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, B = `vecB_(0)hati`
a) Given `vecV_(1)=(10^(6))/(sqrt2)hati+(10^(6))/(sqrt2)hatj` and `hatvecF_(1)=-5sqrt2xx10^(3)hatk`
From the equation, `vecF=q(vecvxxvecB)`, we have
`(-5sqrt2xx10^(-3))hatk=(10^(-6))[((10^(6))/(sqrt2)hati+(10^(6))/(sqrt2)hatj)xx(B_(0)hati)]=-(B_(0))/(sqrt2)hatk`
`therefore(B_(0))/(sqrt2)=5sqrtxx10^(-3)` or `B_(0)=10^(-2)T`
Therefore, the magnetic field is, `vecB=(10^(-2)hati)T`
b) `F_(2)=B_(0)qv_(2)sin90^(@)`
As the angle between `vecB` and `vecv` in this case is 90°.
`thereforeF_(2)=(10^(-2))(10^(-6))(10^(6))=10^(-2)N`