Question

An ion with a charge of $ + 32 \times 10^{-19}C $ is in a region where a uniform electric field of $ 5 \times 10^4\, V/m $ is perpendicular to a uniform magnetic field of $ 0.8 \,T $ . If its acceleration is zero, then its speed must be

• A. $ 0 $
• B. $ 1.6 \times10^{4}\, m/s $
• C. $ 4.0 \times10^{4}\, m/s $
• D. $ 6.3 \times10^{4}\, m/s $

Answer 1

Answer: (D)

The force experienced by the charge
$q = 3.2 \times 10^{-19}\, C$ in electric field $F_e = qE$
The force experienced by the charge in magnetic
field,
$F_m = q ( v \cdot B)$
$= qvB\, \sin\, \theta$
Here, $\theta = 90^{\circ}$
$F_m = qvB$
As the charge particle is not accelerated, so the resultant forces
$F_e = F_m$
$qE = qvB$
$ \Rightarrow v = \frac{E}{B}$
We have $E = 5 \times 10^4 \,V/m$
and $B = 0.8 \,T$
$\therefore $ speed $v = \frac{5\times 10^4}{0.8} $
$ = \frac{50}{8} \times 10^4$
$ = 6.3 \times 10^4\,m/s$