Find the force on a proton moving with velocity (2i+3j+4k)10^5m/s in a uniform magnetic field of 0.5k T. What is the angle between the magnetic field lines and the velocity?

Explanation:

Force on charge particles

F = q ( v x B )

= 1.6 x 10⁻¹⁹ x [ ( 2i+3j+4k) x .5k ] x 10⁵

=  1.6 x 10⁻¹⁴ x  (  1.5 i - j )

= (2.4 i - 1.6 j ) x 10⁻¹⁴ N

magnitude of this vector

= 2.88 x 10⁻¹⁴ N

Angle between B and v

cosθ =

=

cosθ = .74

θ  = 42° .  

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