Question

A positively charged particle of mass m is passed through a velocity selector. It moves horizontally rightward without deviation along the line y = $\frac{2mv}{qB}$ with a speed v. The electric field is vertically downwards and magnetic field is into the plane of the paper. Now, the electric field is switched off at t = 0. The angular momentum of the charged particle about origin O at t = $\frac{\pi m}{qB}$ is

• A. $\frac{2mE^2}{qB^3}$
• B. zero
• C. $\frac{mE^3}{qB^2}$
• D. $\frac{mE^2}{qB^3}$

Answer 1

Answer: (B)

zero

Answer 2

The correct answer is (B) : zero.