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 $0.8T$ . If its acceleration is zero, then its speed must be

• A. $0$
• B. $1.6\times 10^{4}m/s$
• C. $4.0\times 10^{4}m/s$
• D. $6.3\times 10^{4}m/s$

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 $B=0.8T$
$\therefore$ speed $v=\frac{5\times 10^4}{0.8}$
$=\frac{50}{8}\times 10^4$
$=6.3\times 10^{4}m/s$