Question

The force on a charged particle due to electric and magnetic fields is given by \[\vec{F} = q \vec{E} + q \vec{\nu} \times \vec{B}\].
Suppose \[\vec{E}\]  is along the X-axis and \[\vec{B}\]  along the Y-axis. In what direction and with what minimum speed ν should a positively charged particle be sent so that the net force on it is zero?

Answer 1

According to the problem, the net electric and magnetic forces on the particle should be zero. 

i.e.,
\[\vec{F} = q \vec{E} + q\left( \vec{\nu} \times \vec{B} \right) = 0\]
\[\Rightarrow E = - \left( \vec{\nu} \times \vec{B} \right)\] 
So, the direction of \[\vec{\nu} \times \vec{B}\]  should be opposite to the direction of \[\vec{E}\] Hence, \[\vec{\nu}\] should be along the positive z-direction.
Again, E = νB sin θ 
\[\Rightarrow \nu = \frac{E}{B} \sin \theta\]
For ν to be minimum, \[\theta = 90 \text{ and }, thus, \nu_\min = \frac{E}{B}\]
So, the particle must be projected at a minimum speed of \[\frac{E}{B}\] along the z-axis.