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Determining the velocity of a charged particle physics help!?

I am having trouble finding the correct answer to this problem, and any help would be appreciated. A particle with a charge of -5.80 nC is moving in a uniform magnetic field given by B = (-1.23 T) k̂. The magnetic force acting on the particle is measured to be F = (-7.60 × 10⁻⁷ N) ĵ. Can you help me calculate the x component of the particle's velocity in m/s? Additionally, can the y and z components of the velocity (Vy and Vz) be determined by measuring the force?

2 Answers

F = q(v x B) (x is vector cross product)

F(y) = q(v(z)B(x) - v(x)B(z))

Since B(x) = 0, v(z) cannot be determined from F(y), and

F(y) = -qv(x)B(z) ==> v(x) = -F(y)/(qB(z))

I get v(x) = +106.5321 m/s

A
Anonymous

Feb 17, 2025

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E = σ /εₒ Upward force on the proton = Ee and the downward force = mg Net acceleration upward = [Ee-mg] /m In the vertical direction the distance moved = 0.01 m Time of travel = 0.04/2e6 =2.e-8 s using s = ½ a t^2 0.01 = 0.5 *{ (σ*1.6e-19/8.854e-12) – 1.67e-27*9.8}(2.e-8)²/1.67e-27 σ = 4.6e-6 c/m^2 =======================

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