Question

An electron starts from rest and travels $0.9m$ in an electric field of $200V/m$. After this, it enters a magnetic field at right angles to its direction of motion. If the radius of circular path of the electron is $9cm$, the magnetic field induction is :
(Given $e=1.6\times {10}^{-19}C$, $m=9\times {10}^{-31}kg$)

• A. $5\times {10}^{-4}Wb/m^2$
• B. $5\times {10}^{-5}Wb/m^2$
• C. $5\times {10}^{-3}Wb/m^2$
• D. $5\times {10}^{-2}Wb/m^2$

Answer 1

Answer: (A)

$5\times {10}^{-4}Wb/m^2$

After travelling $0.9m$ in electric field,

Kinetic energy gained by electron $=eVd[whered=0.9m]$
Velocity at the time of entering magnetic field $=V_1$
So, $\frac{1}{2}m{v}^2=eVd$
$\Rightarrow V_1=\sqrt{\frac{2eVd}{m}}$--------------------------(I)
Now, Radius of Magnetic field $=\frac{mV}{Bq}$ -------------------------(II)
$r=\frac{m{V}_1}{Bq}\Rightarrow B=\frac{m{V}_1}{re}$
From (I), $B$ = $\frac{m}{re}\sqrt{\frac{2eVdm}{em}}$ = $\frac{1}{r}\sqrt{\frac{2Vdm}{e}}$ = $\frac{100}{9}\sqrt{\frac{2\times 200\times 0.9\times 9\times {10}^{-31}}{1.6\times {10}^{-19}}}$
$B=5\times {10}^{-4}Wb/m^2$