Question

An electron is released from the origin at a place where a uniform electric field

E

and a uniform magnetic field

B

exist along the negative

y

-axis and the negative

z

-axis respectively. Find the displacement of the electron along the

y

-axis when its velocity becomes perpendicular to the electric field for the first time.

Accepted Answer

According to the right-handed system, the $Z$ -axis is upward and hence the magnetic field is downwards. At any time, the velocity of the electron may be written as $\vec{u} = u_{x} h \vec{i} + u_{y} \vec{j}$

The electric and magnetic fields may be written as $\vec{E} = - E \vec{j}$ and $\vec{B} = - B \vec{k}$ respectively.The force on the electron is $\vec{F} = e (\vec{E} + \vec{u} \times \vec{B})$

$= e E \vec{j} + e B (u_{y} \vec{i} - u_{x} \vec{j})$

Thus, $F_{x} = e u_{y} B$ and $F_{y} = e (E - u_{x} B)$ .

The components of the acceleration are $a_{x} = \frac{d u_{x}}{d t} = \frac{e B}{μ_{y}}$ and $a_{y} = \frac{d u_{y}}{d t} = \frac{e}{m} (E - u_{x} B)$

We have, $\frac{d^{2} u_{y}}{d t^{2}} = - \frac{e B}{m} \frac{d u_{x}}{d t} = - \frac{e B}{m} \frac{e B}{m} u_{y}$ , where, $ω = \frac{e B}{m}$

This equation is similar to that for a simple harmonic motion.

Thus, $u_{y} = A \sin (ω t + δ) \dots (4)$

And hence $\frac{d u_{y}}{d t} = A ω \cos (ω t + δ) \dots (5)$

At $t = 0, u_{y}$ and $\frac{d u_{y}}{d t} = \frac{F_{y}}{d t} = \frac{e E}{m}$ .

Putting in (4) and (5)

$δ = 0$ and $A = \frac{e E}{m ω} \frac{E}{B}$ .

Thus, $u_{y} = \frac{E}{B} \sin ω t$

The path of the electron will be perpendicular to the $Y$ -axis when $u_{y} = 0$ . This will be the case for the first time at $t$ where $\sin ω t = 0$

Or $ω t = π$ $\Rightarrow t = \frac{π}{ω} = \frac{π m}{e B}$

Also, $u_{y} = \frac{d y}{d t} = \frac{E}{B} \sin ω t$

or $\frac{y}{d y} = \frac{E}{B} \sin ω t d t$

Or $y = \frac{E}{B ω} (1 - \cos ω)$

At $t = \frac{π}{ω}, y = \frac{E}{B ω} (1 - \cos π) = \frac{2 E}{B ω}$

Thus, the displacement along the $Y$ -axis is $\frac{2 E}{B ω} = \frac{2 E m}{B e B} = \frac{2 E m}{e B^{2}}$ .

$(1) \oint \vec{E} . \vec{d A}$	$(1) μ_{o} i_{c}$
$(2) \oint \vec{B} . \vec{d A}$	$(2) - \frac{d ϕ_{B}}{d t}$
$(3) \oint \vec{E} . \vec{d l}$	$(3) \frac{Q}{ε_{o}}$
$(4) \oint \vec{B} . \vec{d l}$	$(4) 0$

Important Questions on Electromagnetic Waves

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