Answer to Question #184667 in Electricity and Magnetism for Aisa

Question

A charged particle is moving without deflection in a space region where an elec- tric field E = 3.5x102V/m and a magnetic field B = 0.25T act in perpendicular directions.

Calculate the speed of the particle.

Answer 1

Given :-

Electric field (E )=$3.5\times 10^2 v/m$

Magnetic field (B)= 0.25 T

and to find

V is speed of particle

Lorentz Force

Lorentz force is the force exerted on a charged particle moving through both electric and magnetic field.

$\boxed{F = qE + qv \times B}$ ……….(1)

where,

• F = Lorentz Force
• q = Charge on the Particle
• E = Electric Field
• B = Magnetic Field
• v = Velocity of the Particle

$\boxed{r ={ mv \over qB}}$ ………..(2)

One can also deflect the trajectory of a charged particle with an electric field, although not into a circular path. If the electric force on the particle is both equal and opposite to the magnetic force, the net force on the particle will be zero. From Eq. (1), this will happen if

$\implies\boxed{V={E \over B}}$ ……….(3)

using .....(3)

$\boxed{V = \frac{E}{B}}$

$V = \frac{3.5\times10^2}{0.25}$

$\boxed{V = 14 \times 10^2 m/s}$ velocity of particle