A charge of 4.5 × 10-5 C is placed in an electric field with a strength of 2.0 × 104 . What is the electric force acting on the
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Answer:
the strength of the magnetic field is 3 x 10⁻⁵ T
Explanation:
Given;
velocity of the cosmic ray, v = 5 x 10⁷ m/s
force experienced by the ray, f = 1.7 x 10⁻¹⁶ N
angle between the ray's velocity and the magnetic field, θ = 45⁰
The strength of the magnetic field is calculated as;
Therefore, the strength of the magnetic field is 3 x 10⁻⁵ T
A charged particle moving in a magnetic field experiences a force equal to:
Thus, the magnitude of the force that the proton experiences is given by:
The magnetic field is perpendicular to the proton's velocity, therefore, we have . Replacing the given values, we obtain: