An electron with kinetic energy 2.5 keV moving along the positive direction of an x axis enters a region in which a uniform electric field of magnitude 10 kV`//`m is in the negative direction of the y-axis. A uniform magnetic field `vecB` is to be set up to keep the electron moving along the x-axis , and the direction `vecB` of is to be chosen to minimize the required magnitude of `vecB`. In unit-vector notation, what `vecB` should be set up?

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Calculate the speed of the electron
The kinetic energy (KE) of the electron is given as 2.5 keV. We can convert this to joules using the conversion factor \(1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}\).

\[ KE = 2.5 \text{ keV} = 2.5 \times 10^3 \text{ eV} = 2.5 \times 10^3 \times 1.6 \times 10^{-19} \text{ J} = 4.0 \times 10^{-16} \text{ J} \]

The speed \(v\) of the electron can be calculated using the formula:

\[ v = \sqrt{\frac{2 \cdot KE}{m}} \]

where \(m\) is the mass of the electron, approximately \(9.1 \times 10^{-31} \text{ kg}\).

Substituting the values:

\[ v = \sqrt{\frac{2 \cdot 4.0 \times 10^{-16}}{9.1 \times 10^{-31}}} = \sqrt{\frac{8.0 \times 10^{-16}}{9.1 \times 10^{-31}}} \approx 2.96 \times 10^7 \text{ m/s} \]

### Step 2: Determine the electric field and its direction
The electric field \(E\) is given as \(10 \text{ kV/m}\), which is equal to \(10 \times 10^3 \text{ V/m} = 10^4 \text{ V/m}\). The direction of the electric field is in the negative y-direction, represented as \(-\hat{j}\).

### Step 3: Apply the condition for velocity selection
To keep the electron moving along the x-axis, the electric force must be balanced by the magnetic force. The relationship can be expressed as:

\[ \vec{F}_E + \vec{F}_B = 0 \]

The electric force \(\vec{F}_E\) is given by:

\[ \vec{F}_E = q\vec{E} \]

where \(q\) is the charge of the electron (\(-1.6 \times 10^{-19} \text{ C}\)). Therefore,

\[ \vec{F}_E = -1.6 \times 10^{-19} \cdot 10^4 \hat{j} = -1.6 \times 10^{-15} \hat{j} \text{ N} \]

The magnetic force \(\vec{F}_B\) is given by:

\[ \vec{F}_B = q(\vec{v} \times \vec{B}) \]

### Step 4: Set up the equation for the magnetic field
For the electron to continue moving along the x-axis, we need:

\[ \vec{F}_E + \vec{F}_B = 0 \Rightarrow \vec{F}_B = -\vec{F}_E \]

Thus, we have:

\[ q(\vec{v} \times \vec{B}) = 1.6 \times 10^{-15} \hat{j} \]

### Step 5: Determine the direction of the magnetic field
Since \(\vec{v}\) is in the positive x-direction (\(\hat{i}\)), and \(\vec{E}\) is in the negative y-direction (\(-\hat{j}\)), we can use the right-hand rule to find the direction of \(\vec{B}\).

To satisfy the equation \(\vec{v} \times \vec{B} = \hat{j}\), we can set \(\vec{B}\) in the positive z-direction (\(\hat{k}\)).

### Step 6: Calculate the magnitude of the magnetic field
Using the relation:

\[ |\vec{F}_B| = |q|vB \]

Setting this equal to the magnitude of the electric force:

\[ 1.6 \times 10^{-15} = 1.6 \times 10^{-19} \cdot (2.96 \times 10^7) \cdot B \]

Solving for \(B\):

\[ B = \frac{1.6 \times 10^{-15}}{1.6 \times 10^{-19} \cdot 2.96 \times 10^7} \approx 3.4 \times 10^{-4} \text{ T} \]

### Final Answer
The magnetic field \(\vec{B}\) in unit vector notation is:

\[ \vec{B} = 3.4 \times 10^{-4} \hat{k} \text{ T} \]