Answer to Question #159800 in Electric Circuits for zain ul abdeen

Question

 A velocity selector consists of electric and magnetic fields described by the expressions and , with B  15.0 mT. Find the value of E such that a 750-eV electron moving along the positive x axis is undeflected. 

Answer 1

We know that electric field and magnetic field are equal and opposite.

which is given by:

$E = Bv$

We also know that the kinetic energy of an electron is equal to the energy of the electron

$W_k = \large\frac{mv^2}{2} =$ 750ev $v=\sqrt{\frac{2W_k}{m}}$

we know that e/m of electron is $1.758 *10^{11}$

$E = B * \sqrt{\frac{2W_k}{m}} = 15*10^{-3}*\sqrt{2*750*1.758*10^{11}} = 275.43\frac{KV}{m}$