Respuesta :
Explanation:
It is given that,
Kinetic energy of the electron, [tex]E_k=10\ keV=10^4\ eV=1.6\times 10^{-15}\ J[/tex]
Let the east direction is +x direction, north direction is +y direction and vertical direction is +z direction.
The magnetic field in north direction, [tex]B_y=19911.5\ nT[/tex]
The magnetic field in west direction, [tex]B_x=-3257.1\ nT[/tex]
The magnetic field in vertical direction, [tex]B_z=48381.8 \ nT[/tex]
Magnetic field, [tex]B=(-3257.1i+19911.5j+48381.8k)\ nT[/tex]
Firstly calculating the velocity of the electron using the kinetic energy formulas as :
[tex]E_k=\dfrac{1}{2}mv^2[/tex]
[tex]v=\sqrt{\dfrac{2E_k}{m}}[/tex]
[tex]v=\sqrt{\dfrac{2\times 1.6\times 10^{-15}}{9.1\times 10^{-31}}}[/tex]
[tex]v=5.92\times 10^7 i\ m/s[/tex] (as it is moving from west to east)
The force acting on the charged particle in the magnetic field is given by :
[tex]F=q(v\times B)[/tex]
[tex]F=1.6\times 10^{-19}\times (5.92\times 10^7 i\times (-3257.1i+19911.5j+48381.8k)\times 10^{-9})[/tex]
Since, [tex]i\times j=k\ \\j\times k=i\\k\times i=j[/tex]
And, [tex]i\times i=j\times j=k\times k=0[/tex]
[tex]F=1.6\times 10^{-19}\times [1178 k-2864.20j][/tex]
[tex]|F|=1.6\times 10^{-19}\times \sqrt{1178^2+2864.20^2}[/tex]
[tex]F=4.95\times 10^{-16}\ N[/tex]
(b) Let a is the acceleration of the electron. It can be calculated as :
[tex]a=\dfrac{F}{m}[/tex]
[tex]a=\dfrac{4.95\times 10^{-16}}{9.1\times 10^{-31}}[/tex]
[tex]a=5.43\times 10^{14}\ m/s^2[/tex]
Hence, this is the required solution.
