Question

A proton is accelerated from rest by a potential difference of \(400 . \mathrm{V}\). The proton enters a uniform magnetic field and follows a circular path of radius \(20.0 \mathrm{~cm} .\) Determine the magnitude of the magnetic field.

Short answer

Solution: The magnitude of the magnetic field is approximately \(5.7\times10^{-4}\) T.

Step-by-step solution

Find the kinetic energy of the proton

First, we need to find the kinetic energy (KE) of the proton after it is accelerated by the potential difference. We can use the equation: \(KE = qV\) where q is the charge of the proton (in coulombs), V is the potential difference (in volts), and KE is the kinetic energy (in joules). We know that the charge of a proton is approximately \(1.6 \times 10^{-19}\) C, and the potential difference is given as 400 V, so we can plug in these values: \(KE = (1.6 \times 10^{-19} \text{ C})(400 \text{ V}) = 6.4 \times 10^{-17} \text{ J}\)

Calculate the velocity of the proton using the kinetic energy

Next, we need to find the velocity of the proton using its kinetic energy. We can use the equation: \(KE = \frac{1}{2}mv^{2}\) where m is the mass of the proton (in kg), v is the velocity (in m/s), and KE is the kinetic energy (in J). We are given the mass of a proton is approximately \(1.67 \times 10^{-27}\) kg. Rearrange the equation to solve for v and plug in the known values: \(v = \sqrt{\frac{2(6.4 \times 10^{-17} \text{ J})}{(1.67 \times 10^{-27} \text{ kg})}} = 1.355\times10^{7} \text{ m/s}\)

Calculate the magnetic force acting on the proton

Now that we have the proton's velocity, we can use the centripetal force formula to find the magnetic force acting on the proton, which is given by: \(F_B = \frac{mv^2}{r}\) where \(F_B\) is the magnetic force, m is the mass of the proton, v is the velocity, and r is the radius of the circular path. We are given the radius as 20.0 cm, which is equal to 0.2 m. Plug in the known values: \(F_B = \frac{(1.67 \times 10^{-27} \text{ kg})(1.355\times10^{7} \text{ m/s})^2}{0.2 \text{ m}} = 1.468\times10^{-11} \text{ N}\)

Calculate the magnetic field strength

Finally, we use the equation for the magnetic force acting on a moving charge: \(F_{B} = qvB\sin{\theta}\) Since the proton is moving perpendicular to the magnetic field, \(\sin{\theta} = 1\). We can rearrange the equation to solve for the magnetic field strength (B): \(B = \frac{F_B}{qv}\) Plug in the known values: \(B = \frac{1.468\times10^{-11} \text{ N}}{(1.6 \times 10^{-19} \text{ C})(1.355\times10^{7} \text{ m/s})} = 5.7\times10^{-4} \text{T}\) Therefore, the magnitude of the magnetic field is approximately \(5.7\times10^{-4}\) T.

Key concepts

Understanding Kinetic Energy

Kinetic energy (KE) is the energy an object possesses due to its motion. For a single particle like a proton, the kinetic energy can be expressed using the equation:

\(KE = \frac{1}{2}mv^2\)

where \(m\) is the mass of the particle and \(v\) is its velocity. This form of energy conversion comes to play when forces act upon an object causing it to accelerate, like a proton being accelerated by an electric potential difference in our problem set. When a potential difference accelerates a charged particle, the electric field does work on the particle, transforming electric potential energy into kinetic energy. The larger the potential difference or the electric charge, the more kinetic energy is imparted to the particle. This concept is crucial to understanding the motion of protons in particle accelerators or in the context of our problem, where a potential difference of 400 V gives the proton its kinetic energy.

Proton Acceleration in Physics

Acceleration in physics refers to any change in an object's velocity, either in magnitude or direction. When a proton is accelerated from rest by a potential difference, like the 400 V in our textbook problem, its acceleration is not constant but happens in an instant as it gains velocity.

It's important to note that in electromagnetic contexts, the proton’s acceleration arises from the force exerted by the electric field associated with the potential difference. However, once the proton is in uniform motion within the magnetic field, it no longer accelerates in speed (velocity magnitude), but in direction, due to the magnetic force causing circular motion, which is a different form of acceleration - centripetal acceleration.

Centripetal Force and Its Role

Centripetal force is required for any type of circular motion. It is the 'center-seeking' force that causes an object to follow a curved path.

In the case of a charged proton moving in a magnetic field, this force is provided by the magnetic Lorentz force, and it keeps the proton moving in a circle. The equation for centripetal force is:

\(F = \frac{mv^2}{r}\)

where \(F\) is the centripetal force, \(m\) is the mass, \(v\) is the velocity, and \(r\) is the radius of the circular path. In our exercise, once the proton with known kinetic energy enters the magnetic field, it begins to move in a circular path with the centripetal force equal to the magnetic Lorentz force.

Circular Motion in a Magnetic Field

Circular motion within a magnetic field is a fundamental principle in electromagnetism. When a charged particle, such as our proton, moves at a velocity perpendicular to a uniform magnetic field, it experiences a force perpendicular to both its velocity and the magnetic field.

This force does no work on the proton (since it acts at a right angle to the direction of motion), but it changes the direction of the proton's velocity, causing it to curve and move in a circular path. The magnetic force acting as a centripetal force is given by the relationship:

\(F_B = qvB\)

where \(q\) is the charge of the proton, \(v\) is its velocity, and \(B\) is the magnetic field's strength. The resulting motion is circular, and its radius can tell us much about the magnitude of the magnetic field, provided we understand the concepts of kinetic energy and centripetal force, which are all at play in this scenario.