An electron accelerated from rest inside the space of potential difference of 200V and then enters in side the magnetic field of strength 0.3T perpendicular to it.

Solution,

Potential difference (p.d)=200V
magnetic field (B)=0.3T 
Radius of path (r)=?

We know that,

• \[ Charge=(q)=e= 1.6×10^{-19}C\]
• \[ mass -of- electron=(m)=9.1×10^{-31}Kg\]
• \[Specific-charge =(e/m)= 1.75×10^{11}cKg^{-1}\]

According to work energy theorem ,

Workdone (w) =∆KE

\[q×v=\frac{1}{2}mv^{2}\]

\[v=\sqrt{\frac{2ev}{m}}\]

Magnetic force( FB ) = Centripetal force( FC )

\[q×v×B=\frac{mv^{2}}{r}\]

Or , \[r=\frac{mv^{2}}{q×v×B}\]

Or \[r=\frac{mv}{q×B}\]

Or,\[r=\frac{m}{e×B}×\sqrt{\frac{2eV}{m}}\]

Or,\[r=\sqrt{\frac{2V}{\frac{e}{m}×B^{2}}}\]

Or,\[r=\sqrt{\frac{2×200}{\frac{1.6×10^{-19}}{9.1×10^{-31}}×(0.3)^{2}}}\]

Therefore Radius of an electron is :

\[r=1.12×10^{-5}m\]