Answer with Explanation:
We are given that
[tex]v_x=2.6\times 10^6 m/s[/tex]
[tex]v_y=2.4\times 10^6 m/s[/tex]
[tex]B_x=0.029 T[/tex]
[tex]B_y=-0.14 T[/tex]
a.We have to find the magnitude of the magnetic force on the electron.
[tex]v\times B=\begin{vmatrix}i&j&k\\2.6\times 10^6&2.4\times 10^6&0\\0.029&-0.14&0\end{vmatrix}[/tex]
[tex]v\times B=(-0.364-0.0696)\times 10^6 k=-0.4336\times 10^6 k[/tex]
Charge on an electron,q=[tex]-1.6\times 10^{-19} C[/tex]
[tex]F=q(v\times B)=\mid -1.6\times 10^{-19}(-0.4336)\times 10^6\mid =6.9\times 10^{-14} N[/tex]
Force act along positive z- direction.
b.Charge on proton=[tex]q=1.6\times 10^{-19} C[/tex]
[tex]F=\mid 1.6\times 10^{-19}(-0.4336)\times 10^6\mid =6.9\times 10^{-14} N[/tex]
