Question

A charged particle moves with a constant velocity $(\hat{i}+\hat{j})\text{m/s}$ in a magnetic field $\overrightarrow{B}=(2\hat{i}+3\hat{k})\text{T}$ and uniform electric field $\overrightarrow{E}=(a\hat{i}+b\hat{j}+c\hat{k})N/C,$ then (assuming all quantities to be in S.I. units)

• A. $b=3$
• B. $a^2+b^2+c^2=22$
• C. $a=–3$
• D. $c=-2$

Answer 1

Answer: (A), (B), (C)

$a=–3$

Since the charged particle is moving with constant velocity, its acceleration will be zero.

$F_{\text{net}}=0\Rightarrow F_{el}+F_{\text{mag}}=0$

In equilibrium, electric force = - magnetic force

So, $q\overrightarrow{E}=-q(\overrightarrow{v}\times \overrightarrow{B})\text{or}a\hat{j}+b\hat{j}+c\hat{k}=-(3\hat{i}-3\hat{j}-2\hat{k})$

So, $a=-3,b=3,c=2\text{and}{a}^2+b^2+c^2=9+9+4=22$