Question

A stream of electrons and protons are directed towards a narrow slit in a screen (see figure). The intervening region has a uniform electric field E (vertically downwards) and a uniform magnetic field B (out of the plane of the figure) as shown. Then:

A. Electron and protons with speed $\dfrac{{\left| E \right|}}{{\left| B \right|}}$ will pass through the slit.
B. Protons with speed $\dfrac{{\left| E \right|}}{{\left| B \right|}}$ will pass through the slit, electrons of the same speed will not.
C. Neither electrons nor protons will go through the slit irrespective of their speed.
D. Electrons will always be deflected upwards irrespective of their speed.

Answer 1

When any charged particle of charge q undergoes a magnetic field B and electric field E, the force on the charged particle is given by:
$\vec F = q\left[ {\vec E + \left( {\vec v \times \vec B} \right)} \right]$
Where v is the velocity of the charged particle.

Complete step-by-step answer:

1. As the electric field is directed along –ve Y axis then,
   $\vec E = - E\hat y$

2. As the electric field is directed along z axis then,

$$\vec B = B\hat z \\\ \vec v = v\hat x \\\$$

3. For protons, charge is +q. Then the Lorentz force on it is

$${{\vec F}_P} = q\left[ { - E\hat y + (v\hat x \times B\hat z)} \right] \\\ {{\vec F}_P} = - q\left[ {E + \left( {vB} \right)} \right]\hat y \\\$$

As the proton will move along the negative Y axis, it can not escape through the slit.

4. For electrons charge is –q. Then the Lorentz force on it is

$${{\vec F}_e} = - q\left[ { - E\hat y + (v\hat x \times B\hat z)} \right] \\\ {{\vec F}_e} = q\left[ {E + \left( {vB} \right)} \right]\hat y \\\$$

As the electrons will move along the +ve Y direction, it also cannot escape through the slit.

Hence, C. Neither electrons nor protons will go through the slit irrespective of their speed is the correct option.

Note: Students must be aware of the positive or negative charge while using the Lorentz equation. The velocity, electric field and magnetic field are mutually perpendicular. We must use this fact while fixing the direction of the vectors.