Question

An electric field of $8 \times 10^{4}\text{ V m}^{-1}$ and a magnetic field of $6 \times 10^{-3}\text{ T}$ are applied simultaneously on an electron beam such that the path of the beam remains undeviated. Then the speed of the electron will be:

• A. $3 \times 10^{7}\text{ m s}^{-1}$
• B. $1.5 \times 10^{7}\text{ m s}^{-1}$
• C. $3 \times 10^{6}\text{ m s}^{-1}$
• D. $1.5 \times 10^{6}\text{ m s}^{-1}$

Hint:

Whenever a problem states that a charged particle passes through crossed fields "undeviated" or "without bending", the mass or charge of the particle does not affect the calculation. The required speed is always the simple ratio $v = \frac{E}{B}$.

Answer 1

The Correct Option is C