Q122 / 170 · Manipal (MET) 2018 · Physics
A charged particle travels along a straight line with a speed $v$ in a region where both electric field $\mathbf{E}$ and magnetic fields $\mathbf{B}$ are present. It follows that
- A$|\mathbf{E}|=v|\mathbf{B}|$ and the two fields are parallel
- Correct answerB$|\mathbf{E}|=v|\mathbf{B}|$ and the two fields are perpendicular
- C$|\mathbf{B}|=v|\mathbf{E}|$ and the two fields are parallel
- D$|\mathbf{B}|=v|\mathbf{E}|$ and the two fields are perpendicular
Solution
Force due to magnetic field is
$\begin{aligned}
F & =q v B \sin \theta \\
\theta & =90^{\circ} \\
F & =q v B....(i)
\end{aligned}$
when
Force due to electric field $E$ is $F=q E$....(ii)
Equating Eqs. (i) and (ii), we get $|\overrightarrow{\mathbf{E}}|=v|\overrightarrow{\mathbf{B}}|$
Force due to electric field $E$ is $F=q E$....(ii)
Equating Eqs. (i) and (ii), we get $|\overrightarrow{\mathbf{E}}|=v|\overrightarrow{\mathbf{B}}|$
Hence, the correct answer is option: 'B'.