Question

A beam of electrons is accelerated through a potential difference of 10 kV before entering a region having uniform electric and magnetic fields that are perpendicular to each other and perpendicular to the direction in which the electron is moving. If the magnetic field in this region has a value of 0.010 T, what magnitude of the electric field is required if the particles are to be undeflected as they pass through the region?

Answer 1

Answer:

The magnitude of the electric field is 592.67 V/m

Explanation:

The kinetic energy of an electron moving with speed v is:

K.E. = (1/2)mv²

Also,

The kinetic energy for an electron in a electrical field:

K.E. = eV

Where e is the charge of the electron and V is the potential difference

Thus,

(1/2)mv² = eV

So,

$v=\sqrt{\frac {2eV}{m}}$

Mass of electron = 9.11*10⁻³¹ kg

Charge on electron = 1.6*10⁻¹⁹ C

Given: voltage = 10 kV = 10000 V

So,

$v=\sqrt{\frac {2\times (1.6\times 10^{-19})\times 10000}{9.11\times 10^{-19}}}$

v = 5.9267*10⁷ m/s

The two fields are the crossed fields, So,

E = v×B

Given B = 0.010 T

E = 5.9267*10⁷×0.010 V/m

E = 5.9267*10⁵ V/m

Or,

E = 592.67 kV/m