Question

In Thomson's experiment, a magnetic field of induction \[{10^{ - 2}}{{{\rm{Wb}}} {\left/{\vphantom {{{\rm{Wb}}} {{{\rm{m}}^2}}}} \right.
} {{{\rm{m}}^2}}}\] is used. For an undeflected beam of cathode rays, a p.d. of \[500{\rm{ V}}\] is applied between the plates which are \[0.5{\rm{ cm}}\] apart. Then the velocity of the electron beam is ________ m/s.
(1) \[4 \times {10^7}\]
(2) \[2 \times {10^7}\]
(3) \[2 \times {10^8}\]
(4) \[{10^7}\]

Answer 1

Hint:Electric field between the plates in Thomson's experiment is given the ratio of potential difference and distance between them. We will write the expression for force due to the electric field and force due to the magnetic field, and on comparing them, we will deduce the final expression for the velocity of the electron beam.

Complete step by step answer:
Given:
The magnetic field of induction is \[B = {10^{ - 2}}{{{\rm{Wb}}} {\left/
{\vphantom {{{\rm{Wb}}} {{{\rm{m}}^2}}}} \right.
} {{{\rm{m}}^2}}}\].
The potential difference applied between plates is \[V = 500{\rm{ V}}\].
The distance between the plates is \[r = 0.5{\rm{ cm}}\].
We have to calculate the velocity of the electron beam.
Let us write the expression for the electric field per unit distance.
\[E = \dfrac{V}{r}\]
On substituting \[500{\rm{ V}}\] for V and \[0.5{\rm{ cm}}\] for r in the above expression, we get:
\[\begin{array}{l}
E = \dfrac{{500{\rm{ V}}}}{{0.5{\rm{ cm}} \times \left( {\dfrac{{\rm{m}}}{{{\rm{100 cm}}}}} \right)}}\\
= {10^5}{{\rm{V}} {\left/
{\vphantom {{\rm{V}} {\rm{m}}}} \right.
} {\rm{m}}}
\end{array}\]
We can write the expression for magnetic force due to magnetic field B as below:
\[{F_1} = qvB\]
Here q is the electron particle's charge, and v is the velocity of the electron beam.
We know that the expression for force due to electric field E can be written as:
\[{F_2} = qE\]
The magnetic force is generated due to the electric field; therefore, we can equate force due to the magnetic field and force due to the electric field.
\[
qvB = qE\\
vB = E
\]
On substituting \[{10^5}{{\rm{V}} {\left/
{\vphantom {{\rm{V}} {\rm{m}}}} \right.
} {\rm{m}}}\] for E and \[{10^{ - 2}}{{{\rm{Wb}}} {\left/
 {\vphantom {{{\rm{Wb}}} {{{\rm{m}}^2}}}} \right.
} {{{\rm{m}}^2}}}\] for B in the above expression, we get:
\[
v\left( {{{10}^{ - 2}}{{{\rm{Wb}}} {\left/
{\vphantom {{{\rm{Wb}}} {{{\rm{m}}^2} \times \dfrac{{{\rm{V}} \cdot {\rm{s}}}}{{{\rm{Wb}}}}}}} \right.
} {{{\rm{m}}^2} \times \dfrac{{{\rm{V}} \cdot {\rm{s}}}}{{{\rm{Wb}}}}}}} \right) = {10^5}{{\rm{V}} {\left/
{\vphantom {{\rm{V}} {\rm{m}}}} \right.
} {\rm{m}}}\\
v = {10^7}{\rm{ }}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}
\]
Therefore, the electron beam's velocity in the given Thomson's experiment is \[{10^7}{\rm{ }}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}\], and option (4) is correct.

Note: The unit voltage gives one Weber in one second. We can remember the conversion of Weber into a volt-second for similar problems. The ray of an electron coming out is the cathode in Thomson's experiment and found that they are negatively charged, termed as the electron beam.