Question

An α-particle of mass 6.65x10^-27 kg enter in the magnetic field with the speed 6x 10^5m/s.The strength of magnetic field is 0.2T. Calculate the force on α-particle

1 point

3.84 x10^14

3.84 x10^-14 N

7.24 x10^14 N

4.84 x10^18 N

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Answer 1

Answer:

Solution :

Here, m=6.65×10−27kgm=6.65×10-27kg,

q=+2e=+2×1.6×10−19Cq=+2e=+2×1.6×10-19C,

v=6×105ms−1v=6×105ms-1, B=0.2TB=0.2T, θ=90∘θ=90∘

Force on the αα-particle in the magnetic field is

F=qvBsinθ=(2×1.6×10−19)×(6×105)×0.2×sin90∘F=qvBsinθ=(2×1.6×10-19)×(6×105)×0.2×sin90∘

=3.84×10−14N=3.84×10-14N

Acceleration of the αα-particle =Fm=3.84×10146.65×10−27=Fm=3.84×10146.65×10-27

=5.77×1012ms−1=5.77×1012ms-1