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A proton experiences a force of 4 Newton when it enters perpendicular to the direction of the magnetic field with a speed 100 m/s. What is the force experienced by the proton if it enters parallel to the direction of the magnetic field with a speed of 200 m/s?

1 Answer

3 votes

Answer:

F = 0 N

Step-by-step explanation:

Force on a moving charge in constant magnetic field is given by the formula


F = q(\vec v* \vec B)

so here it depends on the speed of charge, magnetic field and the angle between velocity of charge and the magnetic field

here when charge is moving with speed 100 m/s in a given magnetic field then the force on the charge is given as


F = 4 N

now when charge is moving parallel to the magnetic field with different speed then in that case


\vec v \time \vec B = 0

so here we have

F = 0

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