A charged particle having electric charge Q = 4.0 e enters a region of magnetic field 1.2 T at an angle of $\theta$ relative to the magnetic field direction.If the speed of the particle upon entering the region is 2.8 $\times$ 10⁴ m/s and the magnitude of the force on it is 1.3 $\times$ 10^-14 N,what is the angle $\theta$? A) 47$^{\circ}$ B) 37$^{\circ}$ C) 53$^{\circ}$ D) 65$^{\circ}$

Question

Quizplus · Accepted Answer

The magnetic force on a charged particle is given by F = QvBsin(θ). Solving for θ using the given values: F = 1.3 x 10^-14 N, Q = 4.0 x 1.6 x 10^-19 C, v = 2.8 x 10^4 m/s, B = 1.2 T, we find θ ≈ 37°.

A Charged Particle Having Electric Charge Q = 4 θ\thetaθ Relative to the Magnetic Field Direction

A Charged Particle Having Electric Charge Q = 4 $\theta$ Relative to the Magnetic Field Direction