Question

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?​

Answer 1

Step-by-step 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 =
`((2i+3j+4k).(.5k))/(√(2^2+3^2+4^2)* .5 )`

=
`(2)/(2.69)`

cosθ = .74

θ = 42° .