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 magnetic field `vecB`.

Options

A. `(10^-3T)(hati+hatj)`
B. `(2xx10^-3T)hati`
C. `(10^-3T)hati`
D. `(2xx10^-3T)(hati+hatj)`

Answer

c

Explanation

For the first case: `vecF=qvecvxxvecB`
`implies -5sqrt2xx10^-3hatk`
`=10^-5xx(10^6)/(sqrt2) (hati+hatj)xx(B_xhati+B_yhatj+B_zhatk)`
`=(10/(sqrt2))[B_zhati-B_zhatj+(B_y+B_x)hatk]`
`implies B_z=0, B_y-B_x=-10^-3T....(i)`
Similarly, for the second case:
`F_2hatj=(10^-5)(10^6hatk)xx[(B_xhati+B_yhatj+B_zhatk)]`
`F_2hatj=10(B_xhatj-B_yhati)`
`F_2=10B_x, B_y=0....(ii)`
Using eqs. (i) and (ii), we get `B_x=10^-3T`
Thus, `vecB=(10^-3T) hati`
Also, `F_2=10B_x=10^-2N`.