Question

what is the magnetic force on a proton that is moving at 4.5×10⁷ m/s to the right through a magnetic field that is 1.6 t and pointing away from you? the charge on a proton is 1.6×10⁻¹⁹ c. use f = qv×b sinθ.

a. 1.2×10⁻¹⁹ n down
b. 7.2×10⁻⁷ n up
c. 1.2×10⁻¹¹ n up
d. 7.2×10⁻⁷ n down

Explanation:

Step1: Determine the angle between velocity and magnetic field

The proton is moving to the right and the magnetic field is pointing away from you. Using the right - hand rule, the angle \(\theta = 90^{\circ}\), and \(\sin\theta=\sin90^{\circ} = 1\)

Step2: Calculate the magnitude of the magnetic force

We know that \(F=qvB\sin\theta\). Given \(q = 1.6\times10^{-19}\text{ C}\), \(v = 4.5\times10^{7}\text{ m/s}\), \(B = 1.6\text{ T}\), and \(\sin\theta = 1\)

Substitute the values into the formula:

\(F=(1.6\times 10^{-19}\text{ C})\times(4.5\times 10^{7}\text{ m/s})\times(1.6\text{ T})\times1\)

First, multiply the numerical coefficients: \(1.6\times4.5\times1.6=11.52\)

Then, use the rule of exponents for powers of 10: \(10^{-19}\times10^{7}=10^{-19 + 7}=10^{-12}\)

\(F = 11.52\times10^{-12}\text{ N}\approx1.2\times10^{-11}\text{ N}\)

Step3: Determine the direction of the force

Using the right - hand rule:

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

Answer:

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