Question Answers

Electrons

1. Beams of electrons and protons having the same initial KE enter normally into an electric field, which beam will be more curved? Justify.

The trajectory of the path of a charged particle in the perpendicular electric field is parabolic given by, $$ y = (\frac{eE}{2mv^2})x^2 = (\frac{eE}{2KE})x^2 $$

i.e., the deflection of the path y is inversely proportional to the KE of the charged particle. Since, the KE of electron and proton are same, the trajectory of electron and proton will be same in the perpendicular electric field.

2. Beams of electrons and protons having the same initial velocity enter normally into an electric field, which beam will be more curved? Justify.

The trajectory of the path of a charged particle in the perpendicular electric field is parabolic given by, $$ y = (\frac{eE}{2mv^2})x^2 $$

i.e., the deflection of the path y is inversely proportional to the mass of the charged particle. Since, the mass of electron is less than mass of proton, the deviation of electron will be more than proton.

3. A charged particle is fired into a cubical region of space where there is uniform magnetic field. Outside this region, there is no magnetic field. Is it possible that the particle will remain inside the cubical region? Explain.

The radius of the circular path of a charged particle having charge q moving in a uniform magnetic field with velocity v perpendicular to the field is given by, $$ \frac{mv^2}{r} = Bqv $$

$$ r = \frac{mv}{Bq} $$

Here, m is the mass of the charged particle and r is the radius of the circular path. The particle will remain in the cube if the diameter of the circular path is less than the length of the cube. Hence, the particle will remain inside the cubical region if the velocity of the particle and the magnetic field is so adjusted that the diameter of the circular path is less than the length of the cubical region.

4. A charged particle moves through a region of space with constant velocity. If the external magnetic field is zero in this region, can we conclude that the external field in the region is also zero? Explain.

Since, the particle is moving with constant velocity, net force acting on the particle is zero. As the external magnetic field in this region is zero, no force due to magnetic field is observed. If there would be external electric field, the force due to electric field would accelerate the particle which is not observed. Hence, we can conclude that the external electric field in the region is zero.

5. An electron and proton move with the same speed in a uniform magnetic field of equal magnitude. Compare the radii of their circular path.

The radius of the circular path of a charged particle having charge q moving in a uniform magnetic field with velocity v perpendicular to the field is given by, $$ \frac{mv^2}{r} = Bqv $$

$$ r = \frac{mv}{Bq} $$

For, electron and proton moving with same speed in a uniform magnetic field of same magnitude, (v,B,q) remains constant and

$$ r \propto m $$

If the radius and mass of the electron be r_e and m_e and that of proton be r_p and m_p, we can write,

$$\frac{r_e}{r_p} = \frac{m_e}{m_p} \approx \frac{1}{1837} $$

Hence, the radius of circular path of proton will be more than electron.

6. Write down expressions for acceleration of a moving charge Q in parallel and perpendicular magnetic fields.

When a charge particle having charge Q moving with velocity enters in a uniform magnetic field , it experiences a force called Lorentz force given by, $$ F = Q (\vec{B} \times \vec{v}) = BQv sin \theta $$

where, theta is the angle between the direction of motion of theparticl and the magnetic field.

When the magnetic field is parallel,

$$ \theta = 0^o $$

and the Force F = 0. No acceleration is produced in the maving charge. When themagnetic field is perpndicular,

$$ \theta = 90^o $$

and the forceF = BQv. This force is directed perpendicular to both B and v and the path of the charge is circular. For circular motion of the charged particle,

$$ \\frac{mv^2}{r}= BQv $$

$$ \\frac{v^2}{r}= \frac{BQv}{m} $$

This gives the centripetal acceleration acting on the particle in the perpendicular magnetic field.

7. Electrons is deviated in electric field and magnetic fields.

a. What path does the electron follow in electric field when the electron is projected normally in the field?

Ans: The path followed by the electron projected normally in electric field is parabolic.

b. An electron passes through a space without deviation. Does it mean, there is no fields?

