Q 12-04-019NEETJEE MainEasy
In a region, a uniform electric field of $3 \times 10^4$ V/m and a uniform magnetic field of $0.02$ T are perpendicular to each other. A charged particle moving perpendicular to both fields passes through undeflected. Its speed is
Answer: (D) $1.5 \times 10^6$ m/s
Undeflected means the electric and magnetic forces cancel: $qE = qvB$.
$v = \dfrac{E}{B} = \dfrac{3 \times 10^4}{0.02} = 1.5 \times 10^6$ m/s. The result does not depend on the charge or mass.
Solution by Sreeraj P, M.Sc Physics