Two infinitely long wires each carrying the same current and pointing in $+ y$ direction are placed in the $x y$ - plane, at $x = - 2 cm$ and $x = 1 cm$ . An electron is fired with speed $U$ from the origin making an angle of $+ 45 °$ from the $x$ - axis. The force on the electron at the instant it is fired is
[ $B_{0}$ is the magnitude of the field at origin due to the wire at $x = 1 cm$ alone].

Question

Two infinitely long wires each carrying the same current and pointing in

+ y

direction are placed in the

x y

- plane, at

x = - 2 cm

and

x = 1 cm

. An electron is fired with speed

U

from the origin making an angle of

+ 45 °

from the

x

- axis. The force on the electron at the instant it is fired is
[

B_{0}

is the magnitude of the field at origin due to the wire at

x = 1 cm

alone].

Aicharya · Accepted Answer

Magnetic field at origin due to wire at $x = 1 cm$ is
$B_{0} = \frac{μ_{0} i}{2 π (0.01)} ⊙$
Magnetic field at origin due to wire at $x = - 2 cm$ is
$B = \frac{μ_{0} i}{2 π (0.02)} \otimes$
Therefore, the net magnetic field at origin will be,
${\vec{B}}_{n e t} = \frac{μ_{0} i}{2 π (0.02)} ⊙ = \frac{B_{0}}{2} \hat{k}$
The velocity of the electron is given as $\vec{v} = \frac{U}{\sqrt{2}} \hat{i} + \frac{U}{\sqrt{2}} \hat{j}$
The force on the electron will be,
$\vec{F} = e \vec{v} \times \vec{B} = e (\frac{U}{\sqrt{2}} \hat{i} + \frac{U}{\sqrt{2}} \hat{j}) \times \frac{B_{0}}{2} \hat{k} \Rightarrow \vec{F} = \frac{- e U B_{0}}{2 \sqrt{2}} (\hat{i} - \hat{j})$

Two infinitely long wires each carrying the same current and pointing in + y direction are placed in the x y…

Solution