An electron moving with a velocity `vecV_(1)=2hati`m/s at a point in a magnetic field experience a force `vecF_(1)=-2hatJN`. If the electron is moving with a velocity `vecV_(2)=2hatj`m/s at the same point, it experiences a force `vecF_(2)=+2hatiN`. The force the electron would experience if it were moving with a velocity `vecV_(3)=2hatk`m/s at the same point is

Answer

`vecB=hatk,vecF=0`

Explanation

`vecF=qvecvxxvecB`, Let `vecB=B_(x)hati+B_(y)hatj+B_(z)hatk`
`vecF_(1)=e(1 hati)xx{B_(x)hati+B_(y)hatj+B_(z)hatk}=-ehatj`
`rArr eB_(y)hatk-eB_(y)hatj=-ehatj`
`rArr B_(y)=0,B_(z)=1`
`vecF_(2)=e(hati-hatj)xx{B_(x)hati+1 hatk}=-e(hati+hatj)`
`rArr e(-hatj+V_(z)hatk-hati)=-e((hati+hatj)`
`rArr B_(x)=0`
`rArr vecB=1hatk =hatk wb//m^(2)`
Now `vecv_(3)=vecv_(1)xxvec_(2)=1hatixx(hati-hatj)=-hatk`
Now `vecF=evecv_(3)xxvecB=e(-hatkxxhatk)=0`