A charged particle carrying charge `q=10 muC` moves with velocity `v_1=10^6 ,s^-1` at angle `45^@` with x-axis in the xy plane and experience a force `F_1=5sqrt2 mN` along the negative z-axis. When the same particle moves with velocity `v_2=10^6 ms^-1` along the z-axis, it experiences a force `F_2` in y-direction.
Find the magnitude of the force `F_2`.

Answer

`F_(2)=10^(-2)N`

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)hat(i)`
`(a)` Given `vec(v)_(1)=(10^(6))/(sqrt(2))hat(i)+(10^(6))/(sqrt(2))hat(j)`
and `vec(F)_(1)=-5sqrt(2)xx10^(3)hat(k)`
From the equation, `vec(F)=q(vec(v)xxvec(B)),` we have
`(-5sqrt(2)xx10^(-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))hat(k)`
`:. " "(B_(0))/(sqrt(2))=5sqrt(2)xx10^(-3)" or "B_(0)=10^(-2)T`
Therefore, the magnetic field is , `vec(B)=(10^(-2)hat(i))T`
`(b) F_(2)=B_(0)qv_(2)sin 90^(@)`
As the angle between `vec(B)` and `vec(v)` in this case is `90^(@)`.
`F_(2)=(10^(-2))(10^(-6))(10^(6))=10^(-2)N`.