Question

An electron (mass = $9.1\times {10}^{-31}\text{kg}$ charge $=-1.6\times {10}^{-19}\text{C}$) experiences no deflection if subjected to an electric field of $3.2\times {10}^5\text{V/m}$ and a magnetic field of $2.0\times {10}^{-3}{\text{Wb/m}}^2$. Both the fields are normal to the path of electron and to each other. If the electric field is removed, then the electron will revolve in an orbit of radius

• A. $45\text{m}$
  
• B. $4.5\text{m}$
• C. $0.45\text{m}$
• D. $0.045\text{m}$
  

Answer 1

The correct option is C $0.45\text{m}$

According to question, no deflection of charge. So force due to magnetic field and due to electric field are equal and opposite.
So, $qE=qvB$

$v=\frac{E}{B}=\frac{3.2\times {10}^5}{2\times \times {10}^{{}^{-3}}}=1.6\times {10}^8\text{m/s}$

When electric field is removed, electron would describe a circular motion due to perpendicular magnetic field.
Radius, $R=\frac{mv}{qB}$

$R=\frac{9.1\times {10}^{-31}\times 1.6\times {10}^8}{1.6\times {10}^{-19}\times 2\times {10}^{{}^{-3}}}$

$R=0.455\text{m}$