A beam of electrons is accelerated through a potential difference of 7.0 kV before entering a velocity selector. If the B-field of the velocity selector is perpendicular to the velocity and has a value of 0.04 T, what value of the E-field is required (in the magnetic field region) if the particles are to be undeflected?

Given Information:  

Potential difference = V = 7 kV  

Magnetic field = B = 0.04 T  

Required Information:  

Electric Field = E = ?

Electric Field = 1.98x10⁶ V/m

Step-by-step explanation:  

The required Electric field can be found using

E = vB

Where v is the velocity of electron beam and B is the magnetic field

But first we need to calculate the velocity

V = E/q

E = Vq

Where V is the potential difference and q is the charge 1.60x10⁻¹⁹ C

E = 7x10³*1.60x10⁻¹⁹

E = 11.2x10⁻¹⁶ J

As we know

E = ½mv²

Where m is the mass of electron 9.11x10⁻³¹ kg

v = √2E/m

v = √2*11.2x10⁻¹⁶/9.11x10⁻³¹

v = 4.958x10⁷ m/s

Finally we can now calculate the Electric field

E = vB

E = 4.958x10⁷*0.04

E = 1.98x10⁶ V/m

Therefore, an electric field of 1.98x10⁶ V/m is required if the particles are to be undeflected.

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