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in the figure, an electron moves at speed v = 153 m/s along an x - axis through uniform electric and magnetic fields without accelerating. the magnetic field b is directed into the page and has magnitude 7.03 t. in unit - vector notation, what is the electric field?
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i ^+
j ^+
k ^) units

Explanation:

Step1: Recall the force - balance condition

For an electron to move without accelerating in crossed electric and magnetic fields, the electric force $\vec{F}_e = q\vec{E}$ and the magnetic force $\vec{F}_b=q\vec{v}\times\vec{B}$ must balance each other, so $q\vec{E}=-q\vec{v}\times\vec{B}$ (the charge of an electron $q = - e$), and $\vec{E}=-\vec{v}\times\vec{B}$.

Step2: Express the vectors in unit - vector notation

Given $\vec{v}=153\hat{i}\text{ m/s}$ and $\vec{B}=- 7.03\hat{k}\text{ T}$. Using the cross - product formula $\vec{A}\times\vec{B}=(A_yB_z - A_zB_y)\hat{i}+(A_zB_x - A_xB_z)\hat{j}+(A_xB_y - A_yB_x)\hat{k}$, for $\vec{v}\times\vec{B}$, we have $A_x = 153$, $A_y = 0$, $A_z = 0$, $B_x = 0$, $B_y = 0$, $B_z=-7.03$. Then $\vec{v}\times\vec{B}=(0 - 0)\hat{i}+(0-153\times(- 7.03))\hat{j}+(0 - 0)\hat{k}=1075.59\hat{j}\text{ N/C}$.

Step3: Calculate the electric field

Since $\vec{E}=-\vec{v}\times\vec{B}$, $\vec{E}=- 1075.59\hat{j}\text{ N/C}$.

Answer:

$-1075.59\hat{j}\text{ N/C}$