Question

A proton is projected with a velocity ${10}^{7}m/s$ at right angle to a uniform magnetic field of induction 100 mT. The time taken by proton to traverse ${90}^{\circ }$ arc is

• A. $0.8\times {10}^{-7}s$
• B. $1.57\times {10}^{-7}s$
• C. $2.43\times {10}^{-7}s$
• D. $3.24\times {10}^{-7}s$

Answer 1

The correct option is B $1.57\times {10}^{-7}s$

$T=\frac{2\pi m}{Bq}=\frac{2\times 3.14\times 1.6\times {10}^{-27}}{0.1\times 1.6\times {10}^{-19}}T=6.28\times {10}^{-7}sect=\frac{T}{4}=\frac{6.28\times {10}^{-7}}{4}=1.57\times {10}^{-7}s$