# *8 and 9 plz The accelerating voltage V is now set to the value V1 which accel
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# *8 and 9 plz
The accelerating voltage V is now set to the value V1 which accelerates electrons to a kinetic energy of 4.8*10^-17J. What is the magnitude of field required to have the electron move on a circular path of diameter a?
Physics 212 Spring 2012 Exam II The next three questions pertain to the situation described below. In the demonstration we did in class, we accelerated electrons through a potential difference of V-300 V and applied a O O O magnetic field in a square region as in the diagram. The mass of an electron is 9.11x10 kg and its charge is k. 1.60x10 C. The diameter of the circular path is 20 cm o o o o o o o o 7. If the potential difference through which the electrons are accelerated was increased by a factor of 2, the speed of the electrons would AV a) increase by a factor of 2 increase by a factor of 2 c) not change 8. The accelerating voltage V is now set to the value Vi which accelerates electrons to a kinetic energy of 4.8x10 IT J. What is the magnitude of the field, B, required to have the electron move on a circular path of diameter a? 4 3x10 a) 8.20x108 T b) 1.71x10 9.61x10-6 T d) 3.84x104 T .01x103 T 9. If the charge of the 'electron' were doubled, how would the magnetic field required to have it follow this circular path a) B would double V2 b) B would decrease by a factor of c) B would decrease by a factor of 2 5 of 11 pagesExplanation / Answer
According to the given problem,
8)Electron is moving in a circular orbit due to the magnetic field,
mv2/r = qvB
B = mv/qr
where q = charge of electron, r = radius , m = mass of electron
velocity can be calcualte using the K.E given,
v = 2K.E/m
v = (2*4.8*10-17)/9.11*10-31
v = 1.026*107m/s
Substitute the value in the equation above,
B = 9.11*10-31*1.026*107/(1.602*10-19*0.1m)
B = 5.837*10-4 T[D]
9.)As from the formula,
B 1/q so, when charge doubles, Magnetic field becomes half
Answer:[C]
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