A charged particle of specific charge -charge-mass- - - is released from origin at time - t-0- with velocity - v-v-0- in a uniform magnetic field - B-B-0- Coordinates of the particle at time - t-B-0- are- -v-0-2B-0- -2-v-0- B-0- -v-0-B-0- - -v-0-2B-0- 0- 0 - -v-0-B-0- 0- -2v-0-B-0- - -0- 2v-0-B-0- v-0-2B-0- -

The correct option is C (v_0π/B_0α, 0, 2v_0/B_0α ) α=qm , path of the particle will be a helix of time period, T=2π mB_0q=2πB_0α The given time ...

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Byju's Answer

Standard XII

Physics

The Circular Motion

A charged par...

Question

A charged particle of specific charge (charge/mass) $α$ is released from origin at time $t = 0$ with velocity $\to v = v_{0} (^i +^j)$ in a uniform magnetic field $\to B = B_{0}^i$ . Coordinates of the particle at time $t = \frac{π}{B_{0} α}$ are

(\frac{v_{0}}{2 B_{0} α}, \frac{\sqrt{2} v_{0}}{α B_{0}}, \frac{- v_{0}}{B_{0} α})

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(\frac{- v_{0}}{2 B_{0} α}, 0, 0)

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(0, \frac{2 v_{0}}{B_{0} α}, \frac{v_{0} π}{2 B_{0} α})

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(\frac{v_{0} π}{B_{0} α}, 0, \frac{- 2 v_{0}}{B_{0} α})

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Solution

The correct option is C $(\frac{v_{0} π}{B_{0} α}, 0, \frac{- 2 v_{0}}{B_{0} α})$
$α = \frac{q}{m}$ , path of the particle will be a helix of time period,

$T = \frac{2 π m}{B_{0} q} = \frac{2 π}{B_{0} α}$

The given time $t = \frac{π}{B_{0} α} = \frac{T}{2}$

$∴$ Coordinates of particle at time $t = T / 2$ would be $(v_{x} T / 2, 0, - 2 r)$

Here, $r = \frac{m v_{0}}{B_{0} q} = \frac{v_{0}}{B_{0} α}$

$∴$ The coordinate are $(\frac{v_{0} π}{B_{0} α}, 0, \frac{- 2 v_{0}}{B_{0} α})$

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A charged particle of specific charge (charge/mass) α is released from origin at time t=0 with velocity →v=v0(^i+^j) in a uniform magnetic field →B=B0^i. Coordinates of the particle at time t=πB0α are

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A charged particle of specific charge (charge/mass) $α$ is released from origin at time $t = 0$ with velocity $\to v = v_{0} (^i +^j)$ in a uniform magnetic field $\to B = B_{0}^i$ . Coordinates of the particle at time $t = \frac{π}{B_{0} α}$ are