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