A small particle of mass 2g and charge 10 muC is passes through the origin with
ID: 2057256 • Letter: A
Question
A small particle of mass 2g and charge 10 muC is passes through the origin with a velocity of 2 m/s in the +x-direction into a uniform magnetic field parallel to the x-axis and a uniform gravitational field of 1 g in the -y direction. (a) If there is no electrical field and the particle continues moving in the straight line at constant speed. what is the magnetic force on the particle? (b) What is the magnitude and direction of the magnetic field? (c) If the particle attained its speed by sliding down a ramp before conducting the magnetic field, what was the height of the ramp? A electron cyclotron has a frequency of 2450 Hz and a magnetic field strength of 50 T.(a) What is the radius of the electron's orbit? (b) What is the speed of the electron? (c) What was the voltage on the electron gun which projected the electron into the cyclotron? Each of the lettered points at the corners of the cube in the figure at right represents a positive charge q moving with a velocity v in the direction indicated. There is a uniform magnetic field it in the +x-direction as shown. Find the magnitude and direction of the magnetic force on each charge. The magnitude field in the figure below is = 5 T i. (a) What is the magnetic flux across the surface abed? (b) Through cdef? (c) Through abfe? (d) Through all five surfaces enclosing the volume? An alpha particle (nucleus of a helium atom) has a speed of 3 Times 106 m/s and there is a circle of cubes 10 cm in the x/y-plane to a uniform magnetic field. (a) What is the magnitude and there atom of the magnetic field? (c) What is the cyclotron frequency of the alpha particle? (d) Through what potential What is its energy in units of electron volts?Explanation / Answer
Since no numbers are given, I am just assuming that we are given a representation of the Force.
The general formula for Force is qvB where the velocity is perpendicular to the field.
The direction of the force is found by the right hand rule where your fingers point in the direction of the magnetic field, your thumb points in the direction of the velocity of the charge, and then your palm indicates the direction of the force
Point a)
The velocity is perpendicular to the field and moving in positive z. Since the B field is positive x, the force is...
F = qvB toward the positive y direction
Point b)
The velcocity has a component in the negative z and a component in the positive x. z is perpendicular but x is parallel, so the x component will not figure in. The force is therefore...
F = qvB (sin 45) toward the negative y direction
or .707qvB in the negative y direction
Point c)
The velocity is positive y and the field is positive x. Since its all perpendicular, the force is...
F = qvB toward the negative z direction
Point d)
The velocity is in the negative x and the field is positive x. There will be no force since they are parallel.
F = 0
Point e)
The velocity has a -z component and a positive y component. Therefore the magnitude of the force =
F = [(qvB sin 45)2 + (qvB sin 45)2]
F = 1.41 qvBsin 45 which is .806qvB
The direction is at a 45o angle pointing in the negative y and negative z direction
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