A particle of mass m and charge q has an initial velocity v-v0- If an electric field E-E0- and magnetic field B-B0- act on the particle- its speed will double after a time

Step 1: Given dataMass of the particle=mCharge =qInitial velocity, v=v_0ĵ,Electric field, E=E_0îMagnetic field, B=B_0îStep 2: Finding the net speed of particles ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Electric Potential at a Point in Space Due to a Point Charge Q

A particle of...

Question

A particle of mass $m$ and charge $q$ has an initial velocity $\vec{v} = v_{0} \hat{j}$ , If an electric field $\vec{E} = E_{0} \hat{i}$ and magnetic field $\vec{B} = B_{0} \hat{i}$ act on the particle, its speed will double after a time

$\frac{\sqrt{3} v_{0} m}{q E_{0}}$
$\frac{\sqrt{2} v_{0} m}{q E_{0}}$
$\frac{3 v_{0} m}{q E_{0}}$
$\frac{2 v_{0} m}{q E_{0}}$

Open in App

Solution

The correct option is A
$\frac{\sqrt{3} v_{0} m}{q E_{0}}$

Step 1: Given data
Mass of the particle $= m$
Charge $= q$
Initial velocity, $\vec{v} = v_{0} \hat{j}$ ,
Electric field, $\vec{E} = E_{0} \hat{i}$
Magnetic field, $\vec{B} = B_{0} \hat{i}$
Step 2: Finding the net speed of particles
We know that the magnetic field can only change the direction of speed as it cannot do any work.
As given,
$\vec{v} = v_{0} \hat{j}$
Since, the speed of the particle doubles after time $t$ in $y$ -direction, the speed will be $v_{0}$ only in $y - z$ plane which is always perpendicular to $x$ -direction so the net speed at time $t$ .
${(v_{n e t})}^{2} = {(v_{0})}^{2} + {(v_{x})}^{2}$
Where $V_{n e t}$ is the net speed of particles, $V_{0}$ is speed in $y - z$ plane and $V_{x}$ in $x$ -direction .
Step 3: Finding the time when the speed double.
We know that the force, $F$ is the product of mass, $m$ and acceleration, $a$ .
$F = m a a = \frac{F}{m} \dots \dots (1)$
Also, force, $F$ is the product of the charge, $q$ and the electric field, $E$
$F = q E$
Putting this value in $(1)$ , we get
$a = \frac{q E}{m}$
$\frac{d v_{x}}{d t} = \frac{q E}{m}$
[ Acceleration is the rate of change of the velocity, $v_{x}$ with respect to the time, $t$ .]
Integrating both sides, we get
$v_{x} = \frac{q E}{m} t$
$v_{x} = \frac{q E_{0}}{m} t$ [ $E = E_{0}$ ]
Also by given condition $v_{n e t} = 2 v_{0}$ then
${(2 v_{0})}^{2} = {(v_{0})}^{2} + {(\frac{q E_{0} t}{m})}^{2}$ $[∵ v_{x} = \frac{q E_{0} t}{m}]$
$\Rightarrow$ $3 v_{0}^{2} = {(\frac{q E_{0} t}{m})}^{2}$
$\Rightarrow$ $t = \frac{\sqrt{3} v_{0} m}{q E_{0}}$
Hence, option (A) is correct.

Suggest Corrections

Similar questions

Q. A particle of mass

m

and charge

q

has an initial velocity

\to v = v_{0}^j

. If an electric field

\to E = E_{0}^i

and magnetic field

\to B = B_{0}^i

act on the particle, its speed will double after a time :

Q. A particle of charge

q

and mass

m

starts moving from the origin under the action of an electric field

\to E = E_{o}^i

and

\to B = B_{0}^i

with a velocity

\to v = v_{0}^j

. The speed of the particle will become

2 v_{0}

after a time

Q. A charge particle moves underacted in a region of crossed electric and magnetic fields. If the electric field is switched off, the particle has an initial acceleration

a

, if the magnetic field is switched off, instead of the electric field, the particle will have an initial acceleration:

Q. A particle of charge

q

and mass

m

starts moving from the origin under the action of an electric field

\to E = E_{0}^i

and

\to B = B_{0}^i

with velocity

\to v = v_{0}^j

. The speed of the particle will become

2 v_{0}

after time

Q. A positively charged particle of

2.0 C

moves upward into an area where both a magnetic field and an electric field are acting. The magnetic field has a magnitude of

4.0 \times 10^{4} T

and the electric field has a magnitude of

0.1 N / C

. At what velocity must the particle be moving if it is not deflected when it enters this area?

Join BYJU'S Learning Program

A particle of mass m and charge q has an initial velocity v→=v0j^, If an electric field E→=E0i^ and magnetic field B→=B0i^ act on the particle, its speed will double after a time3v0mqE02v0mqE03v0mqE02v0mqE0