Question

A charged particle with charge $q$ enters a region of constant, uniform and mutually orthogonal fields $\vec{E}$ and $\vec{B}$ with a velocity $\vec{V}$ perpendicular to both $\vec{E}$ and $\vec{B}$, and comes out without any change in its magnitude or direction Then

• A. $\vec{V} =\vec{B}\times\vec{E} / E^{2}$
• B. $\vec{V} =\vec{E} \times\vec{B} / B^{2}$
• C. $\vec{V} =\vec{B}\times\vec{E} / B^{2}$
• D. $\vec{V} =\vec{E} \times\vec{B} /E^{2}$

Answer 1

Answer: (B)

$\vec{V} =\vec{E} \times\vec{B} / B^{2}$

Answer 2

When $\vec{E}$ and $\vec{B}$ are perpendicular and velocity has no change then $qE=qvB$ i.e., $v=\frac{E}{B}$. The two forces oppose each other so, $v$ is along $\vec{E} \times\vec{B}$ i.e., $\vec{v}=\frac{\vec{E}\times\vec{B}}{B^{2}}$ As $\vec{E}$ and $\vec{B}$ are perpendicular to each other $\frac{\left|\vec{E}\times\vec{B}\right|}{B^{2}}=\frac{EB\,sin\,90^{\circ}}{B^{2}}=\frac{E}{B}$