Question

A particle of charge $q$ and mass $m$ is moving through a region of space where crossed magnetic and electric fields produce a zero net force on the charge. At a later time, the Electric field is reduced by a factor of two.
What effect will this have on the motion of the particle? Select all that apply.

• A. The particle will continue to move in a circle of the same radius.
• B. The particle will eventually move in a circle of twice the radius.
• C. The particle will eventually move in a circle of half the radius.
• D. The particle will eventually follow a curved-noncircular path.
• E. The particle will follow a straight line path.

Answer 1

The correct option is C The particle will eventually move in a circle of half the radius.

The force on a charge q of mass m moving with a velocity v in a magnetic field perpendicularly is given by ,

$F_B=qvB$ ,

electric force on charge q by a field is given by ,

$F_E=qE$ ,

as net force on charge q is zero therefore both the fields should be equal and opposite ,

$F_B=F_E$ ,

or $qvB=qE$ ,

or $v=E/B$ ,

radius of the circular pah is ,

$r=mv/qB=m\left(E/B\right)/qB$ ,

now electric field is reduced by a factor of 2 , therefore

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

and $r^{\prime }=m\left(E/2B\right)/qB=\left(1/2\right)mE/q{B}^2=r/2$ ,

the particle moves in a circle of half the radius .