give answer in 2 step with explanation at the end of each step and final answer at the end:1. A proton moving with a velocity v = 1.00 × 104 m/s enters a magnetic field region as shown in the figure.

Question

give answer in 2 step with explanation at the end of each step and final answer at the end:1. A proton moving with a velocity v = 1.00 × 104 m/s enters a magnetic field region as shown in the figure. The magnitude of the magnetic field is B = 1.00 × 10-3 T and its direction is into the page. The charge of the proton is q = +1.60 × 10-19 C. The magnetic field will bend the proton around a semicircular path and when the proton comes back to the lower edge of the magnetic field it will exit the region with a velocity equal to the one it had when it entered but in the opposite direction. Note: The magnetic force does not change the magnitude of the velocity, it merely changes its direction.

(a) Draw a diagram showing which way the magnetic field bends the proton (CW or CCW). On the diagram show the direction of the force at 4 locations. Also show the directions of the velocity and acceleration. Label your diagram clearly.

(b) Starting from Newton's Second Law, F = ma, calculate the magnitude and give the direction of the acceleration of the proton at the instant it enters the magnetic field region.

(c) Calculate the time it takes the proton to move through this semicircular path.

VIPSolutions · Accepted Answer

# (a) Diagram: Direction of Bending, Forces, Velocity, and Acceleration

Since I can't draw images directly, here's a **structured description** of the diagram:

- **Magnetic Field ($ \vec{B} $)**: Into the page (represented by "×" symbols).
- **Proton Initial Velocity ($ \vec{v} $)**: To the right (positive x-direction).
- **Using the Right-Hand Rule**: Thumb (velocity to right), Fingers (magnetic field into page), Palm push (force direction).
- **Result:** The proton curves **upward** in a **counterclockwise (CCW)** semicircle.

### Key Points on the Diagram

1. **Entry Point (Bottom Edge)**
    - **Velocity ($ \vec{v} $)**: Right (+x)
    - **Magnetic Force ($ \vec{F} $)**: Up (+y)
    - **Acceleration ($ \vec{a} $)**: Up (+y)

2. **Quarter Point (Right Edge)**
    - **Velocity**: Up (+y)
    - **Force**: Left (−x)
    - **Acceleration**: Left (−x)

3. **Top Point**
    - **Velocity**: Left (−x)
    - **Force**: Down (−y)
    - **Acceleration**: Down (−y)

4. **Quarter Point (Left Edge)**
    - **Velocity**: Down (−y)
    - **Force**: Right (+x)
    - **Acceleration**: Right (+x)

> **Summary**: The proton moves CCW in a semicircle. At each point, the force (and thus acceleration) is perpendicular to velocity, always pointing to the center of the circular path.

---

# (b) Find Acceleration Magnitude and Direction as Proton Enters

## Step 1: Write Magnetic Force and Newton’s Second Law

- **Magnetic Force:** $ F = qvB $
- **Centripetal Acceleration:** $ F = ma = m a_c $, $ a_c = \frac{v^2}{r} $

Set magnetic force equal to centripetal force:
$$
qvB = m a
$$
$$
a = \frac{qvB}{m}
$$

**Explanation:** The magnetic force provides the centripetal force needed for the circular motion. At entry, this force (and hence the acceleration) is perpendicular to velocity (upward if velocity is rightward).

---

# (c) Calculate Time for Semicircular Path

## Step 2: Find Radius and Time

### Find Radius

$$
qvB = \frac{mv^2}{r} \implies r = \frac{mv}{qB}
$$

Plug in values:
- $ m = 1.67 \times 10^{-27} $ kg (proton mass)
- $ v = 1.00 \times 10^4 $ m/s
- $ q = 1.60 \times 10^{-19} $ C
- $ B = 1.00 \times 10^{-3} $ T

$$
r = \frac{(1.67 \times 10^{-27}) \times (1.00 \times 10^4)}{(1.60 \times 10^{-19}) \times (1.00 \times 10^{-3})}
= \frac{1.67 \times 10^{-23}}{1.60 \times 10^{-22}}
= .104 \text{ m}
$$

### Find Time

- **Semicircular path length:** $ \pi r $
- **Time:** $ t = \frac{\pi r}{v} $

$$
t = \frac{\pi \times .104}{1.00 \times 10^4}
= \frac{.327}{1.00 \times 10^4}
= 3.27 \times 10^{-5} \text{ s}
$$

**Explanation:** The proton moves at constant speed along a semicircle of radius $ r $; the time is arc length divided by speed.

---

# Final Answers

- **(a)** The proton bends **counterclockwise (CCW)**; at each point, force and acceleration are perpendicular to velocity and point to the center of the circle.
- **(b)** **Magnitude of acceleration at entry:**  
  $$
  a = \frac{qvB}{m} = \frac{(1.60 \times 10^{-19})(1.00 \times 10^4)(1.00 \times 10^{-3})}{1.67 \times 10^{-27}} \approx 9.6 \times 10^8 \ \text{m/s}^2
  $$
  **Direction:** Upward (perpendicular to velocity, toward center).
- **(c)** **Time through semicircular path:**  
  $$
  t = 3.27 \times 10^{-5} \ \text{seconds}
  $$

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