A particle of charge per unit mass -is released from origin with a velocity V- -V o- in a uniform magnetic field B- -B 0k- If the particle passes through 0-y-0 then y is equal to

The correct option is A 2v_ o / B_ oα [ R= m v_ 0 / qB #160;; ly=2R= 2m v_ 0 / q B_ 0 ...

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Standard XII

Physics

Measuring Mass

A particle of...

Question

A particle of charge per unit mass \alpha is released from origin with a velocity $¯ V = V_o^i$ in a uniform magnetic field $¯ B = B_0^k$ . If the particle passes through (0,y,0) then y is equal to

- \frac{2 v_o}{B_{o} α}

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\frac{v_o}{B_{o} α}

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\frac{2 v_o}{B_{o} α}

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- \frac{v_o}{B_{o} α}

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Solution

The correct option is A $- \frac{2 v_o}{B_{o} α}$
$\begin{matrix} R = \frac{m v_{0}}{q B} l y = 2 R = \frac{2 m v_{0}}{q B_{0}} = \frac{2 v_{o}}{α B_{0}} ∴ y = \frac{- 2 v_{o}}{α B_{o}} H e n c e, o p t i o n A i s c o r r e c t a n s w e r . \end{matrix}$

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A particle of charge per unit mass \alpha is released from origin with a velocity ¯V=V_o^i in a uniform magnetic field ¯B=B_0^k. If the particle passes through (0,y,0) then y is equal to

The correct option is A −2v_oBoαR=mv0qBly=2R=2mv0qB0=2voαB0∴y=−2voαBoHence,optionAiscorrectanswer.

A particle of charge per unit mass \alpha is released from origin with a velocity $¯ V = V_o^i$ in a uniform magnetic field $¯ B = B_0^k$ . If the particle passes through (0,y,0) then y is equal to

The correct option is A $- \frac{2 v_o}{B_{o} α}$
$\begin{matrix} R = \frac{m v_{0}}{q B} l y = 2 R = \frac{2 m v_{0}}{q B_{0}} = \frac{2 v_{o}}{α B_{0}} ∴ y = \frac{- 2 v_{o}}{α B_{o}} H e n c e, o p t i o n A i s c o r r e c t a n s w e r . \end{matrix}$