The image contains a physics problem related to the motion of an electron in a magnetic field. The key concepts used in the solution involve the Lorentz force, centripetal acceleration, and the period of revolution of an electron. Let me walk you through the steps and explain each part in detail.

Given Data:

• Magnetic field (B) = 0.01 T (tesla)
• Velocity of the electron (v) = $10^6 \, \text{m/s}$
• Mass of the electron (m) = $9.1 \times 10^{-31} \, \text{kg}$
• Charge of the electron ($|q|$) = $1.6 \times 10^{-19} \, \text{C}$

We are asked to find:

• Radius (R) of the electron’s circular trajectory.
• Period (T) of the electron's circular motion.

Step-by-Step Solution:

1. Find the Lorentz force acting on the electron:

The Lorentz force $F$ on a charged particle moving in a magnetic field is given by:

$F = |q| v B \cdot \cos(90^\circ) = |q| v B$

Here, the angle between the velocity vector and the magnetic field is $90^\circ$, so $\cos 90^\circ = 1$. Therefore, the formula simplifies to:

$F = |q| v B$

Substitute the known values:

$F = (1.6 \times 10^{-19} \, \text{C}) \cdot (10^6 \, \text{m/s}) \cdot (0.01 \, \text{T})$ $F = 1.6 \times 10^{-14} \, \text{N}$

2. Relating the Lorentz force to centripetal force:

In circular motion, the Lorentz force acts as the centripetal force. The formula for centripetal force is:

$F = \frac{mv^2}{R}$

Equating the two forces:

$\frac{mv^2}{R} = |q| v B$

Solving for the radius $R$:

$R = \frac{mv}{|q|B}$

3. Calculating the radius (R):

Substitute the known values:

$R = \frac{(9.1 \times 10^{-31} \, \text{kg}) \cdot (10^6 \, \text{m/s})}{(1.6 \times 10^{-19} \, \text{C}) \cdot (0.01 \, \text{T})}$

$R = \frac{9.1 \times 10^{-25}}{1.6 \times 10^{-21}}$

$R \approx 5.7 \times 10^{-4} \, \text{m} = 0.57 \, \text{mm}$

4. Finding the period of revolution (T):

The period of revolution for a particle in circular motion is given by:

$T = \frac{2 \pi R}{v}$

Substitute the values:

$T = \frac{2 \cdot 3.14 \cdot (5.7 \times 10^{-4} \, \text{m})}{10^6 \, \text{m/s}}$

$T \approx \frac{3.14 \cdot 11.4 \times 10^{-4}}{10^6}$

$T \approx 3.6 \times 10^{-9} \, \text{s}$

Final Answers:

• Radius (R) ≈ 0.57 mm
• Period (T) ≈ $3.6 \times 10^{-9}$ s

---

Do you have any questions or would you like further details?

Here are 5 related questions to expand your understanding:

1. What happens to the radius if the magnetic field increases?
2. How does the period of revolution change if the velocity of the electron increases?
3. How would the result change if we used a proton instead of an electron?
4. What would happen if the magnetic field were oriented at an angle other than $90^\circ$?
5. How does the centripetal force relate to the Lorentz force in this scenario?

Tip: The Lorentz force is always perpendicular to both the velocity of the particle and the magnetic field, which is why it causes circular motion for charged particles.