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An +9.7 C charge moving at 0.75 m/s makes an angle of 45∘ with a uniform, 1.5 T magnetic field. What is the magnitude of the magnetic force F that the charge experiences?

Respuesta :

hamzaahmeds

Answer:

F = 7.72 N

Explanation:

The magnetic force on the charge can be given by the following formula:

[tex]F = qvB Sin\theta[/tex]

where,

F = magnetic force = ?

q = magnitude of charge = 9.7 C

v = speed of charge = 0.75 m/s

B = magnetic field = 1.5 T

θ = angle = 45°

Therefore,

[tex]F = (9.7\ C)(0.75\ m/s)(1.5\ T)Sin45^{o}[/tex]

F = 7.72 N

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