Answer :
Sure! Let's break down the problem step-by-step.
### Given Data:
1. Potential difference ([tex]\(V\)[/tex]) = [tex]\(1.36 \times 10^4 \, \text{V}\)[/tex]
2. Distance between plates ([tex]\(d\)[/tex]) = [tex]\(0.04 \, \text{m}\)[/tex] (since 4 cm = 0.04 m)
3. Magnetic field ([tex]\(B\)[/tex]) = [tex]\(2 \times 10^{-3} \, \text{T}\)[/tex]
4. Mass of electron ([tex]\(m\)[/tex]) = [tex]\(9.1 \times 10^{-31} \, \text{kg}\)[/tex]
5. Charge of electron ([tex]\(e\)[/tex]) = [tex]\(1.6 \times 10^{-19} \, \text{C}\)[/tex]
### Step 1: Calculate the Electric Field ([tex]\(E\)[/tex])
The electric field ([tex]\(E\)[/tex]) between the plates can be calculated using the potential difference ([tex]\(V\)[/tex]) and the distance ([tex]\(d\)[/tex]):
[tex]\[ E = \frac{V}{d} \][/tex]
Substitute the known values:
[tex]\[ E = \frac{1.36 \times 10^4 \, \text{V}}{0.04 \, \text{m}} \][/tex]
[tex]\[ E = 340000 \, \text{V/m} \][/tex]
### Step 2: Calculate the Velocity of Electrons ([tex]\(v\)[/tex])
The electrons experience no deflection, which means the electric force is balanced by the magnetic force. Therefore,
[tex]\[ eE = evB \][/tex]
Where [tex]\(e\)[/tex] is the charge of the electron, [tex]\(E\)[/tex] is the electric field, [tex]\(v\)[/tex] is the velocity of the electrons, and [tex]\(B\)[/tex] is the magnetic field. Solving for the velocity ([tex]\(v\)[/tex]):
[tex]\[ v = \frac{E}{B} \][/tex]
Substitute the known values:
[tex]\[ v = \frac{340000 \, \text{V/m}}{2 \times 10^{-3} \, \text{T}} \][/tex]
[tex]\[ v = 170000000 \, \text{m/s} \][/tex]
[tex]\[ v = 1.7 \times 10^8 \, \text{m/s} \][/tex]
### Step 3: Calculate the Radius of the Orbit ([tex]\(r\)[/tex]) if the Electric Field is Made Zero
When the electric field is made zero, the magnetic field provides the centripetal force needed for circular motion. The centripetal force ([tex]\(F_{\text{centripetal}}\)[/tex]) is given by:
[tex]\[ F_{\text{centripetal}} = \frac{mv^2}{r} \][/tex]
At the same time, the magnetic force ([tex]\(F_{\text{magnetic}}\)[/tex]) on the moving electron is:
[tex]\[ F_{\text{magnetic}} = evB \][/tex]
Equate the centripetal force to the magnetic force:
[tex]\[ evB = \frac{mv^2}{r} \][/tex]
Solve for the radius ([tex]\(r\)[/tex]):
[tex]\[ r = \frac{mv}{eB} \][/tex]
Substitute the known values:
[tex]\[ r = \frac{(9.1 \times 10^{-31} \, \text{kg}) \times (1.7 \times 10^8 \, \text{m/s})}{(1.6 \times 10^{-19} \, \text{C}) \times (2 \times 10^{-3} \, \text{T})} \][/tex]
[tex]\[ r = \frac{1.547 \times 10^{-22} \, \text{kg} \cdot \text{m/s}}{3.2 \times 10^{-22} \, \text{C} \cdot \text{T}} \][/tex]
[tex]\[ r = 0.4834375 \, \text{m} \][/tex]
Therefore, the final answers are:
(i) The velocity of electrons is [tex]\(1.7 \times 10^8 \, \text{m/s}\)[/tex].
(ii) The radius of the orbit is approximately [tex]\(0.483 \, \text{m}\)[/tex].
