Consider a charged particle of mass m and charge q released from the origin in a region of magnetic field- B - B 0- and electric field E - E 0k- The speed of the particle as a function of its z - coordinate will be Neglect gravityA- -2 qE 0 Z - m B- -3 qE 0 Z - m C- 2 -q E0 Z-mD- - qE 0 z - m

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Standard XII

Physics

Circular Kinematics

Consider a ch...

Question

Consider a charged particle of mass $m$ and charge $q$ released from the origin in a region of magnetic field, $\to B = B_{0}^j$ and electric field $\to E = E_{0}^k$ . The speed of the particle as a function of its $z -$ coordinate will be (Neglect gravity)

2 \sqrt{\frac{q E_{0} z}{m}}

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\sqrt{\frac{3 q E_{0} z}{m}}

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\sqrt{\frac{2 q E_{0} z}{m}}

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\sqrt{\frac{q E_{0} z}{m}}

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Solution

The correct option is C $\sqrt{\frac{2 q E_{0} z}{m}}$
The speed changes only due to electric field and hence work done is only due to the electric field. Applying kinetic energy - work theorem we get,
$△ K = W_{n e t} \Rightarrow K_{f} - K_{i} = W_{e l e c} \Rightarrow \frac{1}{2} m v^{2} - 0 = q E_{0} z \Rightarrow V = \sqrt{\frac{2 q E_{0} z}{m}}$

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Consider a charged particle of mass m and charge q released from the origin in a region of magnetic field, →B=B0^j and electric field →E=E0^k . The speed of the particle as a function of its z−coordinate will be (Neglect gravity)

The correct option is C √2qE0zmThe speed changes only due to electric field and hence work done is only due to the electric field. Applying kinetic energy - work theorem we get, △K=Wnet⇒Kf−Ki=Welec⇒12mv2−0=qE0z⇒V=√2qE0zm

Consider a charged particle of mass $m$ and charge $q$ released from the origin in a region of magnetic field, $\to B = B_{0}^j$ and electric field $\to E = E_{0}^k$ . The speed of the particle as a function of its $z -$ coordinate will be (Neglect gravity)