Answer:
[tex]B_x=-0.56 T[/tex]
Step-by-step explanation:
We are given that an electron moves through a uniform magnetic field given by
[tex]B=B_x\hat{i}+3..39 B_x\hat{j}[/tex]
Velocity of electron=[tex]2.59\hat{i}+4.09\hat{j}[/tex]m/s
Magnetic force acting on electron, F=[tex]4.22\times 10^{-19} N[/tex]
We have to find [tex]B_x[/tex]
Charge on electron=q=[tex]-1.6\times 10^{-19}C[/tex]
We know that magnetic force on electron is given by
[tex]F=q\vec{v}\times \vec{B}[/tex]
Substitute the value then we get
[tex]4.22\times 10^{-19}=-1.6\times 10^{-19}(2.59\cdot 3.39B_x-4.09B_x)[/tex]
[tex](8.7801B_x-4.09B_x)=\frac{4.22\times 10^{-19}}{-1.6\times 10^{-19}}[/tex]
[tex]4.6901B_x=-\frac{4.22}{1.6}[/tex]
[tex]B_x=-\frac{4.22}{1.6\times 4.6901}[/tex]
[tex]B_x=-0.56 T[/tex]
