Problem 16 - Physics

Question

An \(\alpha\) -particle has a charge of \(+2 e\) and a mass of \(6.64 \times 10^{-27} \mathrm{~kg} .\) It is accelerated from rest through a potential difference that has a value of \(1.20 \times 10^{6} \mathrm{~V}\) and then enters a uniform magnetic field whose magnitude is \(2.20 \mathrm{~T}\). The \(\alpha\) -particle moves perpendicular to the magnetic field at all times. What is (a) the speed of the \(\alpha\) -particle, (b) the magnitude of the magnetic force on it, and (c) the radius of its circular path?

Step-by-Step Solution

Answer

(a) Speed: \(1.54 \times 10^7 \text{ m/s}\). (b) Magnetic Force: \(1.08 \times 10^{-11} \text{ N}\). (c) Radius: \(0.071 \text{ m}\).

Step 1: Calculate the Speed of the Alpha-Particle

We begin by using the conservation of energy principle to determine the speed of the alpha particle. The electric potential energy is converted into kinetic energy. The equation for kinetic energy is \( KE = \frac{1}{2}mv^2 \), and the electric potential energy is \( qV \), where \( q = +2e \). Setting these equal gives \( \frac{1}{2}mv^2 = qV \).Substitute the given values: \( q = 2e = 2 \times 1.60 \times 10^{-19} \text{ C} \), \( V = 1.20 \times 10^{6} \text{ V} \), \( m = 6.64 \times 10^{-27} \text{ kg} \).\[\frac{1}{2} \left(6.64 \times 10^{-27} \right) v^2 = (2 \times 1.60 \times 10^{-19})(1.20 \times 10^6)\]Solving for \( v \): \[ v = \sqrt{\frac{2 \times (2 \times 1.60 \times 10^{-19} \times 1.20 \times 10^6)}{6.64 \times 10^{-27}}}\]\( v \approx 1.54 \times 10^7 \text{ m/s} \). Thus the speed of the \( \alpha \)-particle is approximately \( 1.54 \times 10^7 \text{ m/s} \).

Step 2: Determine the Magnetic Force on the Alpha-Particle

The magnetic force on a charged particle moving in a magnetic field is given by \( F = qvB \sin \theta \). Here, \( \theta \) is the angle between the velocity and magnetic field. As the \( \alpha \)-particle moves perpendicular to the magnetic field, \( \theta = 90^\circ \) and \( \sin 90^\circ = 1 \).Substituting the known values: \( q = 2 \times 1.60 \times 10^{-19} \text{ C} \), \( v = 1.54 \times 10^7 \text{ m/s} \), \( B = 2.20 \text{ T} \).\[ F = (2 \times 1.60 \times 10^{-19})(1.54 \times 10^7)(2.20) \]\( F \approx 1.08 \times 10^{-11} \text{ N} \). Thus, the magnitude of the magnetic force on the \( \alpha \)-particle is approximately \( 1.08 \times 10^{-11} \text{ N} \).

Step 3: Calculate the Radius of the Circular Path

To find the radius of the circular path, we use the centripetal force equation, which is provided by the magnetic force: \( F = \frac{mv^2}{r} \).We equate this with the magnetic force \( F = qvB \), solving for \( r \).\[ \frac{mv^2}{r} = qvB \]Rearranging the formula to solve for \( r \):\[ r = \frac{mv}{qB} \]Substitute the known values: \( m = 6.64 \times 10^{-27} \text{ kg} \), \( v = 1.54 \times 10^7 \text{ m/s} \), \( q = 2 \times 1.60 \times 10^{-19} \text{ C} \), \( B = 2.20 \text{ T} \).\[ r = \frac{6.64 \times 10^{-27} \times 1.54 \times 10^7}{(2 \times 1.60 \times 10^{-19}) \times 2.20} \]\( r \approx 0.071 \text{ m} \). Hence, the radius of the \( \alpha \)-particle's circular path is approximately \( 0.071 \text{ m} \).

Key Concepts

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. For an alpha particle moving through a magnetic field, its kinetic energy defines how fast it will travel. It is given by the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) represents the mass and \( v \) signifies its velocity. Kinetic energy plays a crucial role here because as the alpha particle moves, its electric potential energy, obtained from an electric field, is converted into kinetic energy. This energy transformation is how the particle begins its motion in the magnetic field and achieves a relatively high speed.

Magnetic Force

Magnetic force is the force experienced by a charged particle when it moves within a magnetic field. For an alpha particle, this force is calculated using the equation \( F = qvB \sin \theta \), where \( q \) is the charge, \( v \) is the velocity, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the particle's velocity and the magnetic field. In our scenario, the particle moves perpendicular to the magnetic field, making \( \theta = 90^\circ \), and \( \sin 90^\circ = 1 \), simplifying the equation to \( F = qvB \). This force is central to determining how the particle behaves as it moves.

Circular Motion

Circular motion occurs when a particle moves in a circular path due to a centering force. In this case, the magnetic force acts as the centripetal force that keeps the alpha particle in a circular trajectory. The formula tying the magnetic force to circular motion is \( F = \frac{mv^2}{r} \), where \( r \) is the radius of the circle. Here, we see how the magnetic force ensures the particle maintains a circular path without deviating. Understanding this relationship is key to predicting the particle's path in a magnetic field and its radius based on speed, mass, charge, and magnetic field strength.

Conservation of Energy

The conservation of energy formula tells us that energy cannot be created or destroyed, only transformed. In the case of an alpha particle, its electric potential energy is transformed into kinetic energy. This transformation occurs when the particle is accelerated by the electric potential difference from rest, converting all its initial potential energy into kinetic energy before entering the magnetic field. The principle of conservation allows us to equate the electric potential energy \( qV \) to kinetic energy \( \frac{1}{2}mv^2 \), thus permitting the calculation of the particle's speed once the potential difference and other values are known.

Electric Potential Energy

Electric potential energy arises when a charged particle is positioned within an electric field. For an alpha particle, initially at rest, this energy potential is provided by an electric potential difference or voltage. When it experiences this difference, the particle's electric potential energy \( qV \) is harnessed to accelerate it, thus transforming into kinetic energy. This is crucial for calculating the alpha particle's speed as it moves into the magnetic field, serving as the energy source for its subsequent motion. Understanding electric potential energy helps explain how particles acquire speed and energy in electric and magnetic fields.