A charged particle carrying charge `q = 1μC` moves in uniform magnetic field with velocity `v_(1) = 10^(6)`m/s at angle 45° with x-axis in x-y plane and experiences a force `F_(1)= 5sqrt2mN` along the negative z-axis. When the same particle moves with velocity, `v_(2) = 10^(6)` mis along the z-axis it experiences a force `F_(2)` in y-direction. Find
a) magnitude 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`