Question

A hydrogen ion moving with a velocity of ${10}^{4}m/s$ enters in a magnetic field ${10}^{-4}T$ perpendicularly, then its radius of circular path is:

• A. $10.01m$
• B. $1.04m$
• C. $101m$
• D. $0.101m$

Answer 1

The correct option is B $1.04m$

Given that,

Velocity $v={10}^{4}m/s$

Magnetic field $B={10}^{-4}T$

Mass of hydrogen ion $m=1.67\times {10}^{-27}kg$

Charge $q=1.6\times {10}^{-19}C$

We know that,

$qvB=\frac{m{v}^2}{r}$

$r=\frac{mv}{qB}$

$r=\frac{1.67\times {10}^{-27}\times {10}^4}{1.6\times {10}^{-19}\times {10}^{-4}}$

$r=1.04m$

Hence, the radius is $1.04m$