Question

A proton with an initial velocity given by \((1.0 \hat{x}+$$2.0 \hat{y}+3.0 \hat{z})\left(10^{5} \mathrm{~m} / \mathrm{s}\right)\) enters a magnetic field given by \((0.50 \mathrm{~T}) \hat{z}\). Describe the motion of the proton

Short answer

Question: Describe the motion of a proton with an initial velocity vector \( \vec{v} = (1.0\hat{x} + 2.0\hat{y} + 3.0\hat{z}) \times 10^5 m/s \) in a magnetic field \( \vec{B} = 0.50\hat{z} T \) and determine the radius of its circular trajectory in the x-y plane. Answer: The motion of the proton in the magnetic field is a combination of a circular motion in the x-y plane and a constant motion in the z-direction. The radius of the circular trajectory can be found by calculating the centripetal force and using the given initial velocity and magnetic field vectors. Upon calculation, the value of r for the circular trajectory can be determined.

Step-by-step solution

Find the charge and mass of a proton

To calculate the magnetic force on the proton, we need its charge (q) and mass (m). A proton has a charge of q = +1.6 x 10^{-19} C and a mass of m = 1.67 x 10^{-27} kg.

Write down the initial velocity and magnetic field vectors

We are given the initial velocity vector of the proton as \( \vec{v} = (1.0\hat{x} + 2.0\hat{y} + 3.0\hat{z}) \times 10^5 m/s \) and the magnetic field vector as \( \vec{B} = 0.50\hat{z} T \).

Calculate the magnetic force on the proton

The magnetic force acting on the proton is given by the Lorentz force formula: \( \vec{F} = q(\vec{v} \times \vec{B}) \). Plug the values for q, v, and B in the formula: \( \vec{F} = 1.6 \times 10^{-19} [ (1.0\hat{x} + 2.0\hat{y} + 3.0\hat{z}) \times 10^5 m/s \times (0.50\hat{z} T) ] \).

Compute the cross product of velocity and magnetic field vectors

To compute the cross product of the velocity and magnetic field vectors, use the cross product formula: $ \vec{v} \times \vec{B} = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ 1.0\cdot10^{5} & 2.0\cdot10^{5} & 3.0\cdot10^{5} \\ 0 & 0 & 0.50 \end{vmatrix} = \Big((2.0\cdot10^{5}\cdot0.5)\hat{x} - (1.0\cdot10^{5}\cdot0.5)\hat{y}\Big)$. Hence, \( \vec{v} \times \vec{B} = (1.0\cdot10^{5}\hat{x} - 0.5\cdot10^{5}\hat{y}) \, m/s \, T \).

Calculate the magnetic force component-wise

Plug the cross product value into the magnetic force formula: \( \vec{F} = 1.6\times10^{-19}[(1.0\cdot10^{5}\hat{x} - 0.5\cdot10^{5}\hat{y}) \, m/s\, T ] \). Therefore, \( \vec{F} = (1.6\cdot10^{-14}\hat{x} - 0.8\cdot10^{-14}\hat{y})\, N \).

Analyze the motion of the proton

The magnetic force is perpendicular to the velocity vector, which means that the proton will move in a circular trajectory in the x-y plane because of the sideways force, while moving in the z-direction at a constant speed due to no force acting in the z-direction.

Calculate the radius of the circular trajectory

The magnetic force provides the centripetal force for the circular motion: \( F_c = \frac{mv^2}{r} \). We can express this equation in vector components, as the magnetic force provides the centripetal force only in the x-y plane: \( F_x^2 + F_y^2 = \frac{(mv_{xy})^2}{r} \). Now substitute the given values: \( {(1.6\cdot10^{-14})}^2 + {(-0.8\cdot10^{-14})}^2 = \frac{ \left[ m (1.0\cdot10^{5})^2 + (2.0\cdot10^{5})^2\right]}{r} \). Solve for r: \( r = \frac{ \left[ m (1.0\cdot10^{5})^2 + (2.0\cdot10^{5})^2\right]}{{(1.6\cdot10^{-14})}^2 + {(-0.8\cdot10^{-14})}^2} \). After plugging in the mass of the proton (m), we find the value of r.

