Problem 21 - University Physics with Modern Physics

Question

A deuteron (the nucleus of an isotope of hydrogen) has a mass of 3.34 \(\times\) 10\(^{-27}\) kg and a charge of \(+e\). The deuteron travels in a circular path with a radius of 6.96 mm in a magnetic field with magnitude 2.50 T. (a) Find the speed of the deuteron. (b) Find the time required for it to make half a revolution. (c) Through what potential difference would the deuteron have to be accelerated to acquire this speed?

Step-by-Step Solution

Answer

The speed is about 3.03 \times 10^{6} m/s. Half-revolution time is about 7.2 \times 10^{-9} s. Potential difference is about 47.3 kV.

Step 1: Identify the Relevant Formula for Speed

To find the speed of the deuteron when it travels in a magnetic field, use the formula for centripetal force: \ F_c = qvB = \frac{mv^2}{r} \ where \( q \) is the charge of the deuteron, \( v \) is the velocity, \( B \) is the magnetic field, \( m \) is the mass, and \( r \) is the radius. Rearranging gives: \ v = \frac{qBr}{m}.

Step 2: Calculate the Speed of the Deuteron

Substitute the given values into the equation: \ q = 1.6 \times 10^{-19} \text{C} (charge of deuteron) \ B = 2.50 \text{T}, \ r = 6.96 \times 10^{-3} \text{m} (convert mm to m), \ m = 3.34 \times 10^{-27} \text{kg}. \ v = \frac{(1.6 \times 10^{-19} \text{C})(2.50 \text{T})(6.96 \times 10^{-3} \text{m})}{3.34 \times 10^{-27} \text{kg}} \ Calculate to find \( v \).

Step 3: Calculate Half Revolution Time

The time for half a revolution is found by calculating the period \( T \) and halving it. The period is given by \( T = \frac{2\pi r}{v} \). Calculate this period and then divide by 2 to get the time for half a revolution.

Step 4: Calculate Potential Difference Required

The kinetic energy gained by the deuteron when accelerated through a potential difference \( V \) can be expressed as: \ KE = \frac{1}{2}mv^2 = qV \ Solve for \( V \) using \( v \) from Step 2: \ V = \frac{\frac{1}{2}mv^2}{q}.

Step 5: Solve for the Potential Difference

Plug in values: \ m = 3.34 \times 10^{-27} \text{kg}, \ v = \text{value from Step 2}, \ q = 1.6 \times 10^{-19} \text{C}. \ Calculate \( V \) to find the potential difference.

Key Concepts

Magnetic Field

In physics, a magnetic field is an invisible field that exerts a force on particles that are charged or moving through it. Magnetic fields surround magnets, electric currents, and changing electric fields. For a charging particle like a deuteron, when it enters a magnetic field, it experiences a force that is perpendicular to both the velocity of the particle and the direction of the magnetic field. This force can be described by the equation \( F = qvB \), where \( F \) is the magnetic force, \( q \) is the charge of the particle, \( v \) is the velocity, and \( B \) is the magnetic field strength.

• This perpendicular force causes the particle to move in a circular path.
• The strength of the magnetic field affects the radius of this circular motion.
• The direction of the magnetic field determines which way the particle will curve.

Centripetal Force

Centripetal force is the force needed to make an object move in a circle. For a deuteron moving in a magnetic field, this force is provided by the magnetic force. The centripetal force ensures that the deuteron remains in circular motion. The formula for centripetal force \( F_c \) is given as \( F_c = \frac{mv^2}{r} \), where \( m \) is the mass of the deuteron, \( v \) is the velocity, and \( r \) is the radius of the circular path.

• The force keeps the particle moving in its circular path and prevents it from flying off tangentially.
• It is directed towards the center of the circle along which the deuteron is moving.
• The magnetic force acting on the deuteron provides the necessary centripetal force.

Potential Difference

Potential difference, also known as voltage, is the energy required to move a charge between two points in a field. For a deuteron, to achieve a specific speed, it needs to be accelerated through a potential difference. This is expressed in the relation between kinetic energy and potential difference:\[ KE = \frac{1}{2}mv^2 = qV \]Here, \( KE \) is kinetic energy, \( m \) is mass, \( v \) is velocity, \( q \) is charge, and \( V \) is potential difference.

• Potential difference gives the deuteron the energy needed to obtain its speed in the magnetic field.
• Voltage is crucial for accelerating the particle to a speed where magnetic and centripetal forces balance.
• This relationship shows the direct link between energy, speed, and voltage.

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. For the deuteron, once it travels through a potential difference, it will gain kinetic energy proportionate to its speed. The equation for kinetic energy is given by:\[ KE = \frac{1}{2}mv^2 \]Kinetic energy depends on both the mass of the particle and the square of its velocity.

• A faster-moving deuteron has higher kinetic energy.
• This energy depends on the speed achieved when the deuteron is accelerated through a potential difference.
• Kinetic energy demonstrates how fast the particle is able to move once it enters the magnetic field.