or 6. z direction An electron in a vacuum is first accelerated by a voltage of 1
ID: 2298306 • Letter: O
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or 6. z direction
An electron in a vacuum is first accelerated by a voltage of 1.019 Times 105 V and then enters a region in which there is a uniform magnetic field of 0.114 T at right angles to the direction of the electron's motion. The mass of the electron is 9.11 Times 10-31 kg and its charge is 1.60218 Times 10-19 C. What is the magnitude of the force on the electron due to the magnetic field? A uniform magnetic field B points vertically downward. At a certain time, a proton passes through a potential difference delta V. and moves with speed v from East to West through the field (see figure). What is the direction of the force acting on the proton at this time? Assume the proton is at rest initially. The charge on the proton Ls 1.602 Times 10-19 C and its mass is 1.67 Times 10-27 kg. If the magnetic field is 1.24 T and the potential difference Ls 1.83 MV, find the magnitude of the force on the proton after the proton passes through the potential difference. Answer in units of N The proton will subsequently move in a circle. How many revolutions will the proton perform per second? A wire carries a steady current of 2.76 A. A straight section of the wire is 0.593 m long and lies along the x axis within a uniform magnetic field of magnitude 1.86 T in the positive z direction. If the current is in the positive x direction, what is the magnetic force on the section of wire? A 0.0993 T magnetic field in a mass spectrometer causes an Isotope of sodium to move in a circular path with a radius of 0.052 m. The charge on an ion is 1.60218 Times 10-19 C. If the ions are moving with a speed of 16500 m/s, what is the isotope's mass? Answer in units of kgExplanation / Answer
1.
KE=qV=(1/2)*m*v^2
(1.60218*10^-19)*(1.019*10^5)=(1/2)*(9.11*10^-31)*v^2
v=1.893*10^8 m/s
so magnetic force
F=q*v*B=(1.60218*10^-19)*1.893*10^8*0.114
F=3.458*10^-12 N
2.
south
3.
a)
KE=qV=(1/2)*m*v^2
(1.602*10^-19)*(1.83*10^6)=(1/2)*(1.67*10^-27)*v^2
v=1.874*10^7 m/s
F=qvB=(1.602*10^-19)*(1.874*10^7)*1.24
F=3.72*10^-12 N
b)
r=mv/qB=(1.67*10^-27)*(1.874*10^7)/(1.602*10^-19)*1.24
r=0.1575 m
W=sqrt[F/mr] =sqrt[3.72*10^-12/1.67*10^-27*0.1575)
W=1.19*10^8 rad/s
in rev/s
W=1.19*10^8/2pi=1.89*10^7 rev/s
4.
a)
F=BIL =1.86*2.76*0.593
F=3.044 N
b)
in negative Y direction
5.
Centripetal force =magnetic force
mv^2/r=qvB
=>m=qBr/v =(1.60218*10^-19)*0.052*0.0993/16500
m=5.014*10^-26 kg
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