QJEEM24Ph8f46: JEE Main 2024 Physics - Class 12 JEEM24Ph8f46 Solution: An electron with kinetic energy enters a region of uniform…

To solve this problem, we first need to understand that we want the electron to move along the same path in the presence of both electric and magnetic fields. This implies that the forces due to the electric field and magnetic field must balance each other.

The force on an electron due to the electric field is given by:

$F_E = eE$

where $e$ is the charge of the electron and $E$ is the electric field.

The force on an electron due to the magnetic field (Lorentz force) is given by:

$F_B = evB$

where $v$ is the velocity of the electron and $B$ is the magnetic field.

For the electron to move in a straight path, the forces due to the electric field and magnetic field must be equal in magnitude:

$eE = evB$

From this, we can solve for the electric field $E$ :

$E = vB$

Next, we need to find the velocity $v$ of the electron. The kinetic energy (KE) of the electron is related to its velocity by the equation:

$KE = \frac{1}{2} mv^2$

Given the kinetic energy (KE) is $5 \, \text{eV}$ , we first convert this energy into joules since the given constants are in SI units:

$5 \, \text{eV} = 5 \times 1.6 \times 10^{-19} \, \text{J} = 8 \times 10^{-19} \, \text{J}$

Now, solving for $v$ :

$8 \times 10^{-19} = \frac{1}{2} \times 9 \times 10^{-31} \times v^2$

Rearrange to solve for $v^2$ :

$v^2 = \frac{2 \times 8 \times 10^{-19}}{9 \times 10^{-31}}$

$v^2 = \frac{16 \times 10^{-19}}{9 \times 10^{-31}}$

$v^2 = \frac{16}{9} \times 10^{12}$

$v = \sqrt{\frac{16}{9} \times 10^{12}}$

$v = \frac{4}{3} \times 10^6 \, \text{m/s}$

Now we can find the electric field $E$ . Using the value of the magnetic field $B$ given as $3 \, \mu \text{T} = 3 \times 10^{-6} \, \text{T}$ :

$E = vB = \left(\frac{4}{3} \times 10^6 \, \text{m/s}\right) \times \left(3 \times 10^{-6} \, \text{T}\right)$

$E = \frac{4}{3} \times 3 \times 10^0 \, \text{N/C}$

$E = 4 \, \text{N/C}$

Therefore, the value of the electric field $E$ required for the electron to move along the same path is:

$\boxed{4 \, \text{N/C}}$