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

The Correct Option is B

The correct answer is (B) : zero.