 10⁻¹⁹ 2 10^7 1.5 1 2 & =2.4 10⁻¹² ~N aligned