A charged particle carrying charge q=1 mC moves in uniform magnetic field with velocity v1=106 m/sat angle 450 with x-axis in the xy-plane and experiences a force F1=52mNalong the negative z-axis. When the same particle moves with velocity v2=106 m/s along the z-axis, it experiences a force F2 in y-direction. Find the magnitude of the force F2 (in N).

F2 it in y-direction when velocity is along z-axis. Therefore, magnetic field should be along x-axis. So let,B=B0i^ Given v1=1062i^+1062j^  and   F1=-52×10-3k^  From equation F=q(v×B) We have -52×10-3k^=10-61062i^+1062j^×B0i^ =-B02k^ ∴  B02=52×10-3 or B0=10-2 T  Therefore, the magnetic field is B=10-2i^T F2=B0qv2sin90°  As the angle between B and v in this case is 90°.  F2=10-210-6106 =10-2 N≈0.01 N