Question

A mono energetic electron beam with the speed of $5.2\times {10}^{6}m{s}^{-1}$ enters into the magnetic field of induction $3\times {10}^{-4}T$, directed normal to the beam. Then the radius of the circle traced by the beam :
(given $e/m=1.76\times {10}^{11}CK{g}^{-1}$)

• A. $0.098m$
• B. $0.98m$
• C. $0.089m$
• D. $9.8m$

Answer 1

The correct option is A $0.098m$

Magnetic field $=3\times {10}^{-4}$
Now, Force required for centripetal acceleration $=\frac{m{v}^2}{r}$ --------------------- (I)
For provided by B towards the center $=qvB$
For the electron to move in a circle,
$\frac{m{v}^2}{r}=qvB$ [(I) = (II)]
$\Rightarrow \frac{mv}{Bq}=r\Rightarrow r=\frac{5.2\times {10}^6}{3\times {10}^{-4}\times 1.76\times {10}^{11}}$
$\Rightarrow r=0.09848m$
$\Rightarrow r=0.098m$