Question

A proton travels at 1.2x10^7 m/s in a plane perpendicular to a uniform 0.020T magnetic field. Give a quantitativ description of its path

Answer 1

Gives

m=$1.67\times10^{-27}kg$

$V=1.2\times10^7m/sec$

Magnetic field (B)=0.020T

$\theta=90°$

We know that

Circuler path traced by the proton

$\frac{mv^2}{r}=q(v\times B)$

$\frac{mv^2}{r}=qvBsin90°$

$\frac{mv^2}{r}=qvB$

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

Put value

$$r=\frac{1.67\times10^{-27}\times{1.2\times10^7}}{{1.6\times10^{-19}\times0.02}}=6.2625m$$

Radius of circuler path (r)=6.2625 m