Question

4. A beam of electrons is accelerated through a potential difference of 9.0 kV before entering a velocity selector. If the B-field of the velocity selector is perpendicular to the velocity and has a value of 0.03 T, what value of the E-field is required (in the magnetic field region) if the particles are to be undeflected?

Answer 1

Answer:

The required electric field is 1.68 x 10⁶ N/C

Explanation:

Given;

Magnitude of the induced e.m.f = 9.0 kV = 9,000 V

Magnetic field strength is given, B = 0.03 T

mass of electron = 9.11 x 10⁻³¹ kg

charge of electron, q = 1.602 x 10⁻¹⁹ C

Electric field strength = ?

qE = qvB

E = vB

Where;

v is the velocity of the electron

K.E = ¹/₂ × mv² = qV

v² = 2qV/m

$v =\sqrt{\frac{2qV}{m}} = \sqrt{\frac{2*1.602*10^{-19} *9000}{9.11*10^{-31}}} \\\\v =5.6 X10^{7} m/s$

E = vB

E = 5.6 x 10⁷ x 0.03 = 1.68 x 10⁶ N/C

The required electric field is 1.68 x 10⁶ N/C