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).

Question

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).

Infinity Learn · Accepted Answer

$$F2$$ 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

Start JEE / NEET / Foundation preparation at rupees 99/day !!

Detailed Solution

Get Expert Academic Guidance – Connect with a Counselor Today!