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

Answer 2

When the electric field encounters the path of a particle, it undergoes deviation. Then the magnitude of electric field required for undeflected pass of the particle is  $5.92 \times 10^{5}\;\rm V/m$.

What is a uniform magnetic field?:

The space where the magnetic field has the same magnitude throughout is known as the uniform magnetic field.

Given data:

The potential difference of an electron is, V' = 10 kV = 10,000 V.

The strength of magnetic field is, B = 0.010 T.

The kinetic energy of electrons through the electric field is expressed as,

$KE = eV'$

here,

e is the charge of an electron. And for moving particles (electrons), the electric field is also expressed as,

$\dfrac{1}{2}mv^{2} =eV'$

Here,

m is the mass of an electron.

v is the speed of an electron.

Solving as,

 $\dfrac{1}{2} \times (9.31 \times 10^{-31})\times v^{2} =(1.6 \times 10^{-19})\times 10000\\\\v = \sqrt\dfrac{2 \times (1.6 \times 10^{-19}) \times 10000}{9.31 \times 10^{-31}}}\\\\v =5.92 \times 10^{7} \;\rm m/s$

Now, the magnitude of an electric field is given as,

$E = v \times B$

Solving as,

$E = (5.92 \times 10^{7}) \times 0.010\\\\E = 5.92 \times 10^{5}\;\rm V/m$

Thus, we can conclude that the magnitude of electric field required for undeflected pass of the particle is $5.92 \times 10^{5}\;\rm V/m$.

Learn more about the magnetic field here:

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