Question

A carbon ion (c^+ ) moves in the xy-plane with a speed of 2.30 ✕ 103 m/s. if a constant magnetic field is directed along the z-axis with a magnitude of 2.75 ✕ 10−5 t, find the magnitude of the magnetic force acting on the ion and the magnitude of the ion's acceleration. hint (a) the magnitude (in n) of the magnetic force acting on the ion n (b) the magnitude (in m/s2) of the ion's acceleration m/s²

A) 6.33 × 10^−2 N; 3.82 × 10^4 m/s²
B) 5.02 × 10^−2 N; 4.13 × 10^4 m/s²
C) 6.66 × 10^−2 N; 2.93 × 10^4 m/s²
D) 4.87 × 10^−2 N; 3.51 × 10^4 m/s²

Answer 1

The magnetic force and acceleration of a carbon ion moving in a magnetic field can be calculated using the formula F = qvBsin(θ) and Newton's second law F = ma, respectively.

Step-by-step explanation:

The question asks for the magnitude of the magnetic force acting on a carbon ion and the magnitude of its acceleration when it moves in the presence of a magnetic field. To solve for the magnetic force (F), we use the formula F = qvBsin(θ), where q is the charge of the ion, v is the speed of the ion, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field. Since the ion moves in the xy-plane and the magnetic field is directed along the z-axis, the angle θ is 90 degrees, and sin(θ) is 1. Once the force is calculated, we can determine the acceleration (a) using Newton's second law F = ma, where m is the mass of the ion.