Question

If a charged particle of charge to mass ratio $\left(q/m\right)=\alpha$ enters in a magnetic field of strength $B$ at a speed $v=\left(2\alpha d\right)\left(B\right)$, then :

• A. angle subtended by the path of charged particle in magnetic field at the center of circular path is $2\pi$
• B. the charge will move on a circular path and then will come out from magnetic field at some distance from the point of insertion
• C. the time for which particle will be in the magnetic field is $\frac{2\pi }{\alpha B}$
• D. angle subtended by the path of charged particle in magnetic field at the center of circular path is $\pi /2$

Answer 1

Answer: (B)

Radius of the portion of the circular track inside magnetic field: $r=\frac{mv}{qB}=\frac{v}{\frac{q}{m}B}=\frac{2\alpha dB}{\alpha B}=2d$
Time period $T=\frac{2\pi m}{qB}=\frac{2\pi }{\frac{q}{m}B}=\frac{2\pi }{\alpha B}$
The particle spends half of $T$ inside magnetic field.
Time spent inside magnetic field is $\frac{T}{2}=\frac{1}{2}\frac{2\pi }{\alpha B}=\frac{\pi }{\alpha B}$
Angle subtended by the path of charged particle in magnetic field at the center of circular path is $\pi$.