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?
Answer:
Θ = 6.56°
Step-by-step 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°