An electron enters a crossed electrical and magnetic field region with a speed of 100 ms^-1 along + x-axis and passes undeviated. The strength of magnetic field is 5.0 T and it acts along + z-axis. Give the electrical field in vector notations indicating its magnitude as well as direction:

The correct option is:

$\left(500\text{V m}{}^{-1}\right)\hat{j}$

Step-by-step Explanation:

Step 1: Understand the condition for undeviated motion
When a charged particle passes through a region containing both electric and magnetic fields, it experiences both electric and magnetic forces. For the particle to pass undeviated (meaning it moves in a straight line at a constant velocity), the net force acting on it must be zero.

The net electromagnetic force (Lorentz force) on a charge $q$ is given by:

${\overrightarrow{F}}_{\text{net}}={\overrightarrow{F}}_e+{\overrightarrow{F}}_m=q\overrightarrow{E}+q(\overrightarrow{v}\times \overrightarrow{B})$

Setting the net force to zero for undeviated motion:

$q\overrightarrow{E}+q(\overrightarrow{v}\times \overrightarrow{B})=\overrightarrow{0}$

Dividing by the charge $q$ (since $q\ne 0$), we get the relationship between the electric field and the magnetic force terms:

$\overrightarrow{E}+(\overrightarrow{v}\times \overrightarrow{B})=\overrightarrow{0}$

$\overrightarrow{E}=-(\overrightarrow{v}\times \overrightarrow{B})$

Step 2: Express the given physical quantities as vectors
- Velocity ($\overrightarrow{v}$) is along the positive x-axis ($\hat{i}$ direction) with a magnitude of 100 m s^-1:

$\overrightarrow{v}=100\hat{i}{\text{m s}}^{-1}$

- Magnetic field ($\overrightarrow{B}$) is along the positive z-axis ($\hat{k}$ direction) with a magnitude of 5.0 T:

$\overrightarrow{B}=5.0\hat{k}\text{T}$

Step 3: Calculate the cross product $\overrightarrow{v}\times \overrightarrow{B}$
Substitute the vectors into the cross product expression (keeping in mind the unit vector cross product rule $\hat{i}\times \hat{k}=-\hat{j}$):

$\overrightarrow{v}\times \overrightarrow{B}=\left(100\hat{i}\right)\times \left(5.0\hat{k}\right)$

$\overrightarrow{v}\times \overrightarrow{B}=500(\hat{i}\times \hat{k})$

$\overrightarrow{v}\times \overrightarrow{B}=-500\hat{j}$

Step 4: Determine the electric field vector
Using the relation from Step 1, substitute the cross product:

$\overrightarrow{E}=-(-500\hat{j})$

$\overrightarrow{E}=500\hat{j}{\text{V m}}^{-1}$

Thus, the required electric field has a magnitude of 500 V m^-1 and acts in the positive y-axis direction ($\hat{j}$).