Question

A charged particle accelerated through a potential difference of $100V$ passes through uniform electric and magnetic fields so as to experience no deflection. $E=15\times {10}^{6}V{m}^{-1}$ and $B=5\times {10}^{3}T$. Then the specific charge $\frac{e}{m}$ is :

• A. $4.5\times {10}^{4}C/kg$
• B. $9\times {10}^{7}C/kg$
• C. $4.5\times {10}^{3}C/kg$
• D. $9\times {10}^{5}C/kg$

Answer 1

The correct option is A $4.5\times {10}^{4}C/kg$

$K.E.=V\times e$
$\frac{1}{2}m{v}^2=100\times e$ --(1)

for no deflection, $vB=E$
$v=\frac{E}{B}$
$=\frac{15\times {10}^6}{5\times {10}^3}$
$=3\times {10}^3\frac{m}{s}$

Now putting value of $v$ in equation --(1)
$\frac{1}{2}m(3\times {10}^3{)}^2=100\times e$
$\frac{(3\times {10}^3{)}^2}{2\times 100}=\frac{e}{m}$
$\Rightarrow \frac{e}{m}=\frac{3\times {10}^3\times 3\times {10}^3}{2\times 100}$
$=4.5\times {10}^4\frac{C}{kg}$.