An electron with a speed of 6.00 * 10⁵ m/s enters a uniform magnetic field of 0.0600 T at an angle $\theta$ to the magnetic field lines.The electron will travel in a helical path of radius 4.80 *10^-5 m.Find the angle $\theta$.

Question

Quizplus · Accepted Answer

We can use the equation for the radius of the helical path:
r = mv⊥/(qB)
where m is the mass of the electron, v⊥ is the component of the velocity perpendicular to the magnetic field (which is equal to v sinθ), q is the charge of the electron, and B is the magnitude of the magnetic field.

We can solve for the angle θ by using the given values for r, v, and B:
sinθ = rB/(mv)
sinθ = (4.80×10^-5 m)(0.0600 T)/[(9.11×10^-31 kg)(6.00×10^7 m/s)]
sinθ = 0.992
θ = sin^-1(0.992)
θ ≈ 57.6°

Therefore, the answer is choice C, 57.6°.

An Electron with a Speed of 6 θ\thetaθ To the Magnetic Field Lines

An Electron with a Speed of 6 $\theta$ To the Magnetic Field Lines