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

Concept: This setup describes a velocity selector configuration. When a charged particle moves through perpendicular electric ($\vec{E}$) and magnetic ($\vec{B}$) fields, it experiences both an electrostatic force ($F_e = qE$) and a magnetic Lorentz force ($F_m = qvB$). For the particle path to remain completely straight and undeviated, these two forces must be equal in magnitude and opposite in direction: \[ qE = qvB \implies v = \frac{E}{B} \]

Step 1: Divide the field magnitudes to isolate speed ($v$).
We are given:
• Electric field, $E = 1.8 \times 10^{4}\text{ V m}^{-1}$
• Magnetic field, $B = 6 \times 10^{-3}\text{ T}$ Calculating the velocity: \[ v = \frac{1.8 \times 10^{4}}{6 \times 10^{-3}} = \left(\frac{1.8}{6}\right) \times 10^{4 - (-3)} \] \[ v = 0.3 \times 10^{7} = 3 \times 10^{6}\text{ m s}^{-1} \]