Question

A beam of ions enters normally into a uniform magnetic field of $4 \times 10^{-2} T$ with velocity of $2 \times 10^{5} m / s$. If the specific charge of the ion is $5 \times 10^{7} C / kg$, then the radius of the circular path described will be

• A. 0.10 m
• B. 0.06 m
• C. 0.20 m
• D. 0.25 m

Answer 1

Answer: (A)

Force due to magnetic field provides the required centripetal force to the ions.
$\therefore $ Centripetal force = force due to magnetic field
$\frac{m v^{2}}{r}=e v B \sin \theta$
Given, $\theta=90^{\circ}, \frac{e}{m}=5 \times 10^{7} C / kg$,
$B =4 \times 10^{-2} T,$
$v=2 \times 10^{5} m / s$
$\therefore r=\frac{m v}{e B}$
$\Rightarrow r=\frac{2 \times 10^{5}}{5 \times 10^{7} \times 4 \times 10^{-2}}$
$=0.1\, m$