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The strength of the Earth’s magnetic field B at the equator is approximately equal to 5 × 10−5 T. The force on a charge q moving in a direction perpendicular to a magnetic field is given by F = q v B, where v is the speed of the particle. The direction of the force is given by the right-hand rule. Suppose you rub a balloon in your hair and your head acquires a static charge of 9 × 10−9 C. If you are at the equator and driving west at a speed of 90 m/s, what is the strength of the magnetic force on your head due to the Earth’s magnetic field? Answer in units of N. What is the direction of that magnetic force?

Respuesta :

skyluke89

1. Magnitude: [tex]4.05\cdot 10^{-11} N[/tex]

The magnitude of the magnetic force exerted on your head by the Earth's magnetic field is:

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

where we have:

[tex]q=9\cdot 10^{-9}C[/tex] is the charge on the head

v = 90 m/s is the magnitude of the velocity

[tex]B=5\cdot 10^{-5}T[/tex] is the strength of the magnetic field

[tex]\theta=90^{\circ}[/tex] is the angle between the direction of B and v (they are perpendicular)

Substituting into the equation, we find

[tex]F=(9\cdot 10^{-9} C)(90 m/s)(5\cdot 10^{-5} T) sin 90^{\circ}=4.05\cdot 10^{-11} N[/tex]

2. Direction: upward

The direction of the magnetic force can be found by using the right hand rule. We have:

- index finger: direction of velocity --> west

- middle finger: direction of magnetic field --> south (the magnetic field of the Earth goes from north pole to south pole)

- thumb: direction of the force --> upward

So, the direction of the magnetic force is upward.