Asked by anton
1.A proton travels at 4 x 10^6 m/s in a direction that is at an angle of 60 degrees to a magnetic field of 0.5 T. What is the magnitude of the force on the proton in Newtons?
2.What is the acceleration of the proton?
2.What is the acceleration of the proton?
Answers
Answered by
Elena
F=qvBsinα=
=1.6•10⁻¹⁹•4•10⁶•0.5•sin60=
=2.77•10⁻¹³ N
a=F/m=2.77•10⁻¹³/1.67•10⁻²⁷=1.66•10¹⁴ m/s²
=1.6•10⁻¹⁹•4•10⁶•0.5•sin60=
=2.77•10⁻¹³ N
a=F/m=2.77•10⁻¹³/1.67•10⁻²⁷=1.66•10¹⁴ m/s²
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