A charge 2 `mu` C is moving at an angle of `60^(@)` with respect to the magnetic field 0.2 T at a speed of `10^(4) ms ^(-1)`. If the electric field `10^(3) "Vm"^(-1)` is at right angles to the magnetic field then find the net field.

Explanation

Given : q = `2 xx 10 ^(-6) C, B = 0.2 T , v = 10 ^(4) "ms"^(-1), E = 10^(3) "Vm"^(-1),` F = ?
We know that , `vecF = q (vecE+vecvxxvecB)`
Now `" " vecvxxvecB = vB sin theta hatk`
`= 0.2 xx 10^(4) xx (sqrt(3))/(2) hatk`
`vecv xx vecB = 1.732 xx 103 hatk`
`vecE_(z) = E cos theta hatk = 10^(3) cos 60^(@) hatk = 0.5 xx 10 ^(3) hatk`
`vecF = 2xx10^(-6)(0.5xx10^(3)hatk+1.732xx10^(3)hatk)`
`vecF = 2xx10^(-6)xx2.232 xx10^(3) hatkN`
i.e.,, `" " vecF = 4.464 xx 10^(-3) veckN`
`|vecF|= 4.464 xx 10^(-3)` N