Question

An electron beam passes through a magnetic field of $2\times {10}^{-3}Wb/m^2$ and an electric field of $1.0\times {10}^{4}V/m$ both acting simultaneously. The path of electron remains undeviating. The speed of electron if the electric field is removed, and the radius of electron path will be respectively.

• A. $10\times {10}^{6}m/s,2.43$
• B. $2.5\times {10}^{6}m/s,0.43$
• C. $5\times {10}^{6}m/s,1.43$
• D. None of these

Answer 1

The correct option is C $5\times {10}^{6}m/s,1.43$

$B=2\times {10}^{-3}Wb/m^2$,

$E=1\times {10}^{4}V/m^2$

Since the path of electron remains undeviated, $qvB=qE$ or

$v=\frac{E}{B}=\frac{1\times {10}^{+4}}{2\times {10}^{-3}}=0.5\times {10}^7$

$=5\times {10}^{6}m/s$

If the electric field is removed, the path of the charged particle is circular and magnetic field provides the necessary centripetal force, i.e.,

$\frac{m{v}^2}{r}=Bev\Rightarrow r=\frac{mv}{Be}$

$=\frac{9.1\times {10}^{-31}\times 5\times {10}^6}{2\times {10}^{-3}\times 1.6\times {10}^{-19}}$

$=14.3\times {10}^{-3}m=1.43cm$