Question

An electron whose mass in 1×10^-31 kg and charge 1.6×10^-19, moving with speed of 3×10^6 parallel to axis ,enters acting in the magnetic field of 0.3 weber/m² parallel to z-axis. Find the force acting on it, acceleration and direction of acceleration.

Answer 1

SOLUTION:-

Given:

Charge = 1.6× 10^-19 Coulomb

Velocity= 3× 10^6 m/sec,

B= 0.3 weber/m²

F= ? & ∅= 90°

From formula of Lorentz force, F=qvsin∅.

Force on electron,

Force on electron,$f = 0.3 \times 1.6 \times {10}^{ - 19} \times 3 \times {10}^{6} \times sin90 \degree \\ \\ = > 1.44 \times {10}^{ - 13} n \: \: \: \: \: \: \: </u><u>[</u><u>sin</u><u> \: 90 \degree = 1</u><u>]</u><u>$

& the direction of force

Again,

$F= 1.44 \times {10}^{ - 13} n \: and \: m = 9 \times {10}^{ - 31} kg$

Therefore,

From formula F= ma

Induced acceleration= F/m

$= > \frac{1.44 \times {10}^{ - 31} }{9 \times {10}^{ - 31} } \\ \\ = > 1.6 \times {10}^{17} m/sec {}^{2}$

Hope it helps ☺️

Answer 2

Answer:

Given:

Charge = 1.6× 10^-19 Coulomb

Velocity= 3× 10^6 m/sec,

B= 0.3 weber/m²

F= ? & ∅= 90°

From formula of Lorentz force, F=qvsin∅.

Force on electron,

Force on electron,

& the direction of force

Again,

Therefore,

From formula F= ma

Induced acceleration= F/m