Problem 4
Question
An alpha particle (charge + 2e) and an electron move in opposite directions from the same point, each with the speed of \(2.50 \times 10^{5} \mathrm{m} / \mathrm{s}\) (Fig. \(28.32 ) .\) Find the magnitude and direction of the total magnetic field these charges produce at point \(P,\) which is 1.75 \(\mathrm{nm}\) from each of them.
Step-by-Step Solution
Verified Answer
The total magnetic field is \(1.81 \times 10^{-9} \text{ T}\) directed into the page.
1Step 1: Identify Charge and Velocities
The problem involves two particles: an alpha particle with a charge of \(+2e\) and an electron with a charge of \(-1e\). Both particles move with a speed of \(2.50 \times 10^5 \text{ m/s}\) in opposite directions.
2Step 2: Compute the Magnetic Field from the Alpha Particle
Using the Biot-Savart Law, the magnetic field due to a moving charge is given by:\[B_{\alpha} = \frac{\mu_0}{4\pi} \frac{qv \sin \theta}{r^2}\]where \(q = +2e\), \(v = 2.50 \times 10^5 \text{ m/s}\), \(\theta = 90^\circ\) (since perpendicular to \(\vec{r}\) from point P), and \(r = 1.75 \times 10^{-9} \text{ m}\). Calculate \(B_{\alpha}\).- Plugging in the values, you find that:\[B_{\alpha} = \frac{10^{-7}}{(1.75 \times 10^{-9})^2} \times 2 \times 1.6 \times 10^{-19} \times 2.50 \times 10^5 \approx 3.63 \times 10^{-9} \text{ T}\]
3Step 3: Compute the Magnetic Field from the Electron
Similarly, compute the magnetic field from the electron using:\[B_{e} = \frac{\mu_0}{4\pi} \frac{qv \sin \theta}{r^2}\]where \(q = -1e\), and other values remain the same. The electron contributes:\[B_{e} = \frac{10^{-7}}{(1.75 \times 10^{-9})^2} \times (-1.6 \times 10^{-19}) \times 2.50 \times 10^5 \approx -1.82 \times 10^{-9} \text{ T}\]
4Step 4: Find the Total Magnetic Field at Point P
Since the alpha particle and electron move in opposite directions, their magnetic fields at point P add up algebraically due to vector directions, knowing that both are perpendicular:\[B_{\text{total}} = B_{\alpha} + B_{e}\]Substitute the calculated values:\[B_{\text{total}} = 3.63 \times 10^{-9} + (-1.82 \times 10^{-9}) = 1.81 \times 10^{-9} \text{ T}\]
5Step 5: Determine the Direction
Using the right-hand rule, the magnetic field from a positive charge moving in a given direction circulates counterclockwise, and for a negative charge, it moves clockwise. At point P, the alpha particle's contribution would point into the page, while the electron's contribution points out of the page. The net result, due to the larger field from the alpha particle, points into the page.
Key Concepts
Biot-Savart LawAlpha ParticleElectron MotionRight-Hand Rule
Biot-Savart Law
The Biot-Savart Law is a fundamental principle used to calculate the magnetic field produced by a moving charge. It describes how the magnetic field \( B \) is established at a point in space, due to a specific charge moving with velocity \( v \). The formula for the magnetic field \( B \) produced by a moving charge is:
- \[B = \frac{\mu_0}{4\pi} \frac{qv \sin \theta}{r^2} \]
- \( \mu_0 \) is the permeability of free space.
- \( q \) represents the charge.
- \( v \) is the speed of the charge.
- \( \theta \) is the angle between the velocity of the charge and the vector pointing from the charge to the point where the field is calculated.
- \( r \) is the distance from the charge to the point of measurement.
Alpha Particle
An alpha particle is a type of nuclear particle consisting of two protons and two neutrons, giving it a net positive charge of \(+2e\). It is significantly more massive compared to a single proton or neutron. In our exercise, the movement of an alpha particle contributes to the magnetic field at a specific point.
- The charge of the alpha particle is \(+2e\) which doubles the effect it might have in comparison to a single charge of \(+1e\).
- The velocity given is \(2.50 \times 10^{5} \, \text{m/s} \), a relatively high speed for subatomic particles.
- The direction is perpendicular to the radius vector pointing to the point where the magnetic field is measured, which means \(\theta = 90^\circ\).
Electron Motion
An electron is a subatomic particle with a negative charge, denoted as \(-1e\). Due to its relatively smaller mass compared to an alpha particle, electrons are highly responsive to electric and magnetic fields.
- In the exercise, the electron moves at a speed of \(2.50 \times 10^{5} \, \text{m/s} \), initially from the same point as the alpha particle but in the opposite direction.
- The contribution to the magnetic field due to the electron is calculated using similar formulas to that of the alpha particle but accounting for its single negative charge.
- This charge means the direction of its magnetic field contribution differs from the alpha particle, being opposite when viewed from the same perspective.
Right-Hand Rule
The right-hand rule is a useful mnemonic for determining the direction of a magnetic field relative to the direction of current or, in this case, the velocity of moving charges. It is applied by following these simple steps:
- Point your thumb in the direction of the conventional current (positive to negative) or velocity of the positive charge.
- Curl your fingers around; the direction in which your fingers curl is the direction of the magnetic field lines.
- For the alpha particle moving in one direction, the right-hand rule showed that its field at point P points into the page.
- The electron, being negatively charged and moving in the opposite direction, produces a magnetic field pointing out of the page.
Other exercises in this chapter
Problem 1
A \(+6.00-\mu C\) point charge is moving at a constant \(8.00 \times 10^{6} \mathrm{m} / \mathrm{s}\) in the \(+y\) -direction, relative to a reference frame. A
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Fields within the Atom. In the Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius \(5.3 \times 10^{-11} \mathrm{m}\) with a speed
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Along, straight wire lies along the \(z\) -axis and carries a \(4.00-\mathrm{A}\) current in the \(+z\) -direction. Find the magnetic field (magnitude and direc
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The Magnetic Field from a Lightning Bolt. Lightning bolts can carry currents up to approximately 20 kA. We can model such a current as the equivalent of a very
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