Question

A proton is projected with a velocity ${10}^{7}m{s}^{-1},$ at right angles to a uniform magnetic field of induction 100 mT. The time (in seconds) taken by the proton to traverse $90{}^{\circ }$ arc is: (Mass of proton $=1.65\times {10}^{-27}kg$ and charge of proton $=1.6\times {10}^{-19}C$ )

• A. $0.81\times {10}^{-7}$
• B. $1.62\times {10}^{-7}$
• C. $2.43\times {10}^{-7}$
• D. $3.24\times {10}^{-7}$

Answer 1

Answer: (B)

When proton enters in a uniform magnetic field at right angle then it moves on a circular path. In this case velocity of proton. $v=\frac{Bqr}{m}$ where r = radius of circular path Distance covered by proton to traverse 90° arc $=\frac{1}{4}$ circumference $d=\frac{1}{4}\times 2\pi r=\pi r/2$ Time taken by proton to cover distance d $t=\frac{\pi r/2}{v}=\frac{\pi r}{2v}$ $t=\frac{\pi }{2}\frac{r}{Bqr/m}$ $t=\frac{\pi m}{2Bq}$ Putting $m=1.65\times {10}^{-27}kg,$ $B=100mT=100\times {10}^{-3}T,q=1.6\times {10}^{-19}C$ $t=\frac{3.14\times 1.65\times {10}^{-27}}{2\times 100\times {10}^{-3}\times 1.6\times {10}^{-19}}$ $t=1.62\times {10}^{-7}\sec$