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Byju's Answer

Standard XII

Physics

Analysis of Force Equation

A particle ha...

Question

A particle having mass m and charge q is released from the origion in a region in which electric field and magnetic field are given by

$\to B = - B_{0} \to J$ and $\to E = E_{o} - \to K .$

Find the speed of the particle as a function of its z-coordinate.

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Solution

Velocity will be along x-z plane - $¯ ¯¯ ¯ B$

$= - {¯ ¯¯ ¯ B}_{0} ¯ J, ¯ ¯¯ ¯ E = E_{0} ¯ ¯¯¯ ¯ K$

$F = q (E + V \times B)$

$= q E_{0} K - q V_{x} B_{0} K + q V_{z} B_{0} i$

Since $V_{x} = 0, F_{z} = q E_{0}$

So $a = \frac{q E_{0}}{m}$

$∵ v^{2} = u^{2} + 2 a s = 2 \frac{q E_{0}}{m} z$

So $v = \sqrt{\frac{2 q E_{0} z}{m}}$

[dist. along z-direction be z]

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A particle having mass m and charge q is released from the origion in a region in which electric field and magnetic field are given by →B=−B0→J and →E=Eo−→K. Find the speed of the particle as a function of its z-coordinate.

Velocity will be along x-z plane - ¯¯¯¯B =−¯¯¯¯B0 ¯J,¯¯¯¯E=E0 ¯¯¯¯¯K F=q (E+V×B) =qE0K−qVx B0 K+qVz B0i Since Vx=0, Fz=qE0 So a=qE0m ∵ v2=u2+2as=2qE0mz So v=√2qE0 zm [dist. along z-direction be z]

A particle having mass m and charge q is released from the origion in a region in which electric field and magnetic field are given by

$\to B = - B_{0} \to J$ and $\to E = E_{o} - \to K .$

Find the speed of the particle as a function of its z-coordinate.