Question

what is the magnetic force on a proton that is moving at 5.2×10⁷ m/s to the right through a magnetic field that is 1.4 t and pointing away from you? the charge on a proton is 1.6×10⁻¹⁹ c. use f = qv×b sin(θ).
a. 8.3×10⁻¹¹ n left
b. 1.2×10⁻¹¹ n down
c. 1.2×10⁻¹¹ n up
d. 0 n

Explanation:

Step1: Calculate the magnitude of the magnetic force

Given \(F = qvB\sin(\theta)\), where \(q = 1.6\times10^{-19}\text{ C}\), \(v = 5.2\times10^{7}\text{ m/s}\), \(B = 1.4\text{ T}\), and \(\theta = 90^{\circ}\) (since the velocity is perpendicular to the magnetic field, \(\sin(90^{\circ})=1\)).

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Step2: Determine the direction of the force

Using the right - hand rule:

• Point the fingers of the right hand in the direction of the velocity (to the right).
• Curl the fingers in the direction of the magnetic field (away from you).
• The thumb points in the direction of the force on a positive charge (up).

Answer:

C. \(1.2\times 10^{-11}\text{ N}\) up