Question

in a region of electric field E=5* 10^3 j NC and a magnetic field of B= 0.1KT are applied a beam of charged particles are projected along x direction find the velocity of charged particles which move un deflected in these crossedn fields​

Answer 1

Answer:

50000

Explanation:

how u have understood

Answer 2

Given:

Electric field (E) = 5×10³ N/C

Magnetic field (B) = 0.1 T

To find:

Velocity (v) of charged particles

Explanation:

• In an electromagnetic wave, the electric and magnetic waves run perpendicular to each other and also perpendicular to the direction of propagation of the wave.

• The speed of an electromagnetic wave is equal to the speed of light in vacuum mathematically which is 3×10⁸ m/s.

• They are related by the formula that is

               speed of light in vacuum $c = \frac{E}{B}$  

Solution:

We know the relation,

velocity $v = \frac{E}{B}$  

Substituting the values, we get

$v = \frac{5*10^{3} }{0.1}$

or  

v = 5 × 10⁴ m/s

Final answer:

Hence, the velocity of the undeflected particles will be 5× 10⁴ m/s along the x-axis.