Question

A particle of charge $q$ and mass $m$ moves in a circular path of radius $r$ in a uniform magnetic field $B$. The magnetic field is reduced by a factor of 2 while the speed is increased by a factor of $\frac{1}{2}$. The radius of the new circular path the particle makes is:

• A. $\frac{1}{4}r$
• B. $\frac{1}{2}r$
• C. $r$
• D. $2r$
• E. $4r$

Answer 1

The correct option is C $r$

The radius r of a charge q of mass m , moving in a magnetic field B with velocity v is given by ,

$r=mv/qB$ ..........................eq1

now if $v^{\prime }=v/2,B^{\prime }=B/2$ ,

hence $r^{\prime }=m{v}^{\prime }/q{B}^{\prime }=\frac{m\left(v/2\right)}{q\left(B/2\right)}=mv/qB$ ..........................eq2

it is clear that new radius $r^{\prime }=r$ (old radius)