### Given Data:
1. Potential difference ([tex]\(V\)[/tex]) = [tex]\(1.36 \times 10^4 \, \text{V}\)[/tex]
2. Distance between plates ([tex]\(d\)[/tex]) = [tex]\(0.04 \, \text{m}\)[/tex] (since 4 cm = 0.04 m)
3. Magnetic field ([tex]\(B\)[/tex]) = [tex]\(2 \times 10^{-3} \, \text{T}\)[/tex]
4. Mass of electron ([tex]\(m\)[/tex]) = [tex]\(9.1 \times 10^{-31} \, \text{kg}\)[/tex]
5. Charge of electron ([tex]\(e\)[/tex]) = [tex]\(1.6 \times 10^{-19} \, \text{C}\)[/tex]
### Step 1: Calculate the Electric Field ([tex]\(E\)[/tex])
The electric field ([tex]\(E\)[/tex]) between the plates can be calculated using the potential difference ([tex]\(V\)[/tex]) and the distance ([tex]\(d\)[/tex]):
[tex]\[ E = \frac{V}{d} \][/tex]
Substitute the known values:
[tex]\[ E = \frac{1.36 \times 10^4 \, \text{V}}{0.04 \, \text{m}} \][/tex]
[tex]\[ E = 340000 \, \text{V/m} \][/tex]
### Step 2: Calculate the Velocity of Electrons ([tex]\(v\)[/tex])
The electrons experience no deflection, which means the electric force is balanced by the magnetic force. Therefore,
[tex]\[ eE = evB \][/tex]
Where [tex]\(e\)[/tex] is the charge of the electron, [tex]\(E\)[/tex] is the electric field, [tex]\(v\)[/tex] is the velocity of the electrons, and [tex]\(B\)[/tex] is the magnetic field. Solving for the velocity ([tex]\(v\)[/tex]):
[tex]\[ v = \frac{E}{B} \][/tex]
Substitute the known values:
[tex]\[ v = \frac{340000 \, \text{V/m}}{2 \times 10^{-3} \, \text{T}} \][/tex]
[tex]\[ v = 170000000 \, \text{m/s} \][/tex]
[tex]\[ v = 1.7 \times 10^8 \, \text{m/s} \][/tex]
### Step 3: Calculate the Radius of the Orbit ([tex]\(r\)[/tex]) if the Electric Field is Made Zero
When the electric field is made zero, the magnetic field provides the centripetal force needed for circular motion. The centripetal force ([tex]\(F_{\text{centripetal}}\)[/tex]) is given by:
[tex]\[ F_{\text{centripetal}} = \frac{mv^2}{r} \][/tex]
At the same time, the magnetic force ([tex]\(F_{\text{magnetic}}\)[/tex]) on the moving electron is:
[tex]\[ F_{\text{magnetic}} = evB \][/tex]
Equate the centripetal force to the magnetic force:
[tex]\[ evB = \frac{mv^2}{r} \][/tex]
Solve for the radius ([tex]\(r\)[/tex]):
[tex]\[ r = \frac{mv}{eB} \][/tex]
Substitute the known values:
[tex]\[ r = \frac{(9.1 \times 10^{-31} \, \text{kg}) \times (1.7 \times 10^8 \, \text{m/s})}{(1.6 \times 10^{-19} \, \text{C}) \times (2 \times 10^{-3} \, \text{T})} \][/tex]
[tex]\[ r = \frac{1.547 \times 10^{-22} \, \text{kg} \cdot \text{m/s}}{3.2 \times 10^{-22} \, \text{C} \cdot \text{T}} \][/tex]
[tex]\[ r = 0.4834375 \, \text{m} \][/tex]
Therefore, the final answers are:
(i) The velocity of electrons is [tex]\(1.7 \times 10^8 \, \text{m/s}\)[/tex].
(ii) The radius of the orbit is approximately [tex]\(0.483 \, \text{m}\)[/tex].
