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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?​

1 Answer

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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°

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