9. Electrons pass through the parallel electrodes shown in FIGURE EX37.9 with a speed of m/s towards left. What magnetic field strength and direction will allow the electrons to pass through without being deflected? Assume that the magnetic field is confined to the region between the electrodes.
Magnetic field will be tesla and is directed into the plane of paper.
Given values are :-Voltage =V, Length=mm, Width= cm ,Speed=m/s towards left.
We have to find the magnetic field and its direction which will cancel the force due to electric field on electrons so that the electron can pass without being deflected.
Electric field E=
The upper plate is at a higher potential so Electric field will be in downward direction .
But electron being negatively charged will have a force in upward direction of magnitude q E .
Magnetic force on electron is given by where B is magnetic field.
Since electron is going left and we need the resulting magnetic force to be in downward direction (opposite to electric field) so that net force is zero.
BY using right hand rule we can see that magnetic field is directed into the plane of paper.
FIGURE Q37.6 shows a magnetic field between two parallel, charged electrodes. An electron with speed passes between the plates, from left to right, with no deflection. If a proton is fired toward the plates with the same speed ,will it be deflected up, deflected down, or pass through with no deflection? Explain.
An electron in a cathode-ray beam passes between long parallel-plate electrodes that are apart. A
, wide magnetic field is perpendicular to the electric field between the plates. The electron passes through the electrodes without being deflected if the potential difference between the plates is .
a. What is the electron’s speed?
b. If the potential difference between the plates is set to zero, what is the electron’s radius of curvature in the magnetic field?
94% of StudySmarter users get better grades.Sign up for free