Ans: If the electron is not deflected in passing through a certain space, it is not necessary that there is no fields. The electron would not be deflected if both the electric and magnetic field are acting normally to the direction of electron and so arranged that the velocity of electron is given by

$$ v = \frac{E}{B} $$

c. Is there any condition that an electron does not experience any force inside the magnetic field?

Ans: We have, magnetic force on the electron as, $$ F = Bev sin \theta $$

where, θ is angle between the direction of electron and magnetic field.

If $$\theta = 0^o $$, then F=0. Hence, the force will be zero when the direction of motion is along the direction of the magnetic field.

8. Why is a magnetic field used to deflect electron beam but not an electric field in a TV picture Tube?

If electric field is used then we have to apply a very high voltage, or we may require a very long tube. However, if a magnetic field is used, even a small field can produce large deflection, on the other hand, the required size of the picture tube will be highly reduced. Due to this reason, the magnetic field is used to deflect the electron beam in T.V. instead of the electric field.

9. Compare the specific charge of an electron with that of a proton.

If mass of the electron be m_e and that of proton be m_p, we can write,

$$Specific Charge of electron = \frac{e}{m_e} and Specific charge of proton = \frac{e}{m_p} $$

Then

$$\frac{Specific Charge of electron}{Specific charge of proton} = \frac{m_p}{m_e} \approx 1837 $$

10. What is the Millikan’s oil drop experiment?

The Millikan’s oil drop experiment is a method for direct determination of electric charge of electron. It is based on the study of an oil drop falling under gravity and in a uniform electric field at constant temperature. This experiment provides the basis for quantization of charge.

11. Explain why electric discharge through a gas takes place at low pressure.

At low pressure, the density of ions is less and the collision of ions becomes less frequent. The mean free path of the ions becomes large. Consequently, the ionized particles are able to reach their respective electrodes. Hence, electric discharge takes place at low pressure.

12. The value of e/m is constant for cathode rays, but not for positive rays. Explain

Cathod rays are fast moving beam of electrons. The charge and mass of an electron are constant. So, the specific charge (e/m) of cathode rays is constant. But, the positive rays are the stream of positive ions which have different mass for different gas. Hence, the specific charge (e/m) of positive rays is not constant.

13. A gas at lower pressure conducts electricity but the same gas at higher pressure does not. Explain.

At low pressure, the density of ions is less and the collision of ions becomes less frequent. The mean free path of the ions becomes large. Consequently, the ionized particles are able to reach their respective electrodes. Hence, conduction of electricity through gas takes place at low pressure. But at high pressure, the density of gas is large and the collision of ions is more frequent. The mean free path of the ions becomes small. Consequently, the ionized particles are not able to reach their respective electrodes. Hence, conduction of electricity through gas does not take place at high pressure.

14. Gases at normal pressure are insulators; how do they become conducting at low pressure?

At normal pressure, the density of gas is large and the collision of ions is more frequent. The mean free path of the ions becomes small. Consequently, the ionized particles are not able to reach their respective electrodes. Hence, gas does not conducts electricity and acts as insulator. But at low pressure, the density of ions is less and the collision of ions becomes less frequent. The mean free path of the ions becomes large. Consequently, the ionized particles are able to reach their respective electrodes. Hence, conduction of electricity through gas takes place at low pressure and the gas becomes conducting.

15. What property of the cathode rays indicates that they consist of electrons?

The value of charge e and specific charge e/m for cathode rays is the same as that for electron which indicates that cathode rays consists of electrons.

16. Why does electric discharge take place at low pressure and high potential difference?

At low pressure, the density of ions is less and the collision of ions becomes less frequent. The mean free path of the ions becomes large. Consequently, if high potential difference is applied between the electrodes, the ionized particles are able to reach their respective electrodes. Hence, conduction of electricity through gas takes place at low pressure.

17. Cathode rays cannot be regarded as electromagnetic waves. Why?

Cathode rays are not regarded as electromagnetic waves because: a) It consist of negative charge where as electromagnetic waves carry no charge. b) The speed of cathode rays varies with the applied potential between electrodes where as electromagnetic waves travel with speed of light. c) Cathode rays are deflected by electric and magnetic field where as electromagnetic field are not deflected. Hence, cathode rays are not electromagnetic waves.