A particle of charge -q and mass m enters a uniform magnetic field vecB at A with speed v1 at an angle propto and leaves the field at C with speed v2 at an angle β as shown. Then

Question

A particle of charge -q and mass m enters a uniform magnetic field vecB at A with speed v1 at an angle propto and leaves the field at C with speed v2 at an angle β as shown. Then (A) α=β (B) v1=v2 (C) Particle remains in the field for time t=(2 m(π-α)/q B), (D) All of these

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/d880bce3e9b246d79e337230db237c16-.png" /> β=α v1=v2(∵ Fm perp v) T =(2(π-α) m/q B)

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

$α = β$

$v_{1} = v_{2}$

Particle remains in the field for time $t = \frac{2 m ( π - α )}{qB}$ ,

All of these

$β = α$
$v_{1} = v_{2} (∵ F_{m} ⊥ v)$
$T = \frac{2 ( π - α ) m}{qB}$

Q. A particle of charge −q and mass m enters a uniform magnetic field B at A with speed v1​ at an angle ∝ and leaves the field at C with speed v2​ at an angle β as shown. Then

α=β

v1​=v2​

Particle remains in the field for time t=qB2m(π−α)​,