Explanation:
It is given that,
The horizontal and vertical component of velocity of an electron is:
[tex]v=(2.6i+4j)\times 10^6\ m/s[/tex]
The magnetic field acting there is given by :
[tex]B=(0.037i-0.17j)\ T[/tex]
(a) The magnitude of the magnetic force on the electron is given by :
[tex]F=q(v\times B)[/tex]
q = e
[tex]F=e(v\times B)[/tex]
[tex]F=1.6\times 10^{-19}\times ((2.6i+4j)\times 10^6\times (0.037i-0.17j))[/tex]
[tex]F=-1.6\times 10^{-19}\times (-0.442\cdot10^{6}-0.1702\cdot10^{6})\ kN[/tex]
[tex]F=9.79\times 10^{-14}\ kN[/tex]
(b) We know that the charge on proton is :
[tex]q=+1.6\times 10^{-19}\ C[/tex]
The magnetic force as same as for electron but the direction is opposite i.e.
[tex]F=-9.79\times 10^{-14}\ kN[/tex]
Hence, this is the required solution.
