An electron with a speed of 6.00 * 10⁵ m/s enters a uniform magnetic field of 0.0600 T at an angle of 61.4 $\circ$ to the magnetic field lines.The electron will travel in a helical path.Find the radius of the helical path.

Question

Quizplus · Accepted Answer

The radius of the helical path is given by:$$r = \frac{mv}{qB\sin	heta}$$where:* m is the mass of the electron* v is the speed of the electron* q is the charge of the electron* B is the magnetic field strength* θ is the angle between the electron's velocity and the magnetic field linesPlugging in the given values, we get:$$r = \frac{(9.11 * 10^{-31} 	ext{ kg})(6.00 * 10^5 	ext{ m/s})}{(1.60 * 10^{-19} 	ext{ C})(0.0600 	ext{ T})\sin(61.4^\circ)}$$$$r = 4.80 * 10^{-5} 	ext{ m}$$

An Electron with a Speed of 6 ∘\circ∘ to the Magnetic Field Lines

An Electron with a Speed of 6 $\circ$ to the Magnetic Field Lines