Answer to Question #112843 in Electricity and Magnetism for anggraeny

Question

The electrons in the beam of television tube have an energy of 12 keV
( ). The tube is oriented so that the electrons move horizontally from
south to north. At MIT, the Earth's magnetic field points roughly vertically down (i.e.
neglect the component that is directed toward magnetic north) and has magnitude B ~
T.
19 1 eV 1.6 10 J − = ×
5 5 10− ×

Answer 1

As per the question,

the energy of the electrons beam $(E)=12KeV=12\times 10^3\times 1.6\times10^{-19}J=19.2\times10^{-16}J$

Magnetic field $B=5.5\times 10^{-5}T$

We know that mass of the electron $m_e=9.1\times 10^{-31}kg$

Now applying the conservation of energy,

$\dfrac{mv^2}{2}=E$

$v=\sqrt{2E/m}=\sqrt{\dfrac{2\times 19.2\times 10^{-16}}{9.1\times 10^{-31}}} m/sec$

$\Rightarrow v = \sqrt{4.22\times 10^{15}}=6.5\times 10^{7}m/sec$

As per the question, the direction of electron from south to north and the direction of magnetic field is vertically downwards, now applying the right hand thumb rule, so force on the electron will be in the east ward, if it is positive charge if it is negative then it will be in the direction of westwards. Here it is given the electron, which have the negative charge, hence it will be in the east to west.

Force on the electron beam $F=evB\sin \theta = 1.6\times 10^{-19}\times 6.5\times 10^7\times 5.5\times 10^{-5}N$

$=57.2\times 10^{-17}N$