A particle of unknown mass orbits in the Earth’s magnetic field. Over the partic

Question

A particle of unknown mass orbits in the Earth’s magnetic field. Over the particle’s orbit the Earth’s field is uniform and the strength of the Earth’s field is B = 4 × 10^5T. The magnetic field is perpendicular to the velocity of the particle. The particle’s trajectory is drawn below with the magnetic field directed into the page. The charge of the particle has magnitude |q| = 1.6 × 10^19C. The radius of the orbit is 4.27m. The particle orbits in the direction drawn with velocity 1/10 th the speed of light or 3 × 10^7 m s (a)What is the sign of the particle’s charge? Justify. (b)What it the mass of the particle? (c)What is the force (magnitude and direction) on the particle at point P? (d)Identify the particle, that is report the name of an elementary particle with features consistent your calculated features of the particle.

Explanation / Answer

part a )

negative , because the particle’s trajectory is drawn below with the magnetic field directed into the page.

part b )

F =qvB

F = mv^2/r

qvB = mv^2/r

qB = mv/r

m = rqB/v

m = 4.27 x 1.6 x 10^-19 x 4 x 10^-5 / 3 x 10^7

m = 9.1 x 10^-31 kg

part c )

F = qvB

F = 1.6 x 10^-19 x 3 x 10^7 x 4 x 10^-5

F = 1.92 x 10^-16 N

direction = upward

part d )

particle is electron

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