Question

In Thomson’s experiment, a magnetic field of induction $${10}^{-2}{\displaystyle \frac{Wb}{m^2}}$$ is used. For an un-deflected beam of cathode rays a potential difference of $$500V$$ is applied between the plates which are $$0.5cm$$ apart. Then the velocity of the cathode ray beam is __________ $${\displaystyle \frac{m}{s}}$$.
A) $$4\times {10}^7$$
B) $$2\times {10}^7$$
C) $$2\times {10}^8$$
D) $${10}^7$$

Answer 1

Hint: J.J Thomson made an experimental set up to find the particles in the cathode rays. There is a cathode ray tube in which the cathode and anode are fixed and the large potential is applied to move the particles from the cathode to anode. Then the particles reach the fields. He created the electric field by placing two plates having a potential difference around the path of the particles. And the particles further flow under the magnetic field force which is created by placing magnets.

Formula used:
(i) Magnetic field force=$$qvB$$
(ii) Electric field force=$$qE$$
Where,
$$q$$= charge
$$v$$=velocity
$$B$$=magnetic field of induction
$$E$$= electric field

Complete step by step answer:
When the cathode ray particles are flowing through the plates, it gets repelled by the negative potential and attracted to the positive potential. And while flowing through the magnets, the particles get deflected. Hence J.J. Thomson concluded that the flowing particles are negatively charged particles.
Hence we can say that the velocity of the particles also depends on the electric field force and magnetic field force.
To find the electric field E value,
$$E={\displaystyle \frac{V}{d}}$$
Where, $$V$$ represents potential difference and $$d$$ represents the distance between the plates. Applying the given values in the formula, we get
$$\Rightarrow E={\displaystyle \frac{500}{0.5\times {10}^{-2}}}$$
$$\therefore E={10}^{5}V{m}^{-1}$$
Equating the electric field force and magnetic field force, therefore
$$qE=qvB$$
$$\Rightarrow E=vB$$
$$\Rightarrow v={\displaystyle \frac{E}{B}}$$
We can substitute the values in the given equations,
$$\Rightarrow v={\displaystyle \frac{{10}^5}{{10}^{-2}}}$$
$$\therefore v={10}^{7}m{s}^{-1}$$

Hence the velocity of the cathode ray particles is $${10}^{7}m{s}^{-1}$$. The correct option is D.

Note:
(i) From the experiment, J.J. Thomson found that the particles in the cathode ray are negatively charged particles. Those particles are called electrons which have the mass very much less than the mass of an atom. And he also found that these negatively charged particles are found in all atoms.
(ii) Thomson found the charge to mass ratio to find the mass of the electrons and found that the mass of an electron is $$\sim {\displaystyle \frac{1}{2000}}$$ of the mass of the hydrogen atom.