Question

What is the radius of the path of an electron (mass 9 x 10-31 kg and charge
1.6 x 10-19 C) moving at a speed of 3 x 10 7m/s in a magnetic field of 6 x 10-4 T
perpendicular to it?​

Answer 1

r =(mv)/(qB)


put all the values given in this and a bit calculation will give the answer

Answer 2

Answer:

The radius of the path of an electron (mass $9.1*10^{-31}kg$ and charge

$1.6*10^{-19}C$) moving at a speed of $3*10^{7}m/s$ in a magnetic field of $6*10^{-4} T$

perpendicular to it is $0.28m$

Explanation:

When a particle with a charge $q$ moving with a speed $v$ enter a magnetic field it will experience Lorentz magnetic force perpendicular to the direction of velocity, so it follows a circular path of radius $r$

The centripetal force is provided by the magnetic force,

$F=q\left(v\times B\right)=\frac{mv^2}{r}$

rearrange this to get the radius of the circle, $r=mv/qB$

from the question, it is given that

$q=1.6*10^{-19} C,m=9.1*10^{-31}kg,v=3*10^{7}m/s,B=6*10^{-4} T$

substitute these values to get the radius

$r=\frac{9.1*10^{-31} *3*10^{7} }{1.6*10^{-19}*6*10^{-4} } =0.28m$