Question

The magnetic field in a certain region of space is given by $\vec{B}=\left(8.35 \times 10^{-2} \hat{i}\right) T$. A proton is shot into the field with velocity $\vec{v}=\left(2 \times 10^{5} \hat{i}+4 \times 10^{5} \hat{j}\right) m / s$. The proton follows a helical path in the field. The distance moved by proton in the $x$-direction during the period of one revolution in the $y z$ - plane will be (Mass of proton $=1.67 \times 10^{27}\, kg$ )

• A. 0.053 m
• B. 0.136 m
• C. 0.157 m
• D. 0.236 m

Answer 1

Answer: (C)

Given, $B=8.35 \times 10^{-2} i T$
$v=\left(2 \times 10^{5} i+4 \times 10^{5} j\right)$
The distance covered by proton
$P=T(v)=2 \pi \frac{m}{q B}$
$(v)=2 \times 3.14 \times \frac{1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 8.35 \times 10^{-2} i}$
$\times\left(2 \times 10^{5} i+4 \times 10^{5} j\right) P =0.157\, m$
(Mass of proton $=1.67 \times 10^{-27}\, kg$)