Question

1. calculate the magnitude and the direction of the magnetic force acting on an alpha particle that is travelling upwards at a speed of 3.00 x 10^5 m/s through a 0.525 t west magnetic field (3 marks). explain all finger directions and the palm direction. (1 mark)

Explanation:

Step 1: Recall the formula for magnetic force on a charged particle

The magnetic force \( F \) on a charged particle with charge \( q \), velocity \( \vec{v} \), and magnetic field \( \vec{B} \) is given by the Lorentz force law: \( F = qvB\sin\theta \), where \( \theta \) is the angle between \( \vec{v} \) and \( \vec{B} \). For an alpha particle, the charge \( q = 2e \), where \( e = 1.60\times10^{-19}\, \text{C} \). The velocity is upward, and the magnetic field is west, so \( \theta = 90^\circ \) (since upward and west are perpendicular), and \( \sin 90^\circ = 1 \).

Step 2: Calculate the charge of the alpha particle

\( q = 2\times1.60\times10^{-19}\, \text{C} = 3.20\times10^{-19}\, \text{C} \)

Step 3: Substitute values into the force formula

Given \( v = 3.00\times10^{5}\, \text{m/s} \), \( B = 0.525\, \text{T} \), \( q = 3.20\times10^{-19}\, \text{C} \), and \( \sin\theta = 1 \):

\( F = qvB = (3.20\times10^{-19}\, \text{C})(3.00\times10^{5}\, \text{m/s})(0.525\, \text{T}) \)

First, multiply \( 3.20\times10^{-19} \times 3.00\times10^{5} \):

\( 3.20\times3.00 = 9.60 \), \( 10^{-19} \times 10^{5} = 10^{-14} \), so this product is \( 9.60\times10^{-14} \)

Then multiply by \( 0.525 \):

\( 9.60\times0.525 = 5.04 \), so \( F = 5.04\times10^{-14}\, \text{N} \)

Step 4: Determine the direction using the right-hand rule

For a positive charge (alpha particle is positive), use the right-hand rule:
- Point your right hand fingers in the direction of the velocity (upward).
- Curl your fingers towards the magnetic field direction (west, so from your fingers (upward) to west, you curl your fingers towards the west).
- The thumb points in the direction of the force. When velocity is up and magnetic field is west, the right-hand rule (for positive charge) gives the force direction as south? Wait, no: Wait, velocity is upward (let's take upward as +y, west as -x). The cross product \( \vec{v} \times \vec{B} \): \( \vec{v} = v\hat{y} \), \( \vec{B} = -B\hat{x} \). The cross product \( \hat{y} \times (-\hat{x}) = \hat{z} \)? Wait, no: \( \hat{y} \times \hat{x} = -\hat{z} \), so \( \hat{y} \times (-\hat{x}) = \hat{z} \). Wait, maybe I messed up the directions. Let's use the right-hand rule properly:
- Hold your right hand so that your fingers point in the direction of the velocity (upward).
- Then, bend your fingers towards the direction of the magnetic field (west).
- Your thumb will point in the direction of the force. Wait, velocity is up (north? Wait, upward could be north? Wait, the problem says "travelling upwards" – maybe upward is north? Wait, west is left, so if velocity is upward (north) and magnetic field is west (left), then using right-hand rule: fingers up (north), curl towards west (left), thumb points south? No, wait, the right-hand rule for \( \vec{F} = q\vec{v} \times \vec{B} \):
- For positive charge, \( \vec{F} = q(\vec{v} \times \vec{B}) \)
- \( \vec{v} \) is upward (let's say +y), \( \vec{B} \) is west (-x)
- \( \vec{v} \times \vec{B} = (v\hat{y}) \times (-B\hat{x}) = vB\hat{y} \times (-\hat{x}) = vB\hat{z} \) (since \( \hat{y} \times \hat{x} = -\hat{z} \), so \( \hat{y} \times (-\hat{x}) = \hat{z} \))
- Wait, \( \hat{y} \times \hat{x} = -\hat{z} \), so \( \hat{y} \times (-\hat{x}) = \hat{z} \). So the force direction is out of the page (positive z-direction)? Wait, maybe the problem considers upward as vertical (like up from the page), but no, probably upward is north (vertical direction is north-south, west-east is horizontal). Wait, maybe I made a mistake in the direction. Let's re-express:
- Velocity: upward (let's take as +y)
- Magnetic field: west (let's take as -x)
- Cross product \( \vec{v} \times \vec{B} = (v\hat{y}) \times (-B\hat{x}) = vB\hat{y} \times (-\hat{x}) = vB\hat{z} \) (because \( \hat{y} \times \hat{x} = -\hat{z} \), so \( \hat{y} \times (-\hat{x}) = \hat{z} \))
- So the force direction is out of the page (if we take z as out of the page). But maybe the problem's "upwards" is vertical (like up from the ground, so north), and west is left, so using right-hand rule:
- Fingers: velocity (north, up)
- Bend fingers towards magnetic field (west, left)
- Thumb points south? No, that's not matching the cross product. Wait, maybe the velocity is upward (vertical, so +z), magnetic field is west (-x). Then \( \vec{v} = v\hat{z} \), \( \vec{B} = -B\hat{x} \). Cross product \( \hat{z} \times (-\hat{x}) = -\hat{y} \) (since \( \hat{z} \times \hat{x} = \hat{y} \), so \( \hat{z} \times (-\hat{x}) = -\hat{y} \)). So force direction is -y (south). Ah, that makes sense. So maybe "upwards" is vertical (z-direction), west is x-direction (negative x). So:
- Velocity: +z (upwards)
- Magnetic field: -x (west)
- Cross product \( \vec{v} \times \vec{B} = \hat{z} \times (-\hat{x}) = -\hat{y} \) (south)
- So the force direction is south.

Answer:

The magnitude of the magnetic force is \( \boldsymbol{5.04\times10^{-14}\, \text{N}} \), and the direction is south (using the right-hand rule: fingers in velocity direction (upwards), curl towards magnetic field direction (west), thumb points south for positive charge (alpha particle)).