A charged particle carrying charge `q=1muc` moves in uniform magnetic with velocity `v_1=10^6m//s` at angle `45^@` with `x`-axis in the `xy`-plane and experiences a force `F_1=5sqrt2N` along the negative `z`-axis. When te same particle moves with velocity `v_2=10^6m//s` along the `z`-axis it experiences a force `F_2` in `y`-direction. Find
a. the magnitude and direction of the magnetic field
b. the magnitude of the force `F_2`.

Explanation

`F_2` is in y-direction when velocity is along z-axis Therefore, magnetic field should be along x-axis.
So let , `vec B = B_0 hati`
a) Given `vec(V_1) = (10^6)/(sqrt(2)) hat i + (10^6)/(sqrt(2)) hat j`
and `vec F_1 = - 5 sqrt(2) xx 10^(-3) hatk`
From the equation, `vec F = q(vec v xx vec B)`
we have `(-5 sqrt(2) xx 10^(-3)) hat k = (10^(-6))[(10^6/sqrt(2) hat i + (10^6)/(sqrt(2)) hat j) xx (B_0 hat i) ] = - (B_0)/(sqrt(2)) hatk`
`therefore (B_0)/(sqrt(2)) = 5sqrt(2) xx 10^(-3) or B_0 = 10^(-2)T`
Therefore , the magnetic field is , `vec B = (10^(-2) hati) T`
`b) F_2 = B_0 qv_2 " sin " 90^@`
as the angle between `vecB` and `vecv` in this case is `90^@`
`therefore F_2 = (10^(-2))(10^(-6))(10^6) = 10^(-2)N`