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Standard XII

Physics

Electric Field

In a certain ...

Question

In a certain region there exist a uniform electric field $\to E$ and a uniform magnetic field $\to B$ both directed along z-axis. A particle of charge $Q$ and mass $m$ is projected at time $t = 0$ with a velocity $v_{0}$ along $x$ -axis. Find the velocity at any instant of time $t$ .

v_{0} cos (\frac{Q B t}{m})^i - v_{0} sin (\frac{Q B t}{m})^j + (\frac{Q E t}{m})^k

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v_{0} cos (\frac{Q B t}{m})^i + v_{0} sin (\frac{Q B t}{m})^j + (\frac{Q E t}{m})^k

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v_{0} sin (\frac{Q B t}{m})^i - v_{0} cos (\frac{Q B t}{m})^j - (\frac{Q E t}{m})^k

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v_{0} cos (\frac{Q B t}{m})^i - v_{0} sin (\frac{Q B t}{m})^j - (\frac{Q E t}{m})^k

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Solution

The correct option is A $v_{0} cos (\frac{Q B t}{m})^i - v_{0} sin (\frac{Q B t}{m})^j + (\frac{Q E t}{m})^k$
The Lorentz's force on a charged particle is,

$\to F = Q (\to E + (\to v \times \to B))$

where $\to E = E^k, \to B = B^k$

Let $\to v = v_{x}^i + v_{y}^j + v_{z}^k$

$\to F = Q [E^k + (v_{x}^i + v_{y}^j + v_{z}^k) \times B^k]$

$m [\frac{d v_{x}}{d t}^i + \frac{d v_{y}}{d t}^j + \frac{d v_{z}}{d t}^k]] = Q [v_{y} B^i - v_{x} B^j + E^k]$

First we consider $z$ -component ; we get

$\frac{d v_{z}}{d t} = \frac{Q E}{m} . . . . . (1)$

Integrating both sides with proper limits we get,

$\int_{0}^{v_{z}} d v_{z} = \int_{0}^{t} \frac{Q E}{m} d t$

Hence $v_{z} = \frac{q E t}{m}$ (as $v_{z} = 0$ when $t = 0$ )

Now considering motion in the $x$ and $y$ -directions. We have

$\frac{d v_{x}}{d t} = \frac{Q B v_{y}}{m} . . . . . (2)$

and $\frac{d v_{y}}{d t} = - \frac{Q B v_{x}}{m} . . . . . (3)$

This is a pair of coupled equations. It can be decoupled by differentiating one of them w.r.t. time and substituting the result into the other equation. We get,

$\frac{d^{2} v_{x}}{d t^{2}} = \frac{Q^{2} B^{2}}{m^{2}} v_{x}$

and $\frac{d^{2} v_{y}}{d t^{2}} = - \frac{Q^{2} B^{2} v_{y}}{m^{2}}$

These equations are similar to equation of S.H.M., with the general solutions ,

$v_{x} = a cos ω t + b sin ω t$

$v_{y} = c cos ω t + d sin ω t$

Where, $ω = \frac{Q B}{m}$

From the given initial condition, $t = 0, v_{x} = v_{0}$ and $v_{y} = 0$

Also $\frac{d v_{x}}{d t} = 0$

From eq (3) we get,

and $\frac{d v_{y}}{d t} = - \frac{Q B v_{0}}{m} = - ω v_{0}$

thus $a = v_{0}, b = 0$

$c = 0, d = - v_{0}$

We can thus finally write the velocity at time $t$ as,

$v = v_{0} cos (\frac{Q B t}{m})^i - v_{0} sin (\frac{Q B t}{m})^j + \frac{Q E t}{m}^k$

Hence, option (a) is the correct answer.

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Electric Field

Standard XII Physics

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In a certain region there exist a uniform electric field →E and a uniform magnetic field →B both directed along z-axis. A particle of charge Q and mass m is projected at time t=0 with a velocity v0 along x-axis. Find the velocity at any instant of time t.

In a certain region there exist a uniform electric field $\to E$ and a uniform magnetic field $\to B$ both directed along z-axis. A particle of charge $Q$ and mass $m$ is projected at time $t = 0$ with a velocity $v_{0}$ along $x$ -axis. Find the velocity at any instant of time $t$ .