A particle having charge $10^{-9}$ C moving in $x$-$y$ plane in fields of $0.4\hat j$ N/C and $4\times10^{-3}\hat k$ T experiences a force of $(4\hat i+2\hat j)\times10^{-10}$ N. The velocity of the particle at that instant is ______ m/s.
Answer: (A) $50\hat i+100\hat j$
Lorentz force: $\vec F=q\vec E+q(\vec v\times\vec B)$.
Electric part: $q\vec E=10^{-9}\times0.4\hat j=4\times10^{-10}\hat j$ N.
So the magnetic part is $q(\vec v\times\vec B)=(4\hat i+2\hat j-4\hat j)\times10^{-10}=(4\hat i-2\hat j)\times10^{-10}$ N.
With $\vec v=v_x\hat i+v_y\hat j$ and $\vec B=B\hat k$: $\vec v\times\vec B=B(v_y\hat i-v_x\hat j)$, and $qB=10^{-9}\times4\times10^{-3}=4\times10^{-12}$.
$4\times10^{-12}v_y=4\times10^{-10}\Rightarrow v_y=100$ m/s; $\ 4\times10^{-12}v_x=2\times10^{-10}\Rightarrow v_x=50$ m/s.
$\vec v=50\hat i+100\hat j$ m/s
Solution by Sreeraj P, M.Sc Physics