A proton with an initial velocity of (1.0x + 2.0y + 3.0z)x 10⁵ m/s enters a magnetic field of 0.50z T.Find the radius of the resulting helix.The mass of the proton is 1.67 * 10²⁷ kg.Where x,y and z are unit vectors in the x,y and z directions,respectively.

Question

Quizplus · Accepted Answer

The radius $r$ of the helical path of a charged particle moving in a magnetic field is given by the formula $r = \frac{mv_{\perp}}{qB}$, where $m$ is the mass of the particle, $v_{\perp}$ is the component of the velocity perpendicular to the magnetic field, $q$ is the charge of the particle, and $B$ is the magnetic field strength. Given:- Initial velocity $v = (1.0\hat{x} + 2.0\hat{y} + 3.0\hat{z}) \times 10^5 \, \text{m/s}$- Magnetic field $B = 0.50\hat{z} \, \text{T}$- Mass of the proton $m = 1.67 \times 10^{-27} \, \text{kg}$- Charge of the proton $q = 1.6 \times 10^{-19} \, \text{C}$Since the magnetic field is in the $z$-direction, the velocity component perpendicular to the magnetic field is in the $xy$-plane, $v_{\perp} = \sqrt{(1.0 \times 10^5)^2 + (2.0 \times 10^5)^2} = \sqrt{1^2 + 2^2} \times 10^5 = \sqrt{5} \times 10^5 \, \text{m/s}$.Plugging the values into the formula for $r$:$$r = \frac{mv_{\perp}}{qB} = \frac{1.67 \times 10^{-27} \times \sqrt{5} \times 10^5}{1.6 \times 10^{-19} \times 0.50} = \frac{1.67 \times \sqrt{5} \times 10^{-22}}{0.8 \times 10^{-19}}$$$$r \approx \frac{1.67 \times \sqrt{5}}{0.8} \times 10^{-3} \, \text{m}$$$$r \approx \frac{1.67 \times 2.236}{0.8} \times 10^{-3} \, \text{m}$$$$r \approx \frac{3.7342}{0.8} \times 10^{-3} \, \text{m}$$$$r \approx 4.67 \times 10^{-3} \, \text{m} = 4.7 \, \text{mm}$$Therefore, the radius of the resulting helix is approximately 4.7 mm.

A Proton with an Initial Velocity of (1