A charge of 0.50 C moves horizontally in a magnetic field of 1.0 T with a speed of 4.0 x 10^2 m/s. The force experienced by the charge is 7.0 N. What is the angle between the charge direction and the magnetic field?​

Question: A charge of 0.50 C moves horizontally in a magnetic field of 1.0 T with a speed of 4.0 x 10^⁻1 m/s. The force experienced by the charge is 7.0 N. What is the angle between the charge direction and the magnetic field?​

Θ = 6.56°

Explanation:

From the question,

Using

F = BvqsinΘ Equation 1

Where F = force experienced by the charge, B = magnetic field, q = Charge, Θ = angle between the charge abd the magnetic field, v = velocity.

Make Θ the subject of the equation

Θ =sin⁻¹ (Bvq/F) Equation 2

Given: v = 4.0×10⁻¹ m/s, B = 1.0 T, q = 0.50 C, F = 7.0 N

Θ = sin⁻¹(1×0.5×4×10⁻¹/7)

Θ = sin⁻¹(0.114)

Θ = 6.56°

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer