Question

A charge having $q/m$ equal to $10^8 \ c/kg$ and with velocity $3×10^5 \ m/s$ enters into a uniform magnetic field $B=0.3 \ tesla$ at an angle $30º$ with direction of field. Then radius of curvature will be:

• A. $0.01\ cm$
• B. $0.5 \ cm$
• C. $1\ cm$
• D. $2\ cm$

Answer 1

Answer: (B)

$0.5 \ cm$

Answer 2

$r=\frac {mV_⊥}{qB}$

$r=(\frac {m}{q})(\frac {3×10^5×sin\ 30°}{0.3})$

$r=\frac {3×10^5}{10^8×0.3×2}$

$r=0.5×10^{-2}m$

$r =0.5\ cm$

So, the correct option is (B): $0.5\ cm$