Conclusion

The motion of the proton in the magnetic field is a combination of a circular motion in the x-y plane and a constant motion in the z-direction. We can calculate the radius of the circular trajectory using the centripetal force formula and the given initial velocity and magnetic field vectors.

Key concepts

Lorentz Force

When studying how charged particles move through magnetic fields, a crucial concept that comes into play is the Lorentz force, named after the Dutch physicist Hendrik Lorentz. This force is responsible for the spectacular spiraling motions of particles under magnetic influences, a phenomenon observable in devices ranging from cyclotrons to the majestic auroras.

The Lorentz force encompasses both electric and magnetic components, but for a proton entering just a magnetic field, we can single out the magnetic part: \( \vec{F}_B = q(\vec{v} \times \vec{B}) \). This equation tells us that the magnetic force (\(\vec{F}_B\)) acting on a proton is the product of its charge (q), the velocity at which it moves (\(\vec{v}\)), and the magnetic field it encounters (\(\vec{B}\)). The most captivating aspect here is the cross product, which dictates that the force will always be perpendicular to the plane formed by the velocity and magnetic field vectors, creating a distinct motion we will explore shortly.

Visualizing the Lorentz Force

Imagine the proton as a tiny but mighty astronaut jetting through space. If the magnetic field is a wall and the proton's path hits this wall at an angle, instead of crashing, the astronaut is nudged into a curved path skimming along the wall's edge, all without slowing down—that's the Lorentz force in action. For protons in magnetic fields, this force doesn't do any work; it changes the direction but not the speed of the particle.

Cross Product in Magnetism

In magnetism, the direction in which a charged particle such as a proton will move can be determined using a mathematical operation known as the cross product. This operation takes two vectors, in this case, the velocity vector of the particle (\(\vec{v}\)) and the magnetic field vector (\(\vec{B}\)), and calculates a third vector that is perpendicular to both. The resulting vector gives both the direction and magnitude of the magnetic force.

For our proton with velocity \(\vec{v} = (1.0\hat{x} + 2.0\hat{y} + 3.0\hat{z}) \times 10^5 \frac{m}{s}\) entering a magnetic field \(\vec{B} = 0.50\hat{z} \, T\), we use the cross product operation to find the direction of the force. In this particular scenario, since the magnetic field is only in the \(\hat{z}\) direction, the cross product operation will result in a force vector that lies in the \(xy\)-plane, indicating that the proton's motion will pivot within this plane.

Performing the Cross Product

Imagine arranging the velocity and magnetic field components into a matrix, with the unit vectors \(\hat{x}\), \(\hat{y}\), and \(\hat{z}\) guiding their directions. Compute the determinant of this matrix to tease out the new, perpendicular force vector. The simplicity of a single non-zero component in \(\vec{B}\) means the resulting force will have only \(\hat{x}\) and \(\hat{y}\) components, and thus the proton pirouettes around the magnetic field's direction, never deviating from its original plane.

Circular Motion in Magnetic Field

Thanks to the Lorentz force's perpendicular push, a proton moving through a magnetic field doesn't just go straight or stop; it dances in circles, executing what we call circular motion. This motion occurs because the magnetic force continually pulls the proton sideways, keeping it on a curved path, much as a string can twirl a ball around in a circle.

The radius of this circle is not a random size but can be precisely calculated using the centripetal force concept. In circular motion, the centripetal force required to maintain the motion is equated to the magnetic Lorentz force. The formula \( F_c = \frac{mv^2}{r} \) captures this relationship, where \(m\) is the mass of the proton, \(v\) its velocity, and \(r\) the radius of the circle.

Diving Into the Proton's Trajectory

Due to the Lorentz force being perpendicular to the proton's velocity and lying in the \(xy\)-plane, a component of the proton's velocity remains unaffected—the one in the direction of the magnetic field (along the \(z\)-axis). In our problem, this means the proton keeps speeding along the \(z\)-axis with its original velocity component, while simultaneously twirling in the plane perpendicular to this axis. Combine these two, and you get a helical path—a corkscrew-like trajectory spiraling forward.