Question

A charged particle is moving on circular path with velocity v in a uniform magnetic field B, if the velocity of the charged particle is doubled and strength of magnetic field is halved, then radius becomes :

• A. 8 times
• B. 4 times
• C. 2 times
• D. 16 times

Answer 1

Answer: (B)

4 times

To perform circular motion required centripetal force would be provided by the magnetic force on the moving charge.

So, $Bqv=\frac{m{v}^2}{r}$ or $r=\frac{mv}{Bq}$
According to the question, $v^{\prime }=2v$ and $B^{\prime }=\frac{B}{2}$
$\therefore r^{\prime }=\frac{m{v}^{\prime }}{B^{\prime }q}=\frac{m\left(2v\right)}{\left(B/2\right)q}=\frac{4mv}{Bq}=4r$