Question

In: Physics

The earth’s north pole is a magnetic south pole. The field above the pole is ~0.05mTesla....

The earth’s north pole is a magnetic south pole. The field above the pole is ~0.05mTesla. A proton from the sun with a velocity of 1,000,000 m/s enters this magnetic field.

a) Find the force on the proton if the velocity is perpendicular to the Magnetic field.

b) This is a centripetal force that makes them take circular orbits. Find the radius of the orbit.

Solutions

Expert Solution

(a)

The magnetic force on charge q moving at velocity v in presence of magnetic field B is

The magnitude of the magnetic force is

where is the angle that velocity makes with the magnetic field. It is given that the velocity is perpendicular to the magnetic field . The magnitude of the magnetic force is

For proton , , . The force on the proton is

(b)

The centripetal force required to move an object of mass m in a uniform circular motion of radius r at speed v is

The magnetic force provides the centripetal force necessary to move the proton in circular motion

The mass of the proton is . The radius of the circular motion is

genius_generous answered 1 year ago

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