Question

A particle of charge $-q$ and mass $m$ enters a uniform magnetic field $\vec{B}$ at $A$ with speed $v_{1}$ at an angle $\propto$ and leaves the field at $C$ with speed $v_{2}$ at an angle $\beta$ as shown. Then

• A. $\alpha=\beta$
• B. $v_{1}=v_{2}$
• C. Particle remains in the field for time $t=\frac{2 m(\pi-\alpha)}{q B}$,
• D. All of these

Answer 1

Answer: (D)

$\beta=\alpha$
$v_{1}=v_{2}\left(\because F_{m} \perp v\right)$
$T =\frac{2(\pi-\alpha) m}{q B}$