Search any question & find its solution
Question:
Answered & Verified by Expert
A particle with charge $q$ moves with a velocity $v$ in a direction perpendicular to the directions of uniform clectric and magnetic ficlds, $E$ and $B$ respectively, which are mutually perpendicular to each other. Which one of the following gives the condition for which the particle moves undeflected in its original trajectory? 
Options:
Solution:
2097 Upvotes
Verified Answer
The correct answer is:
$v=\frac{E}{B}$
According to the question,
Charge on particle is $=q$
Velocity of particle $=v$
Due to uniform electric field,
electric force on particle
$F_{\text {electric }}=q E$
Due to a uniform magnetic field, magnetic force on particle is given by,
$$
F_{\text {magnetic }}=q(v \times B)
$$
When $v$ is perpendicular to $E$ and $B,$ which are mutually perpendicular to each other, $F_{\text {electric }}=F_{\text {magnetic }}$
From Eqs. (i) and (ii)
$$
q E=q v B
$$
$\therefore$
$$
v=\frac{E}{B}
$$
So, $v=\frac{E}{B}$ is the condition for which the particle moves undeflected in its original trajectory.
Charge on particle is $=q$
Velocity of particle $=v$
Due to uniform electric field,
electric force on particle
$F_{\text {electric }}=q E$
Due to a uniform magnetic field, magnetic force on particle is given by,
$$
F_{\text {magnetic }}=q(v \times B)
$$
When $v$ is perpendicular to $E$ and $B,$ which are mutually perpendicular to each other, $F_{\text {electric }}=F_{\text {magnetic }}$
From Eqs. (i) and (ii)
$$
q E=q v B
$$
$\therefore$
$$
v=\frac{E}{B}
$$
So, $v=\frac{E}{B}$ is the condition for which the particle moves undeflected in its original trajectory.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.