An electron moving with velocity `vecv_(1)=1hati m//s` at a point in a magnetic field experiences a force `vecF_(1)=-ehatjN` where `e` is the charge of electron.If the electron is moving with a velocity `vecv_(2)=hati-hatj m//s` at the same point, it experiences a force `vecF_(2)=-e(hati+hatj)N`. Find the magnetic field and the force the electron would experience if it were moving with a velocity `vecv_(3)=vecv_(1)xxvecv_(2)` at the same point.

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`