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A proton travels with a speed of 4.2×106 m/s at an angle of 30◦ west of north. A magnetic field of 2.5 T points to the north. Find the magnitude of the magnetic force on the proton. (The magnetic force experienced by the proton in the magnetic field is proportional to the component of the proton’s velocity that is perpendicular to the magnetic field.)

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skyluke89

Answer:

[tex]8.4\cdot 10^{-13} N[/tex]

Explanation:

The magnitude of the magnetic force on the proton is given by:

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

where:

[tex]q=1.6\cdot 10^{-19} C[/tex] is the proton charge

[tex]v=4.2\cdot 10^6 m/s[/tex] is the proton velocity

[tex]B=2.5 T[/tex] is the magnetic field

[tex]\theta=30^{\circ}[/tex] is the angle between the direction of v and B

Substituting into the formula, we find

[tex]F=(1.6\cdot 10^{-19}C)(4.2\cdot 10^6 m/s)(2.5 T) sin 30^{\circ}=8.4\cdot 10^{-13} N[/tex]

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