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A proton is entering magnetic field B = 0.2i - 0.5j + 0.4k. If the proton...

A proton is entering magnetic field B = 0.2i - 0.5j + 0.4k. If the proton moves at v = 4000j - 5000k in meters per second, find the force vector acting on the proton. Proton charge is +1.6 x 10^-19 C.

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Answer #1

The magnetic force on a proton is

F = q vxB

    = 1.6 x 10^-19 C (4000j - 5000k )( 0.2i - 0.5j + 0.4k.)

= 1.6 x 10^-19 C (4000 ( 0.2 ( jxi) - 4000 ( 0.5 ) jxj + 4000 ( 0.4) jxk - 5000 ( 0.2) kxi + 5000(0.5) kxj-5000(0.4)kxk

= 1.6 x 10^-19 C (4000 ( 0.2 ( -k) - 4000 ( 0.5 ) 0+ 4000 ( 0.4) i - 5000 ( 0.2) j+ 5000(0.5) (-i)-5000(0.4)0

= 1.6 x 10^-19 C (800(-k)+1600 i-1000 j-2500i)

=1.6 * 10 ^-19 C ( -900 i-1000 j- 800 k)

=(-1440 * 10 ^-19 i -1600 * 10 ^-19 j -1280 * 10 ^-19 k)N

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