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General
Easy
Question
A velocity filter uses the properties of electric and magnetic fields to select charged particles that are moving with a specific velocity. Charged particles with varying speeds are directed into the filter as shown in figure. The filter consists of an electric field E and a magnetic field B, each of constant magnitude, directed perpendicular to each other as shown. The particles that move straight through the filter with their direction unaltered by the fields have the specific filter speed, 
. Those with speeds to may experience sufficiently little deflection that they also enter the detector.
The charged particle will experience a force due to the electric field given by the relationship 
where q is the charge of the particle and 
is the electric field. The moving particle will also experience a force due to the magnetic field. This force acts to oppose the force due to the electric field. The strength of the force due to the magnetic field is given by the relationship 
where q is the charge of the particle, 
is the speed of the particle, and 
is the magnetic field strength. When the forces due to the two fields are equal and opposite, the net force on the particle will be zero, and the particle will pass through the filter with its path unaltered. The electric and magnetic field strengths can be adjusted to choose the specific velocity to be filtered. The effects of gravity can be neglected.
The electric and magnetic fields in the filter of figure are adjusted to detect particles with positive charge q of a certain speed, . Which of the following expressions is equal to this speed ?
- B/(q2E)
- E/(q2B)
- B/E
- E/B
The correct answer is: E/B
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Curves in the graph shown give, as functions of radial distance r, the magnitude B of the magnetic field inside and outside four long wires a, b, c and d, carrying currents that are uniformly distributed across the cross– sections of the wires. Overlapping portions of the plots are indicated by double labels
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Which wire has the greatest magnitude of the magnetic field on the surface ?
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Which wire has the greatest radius?
Curves in the graph shown give, as functions of radial distance r, the magnitude B of the magnetic field inside and outside four long wires a, b, c and d, carrying currents that are uniformly distributed across the cross– sections of the wires. Overlapping portions of the plots are indicated by double labels
Which wire has the greatest radius?
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The following experiment was performed by J.J.Thomson in order to measure the ratio of the charge e to the mass m of an electron. Figure shows a modern version of Thomson's apparatus. Electrons emitted from a hot filament and accelerated by a potential difference V. As the electrons pass through the deflector plates, they encounter both electric and magnetic fields. When the electrons leave the plates they enter a field–free region that extends to the fluorescent screen. The beam of electrons can be observed as a spot of light on the screen. The entire region in which the electrons travel is evacuated with a vacuum pump.
Thomson's procedure was to first set both the electric and magnetic fields to zero, note the position of the undeflected electron beam on the screen, then turn on only the electric field and measure the resulting deflection. The deflection of an electron in an electric field of magnitude E is given by 
, where L is the length of the deflecting plates, and v is the speed of the electron. The deflection d1
can also be calculated from the total deflection of the spot on the screen, 
, and the geometry of the apparatus. In the second part of the experiment Thomson adjusted the magnetic field so as to exactly cancel the force applied by the electric field, leaving the electron beam undeflected. This gives 
By combining this relation with the expression for d1 one can calculate the charge to mass ratio of the electron as a function of the known quantities. The result is 
Why was it important for Thomson to evacuate the air from the apparatus ?
The following experiment was performed by J.J.Thomson in order to measure the ratio of the charge e to the mass m of an electron. Figure shows a modern version of Thomson's apparatus. Electrons emitted from a hot filament and accelerated by a potential difference V. As the electrons pass through the deflector plates, they encounter both electric and magnetic fields. When the electrons leave the plates they enter a field–free region that extends to the fluorescent screen. The beam of electrons can be observed as a spot of light on the screen. The entire region in which the electrons travel is evacuated with a vacuum pump.
Thomson's procedure was to first set both the electric and magnetic fields to zero, note the position of the undeflected electron beam on the screen, then turn on only the electric field and measure the resulting deflection. The deflection of an electron in an electric field of magnitude E is given by , where L is the length of the deflecting plates, and v is the speed of the electron. The deflection d1
can also be calculated from the total deflection of the spot on the screen, , and the geometry of the apparatus. In the second part of the experiment Thomson adjusted the magnetic field so as to exactly cancel the force applied by the electric field, leaving the electron beam undeflected. This gives By combining this relation with the expression for d1 one can calculate the charge to mass ratio of the electron as a function of the known quantities. The result is
Why was it important for Thomson to evacuate the air from the apparatus ?
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Figure shows a conducting rod of negligible resistance that can slide on smooth U-shaped rail made of wire of resistance 1 /m. Position of the conducting rod at t = 0 is shown. A time t dependent magnetic field B = 2t Tesla is switched on at t = 0
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