Question

An electron passes undeflected through perpendicular electric and magnetic fields of intensity $3.4\times {10}^{3}V/m$ and $2\times {10}^{-3}Wb/m^2$ respectively. Then its velocity is:

• A. $1.7\times {10}^6$m/s
• B. $6.8\times {10}^6$ m/s
• C. $6.8$m/s
• D. $1.7\times {10}^8$ m/s

Answer 1

Answer: (A)

$1.7\times {10}^6$m/s

For undeflected path
$F=q(V\times B+E)=0$
or $VB=E$
$V=\frac{E}{B}$
= $\frac{3.4\times {10}^3}{2\times {10}^{-3}}$
= $1.7\times {10}^6\frac{m}{s}$.