1.

A proton enters in magnetic field of 2500 N/A-m withvelocity 4.0 x105 m/s perpendicularly Calculate the force act-[Ans. 1.6 x 10-10 N]ing on proton.

Answer»

F= qvB charge on protons = 1.602×10−19B= 2500 N/A-mvelocity= 4*10^5then F = 1.602*4*2500*10^-14= 1.6*10^-10 N



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