Let's analyze and solve the problem in detail.

Problem Analysis

The problem provides us with:

• Mass of the particle $m = 6 \times 10^{-25} \, \text{kg}$
• Charge of the particle $q = 2 \times 10^{-16} \, \text{C}$
• Velocity of the particle $v = 7 \times 10^5 \, \text{m/s}$ along the $x$-axis
• Magnetic field $B = 4 \, \text{T}$, directed along the $-z$-axis (into the page or screen)

The task is to determine the radius of the circular trajectory in millimeters (mm) and sketch the trajectory.

Step 1: Determine the Radius of the Circular Path

In a magnetic field, a charged particle will move in a circular path if its velocity is perpendicular to the magnetic field. The centripetal force required for circular motion is provided by the magnetic force, given by: $F_{\text{magnetic}} = q \cdot v \cdot B$ This force acts as the centripetal force: $F_{\text{centripetal}} = \frac{m \cdot v^2}{r}$ Equating the two forces: $q \cdot v \cdot B = \frac{m \cdot v^2}{r}$ Solving for $r$: $r = \frac{m \cdot v}{q \cdot B}$

Step 2: Substitute the Values

Let's plug in the values provided:

• $m = 6 \times 10^{-25} \, \text{kg}$
• $v = 7 \times 10^5 \, \text{m/s}$
• $q = 2 \times 10^{-16} \, \text{C}$
• $B = 4 \, \text{T}$

$r = \frac{(6 \times 10^{-25} \, \text{kg}) \cdot (7 \times 10^5 \, \text{m/s})}{(2 \times 10^{-16} \, \text{C}) \cdot 4 \, \text{T}}$

Calculating the numerator: $6 \times 10^{-25} \times 7 \times 10^5 = 4.2 \times 10^{-19}$

Calculating the denominator: $(2 \times 10^{-16}) \times 4 = 8 \times 10^{-16}$

Now, calculating $r$: $r = \frac{4.2 \times 10^{-19}}{8 \times 10^{-16}} = 5.25 \times 10^{-4} \, \text{m}$

Step 3: Convert to Millimeters

To convert meters to millimeters, multiply by $10^3$: $r = 5.25 \times 10^{-4} \, \text{m} \times 10^3 = 0.525 \, \text{mm}$

Answer

The radius of the trajectory is: $r = 0.525 \, \text{mm}$

Step 4: Sketch the Trajectory

Since the magnetic field is directed along the $-z$-axis (into the page), and the velocity is along the $x$-axis, the particle will experience a force perpendicular to both the velocity and the magnetic field due to the right-hand rule. This will cause the particle to move in a circular path in the $xy$-plane.

Would you like more details on any of these calculations or steps?

Related Questions

1. What would happen to the radius if the velocity of the particle were doubled?
2. How would the trajectory change if the charge of the particle were negative?
3. What is the significance of the direction of the magnetic field in determining the trajectory?
4. How does the mass of the particle affect the radius of the circular path?
5. If the magnetic field strength were halved, what would be the new radius?

Tip

When working with forces in a magnetic field, always use the right-hand rule to determine the direction of the force on a positively charged particle.